IB Analysis and Approaches HL · Integral calculus · End of topic

IB exam questions on integral calculus

312 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

Level

Paper

Solutions

Antiderivatives: substitution, parts and partial fractions

Exercise 1 BasicIB · P1 TZ2 May 2013

Find the exact value of $\int_{1}^{2} ((x - 2)^{2} + \frac{1}{x} + sin π x) d x$ .

Show answer

Answer: $ln 2 + \frac{1}{3} - \frac{2}{π}$

Show working and answer

We integrate term by term.
First term: $\int_{1}^{2} (x - 2)^{2} d x = [\frac{( x - 2 ) ^{3}}{3}]_{1}^{2} = 0 - \frac{( - 1 ) ^{3}}{3} = \frac{1}{3}$ .
Second term: $\int_{1}^{2} \frac{1}{x} d x = [ln x]_{1}^{2} = ln 2 - ln 1 = ln 2$ .
Third term (adjusting by the chain rule, since there is a $π$ in the argument): $\int_{1}^{2} sin π x d x = [- \frac{cos π x}{π}]_{1}^{2} = - \frac{cos 2 π}{π} + \frac{cos π}{π} = - \frac{1}{π} - \frac{1}{π} = - \frac{2}{π}$ .
We add the three results.

Answer: $ln 2 + \frac{1}{3} - \frac{2}{π}$

Exercise 2 BasicIB · P1 May 2001

Let $f (t) = t^{\frac{1}{3}} (1 - \frac{1}{2 t ^{\frac{5}{3}}})$ . Calculate $\int f (t) d t$ .

Show answer

Answer: $\int f (t) d t = \frac{3}{4} t^{\frac{4}{3}} + \frac{3}{2} t^{- \frac{1}{3}} + C$

Show working and answer

We expand $f (t) = t^{\frac{1}{3}} - \frac{1}{2} t^{\frac{1}{3}} \cdot t^{- \frac{5}{3}} = t^{\frac{1}{3}} - \frac{1}{2} t^{- \frac{4}{3}}$ .
We integrate each power using $\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}$ :
$\int t^{\frac{1}{3}} d t = \frac{t ^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4} t^{\frac{4}{3}}$ .
$- \frac{1}{2} \int t^{- \frac{4}{3}} d t = - \frac{1}{2} \cdot \frac{t ^{- \frac{1}{3}}}{- \frac{1}{3}} = \frac{3}{2} t^{- \frac{1}{3}}$ .

Answer: $\int f (t) d t = \frac{3}{4} t^{\frac{4}{3}} + \frac{3}{2} t^{- \frac{1}{3}} + C$

Exercise 3 BasicIB · P1 TZ2 May 2019

Using the substitution $u = sin x$ , find $\int \frac{cos ^{3} x}{sin x} d x$ .

Show answer

Answer: $2 sin x - \frac{2}{5} sin^{5} x + C$

Show working and answer

Substitute $u = sin x$ , $d u = cos x d x$ .
We split $cos^{3} x = cos^{2} x cos x = (1 - sin^{2} x) cos x$ .
The integral becomes $\int \frac{1 - u ^{2}}{u} d u = \int (u^{- \frac{1}{2}} - u^{\frac{3}{2}}) d u$ .
Integrating: $2 u^{\frac{1}{2}} - \frac{2}{5} u^{\frac{5}{2}} + C$ .
Undo the substitution.

Answer: $2 sin x - \frac{2}{5} sin^{5} x + C$

Exercise 4 BasicIB · P1 TZ0 November 2015

Using integration by parts, find $\int x sin x d x$ .

Show answer

Answer: $- x cos x + sin x + C$

Show working and answer

Integration by parts: $\int u d v = uv - \int v d u$ .
We choose $u = x$ (it simplifies when differentiated) and $d v = sin x d x$ . Then $d u = d x$ and $v = - cos x$ .
$\int x sin x d x = - x cos x + \int cos x d x = - x cos x + sin x + C$ .
Check: $\frac{d}{d x} (- x cos x + sin x) = - cos x + x sin x + cos x = x sin x$ .

Answer: $- x cos x + sin x + C$

Exercise 5 BasicIB · P1 TZ2 May 2010

Find the value of $\int_{0}^{1} t ln (t + 1) d t$ .

Show answer

Answer: $\frac{1}{4}$

Show working and answer

Integration by parts with $u = ln (t + 1)$ and $d v = t d t$ . Then $d u = \frac{1}{t + 1} d t$ and $v = \frac{t ^{2}}{2}$ .
$\int_{0}^{1} t ln (t + 1) d t = [\frac{t ^{2}}{2} ln (t + 1)]_{0}^{1} - \frac{1}{2} \int_{0}^{1} \frac{t ^{2}}{t + 1} d t$ .
The first term equals $\frac{1}{2} ln 2$ .
We divide: $\frac{t ^{2}}{t + 1} = t - 1 + \frac{1}{t + 1}$ .
$\int_{0}^{1} (t - 1 + \frac{1}{t + 1}) d t = [\frac{t ^{2}}{2} - t + ln (t + 1)]_{0}^{1} = \frac{1}{2} - 1 + ln 2 = ln 2 - \frac{1}{2}$ .
Total: $\frac{1}{2} ln 2 - \frac{1}{2} (ln 2 - \frac{1}{2}) = \frac{1}{4}$ .

Answer: $\frac{1}{4}$

Exercise 6 BasicIB · P1 November 2002

Find $\int (θ cos θ - θ) d θ$ .

Show answer

Answer: $θ sin θ + cos θ - \frac{θ ^{2}}{2} + C$

Show working and answer

We integrate each term separately.
$\int θ cos θ d θ$ by parts with $u = θ$ and $d v = cos θ d θ$ : $d u = d θ$ , $v = sin θ$ . This gives $θ sin θ - \int sin θ d θ = θ sin θ + cos θ$ .
$\int θ d θ = \frac{θ ^{2}}{2}$ .
Subtract.

Answer: $θ sin θ + cos θ - \frac{θ ^{2}}{2} + C$

Exercise 7 BasicIB · P1 May 1996

Calculate $\int \frac{d x}{8 x - x ^{2} - 7}$ .

Show answer

Answer: $arcsin (\frac{x - 4}{3}) + C$

Show working and answer

Complete the square: $8 x - x^{2} - 7 = - (x^{2} - 8 x) - 7 = - ((x - 4)^{2} - 16) - 7 = 9 - (x - 4)^{2}$ .
The integral is $\int \frac{d x}{9 - ( x - 4 ) ^{2}}$ .
With $u = x - 4$ , this is the form $\int \frac{d u}{a ^{2} - u ^{2}} = arcsin \frac{u}{a}$ with $a = 3$ .

Answer: $arcsin (\frac{x - 4}{3}) + C$

Exercise 8 BasicIB · P1 November 1996

Calculate $\int \frac{d x}{x ^{2} + 6 x + 13}$ .

Show answer

Answer: $\frac{1}{2} arctan (\frac{x + 3}{2}) + C$

Show working and answer

Complete the square: $x^{2} + 6 x + 13 = (x + 3)^{2} + 4$ .
The integral is $\int \frac{d x}{( x + 3 ) ^{2} + 2 ^{2}}$ . With $u = x + 3$ , the form $\int \frac{d u}{u ^{2} + a ^{2}} = \frac{1}{a} arctan \frac{u}{a}$ with $a = 2$ gives the result.

Answer: $\frac{1}{2} arctan (\frac{x + 3}{2}) + C$

Exercise 9 BasicIB · P1 TZ1 May 2017

Calculate $\int arcsin x d x$ .

Show answer

Answer: $x arcsin x + 1 - x^{2} + C$

Show working and answer

Integration by parts with $u = arcsin x$ and $d v = d x$ . Then $d u = \frac{1}{1 - x ^{2}} d x$ and $v = x$ .
$\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x$ .
With $w = 1 - x^{2}$ , $d w = - 2 x d x$ : $\int \frac{x}{1 - x ^{2}} d x = - 1 - x^{2}$ .
Therefore $x arcsin x + 1 - x^{2} + C$ .

Answer: $x arcsin x + 1 - x^{2} + C$

Exercise 10 BasicIB · P1 May 1997

Calculate the indefinite integral $\int (x^{2} - 1)^{3} x d x$ .

Show answer

Answer: $\frac{( x ^{2} - 1 ) ^{4}}{8} + C$

Show working and answer

Substitution $u = x^{2} - 1$ , $d u = 2 x d x$ , that is, $x d x = \frac{1}{2} d u$ .
$\int u^{3} \cdot \frac{1}{2} d u = \frac{u ^{4}}{8} + C$ .
Undo the substitution.

Answer: $\frac{( x ^{2} - 1 ) ^{4}}{8} + C$

Exercise 11 BasicIB · P1 November 1997

Calculate $\int \frac{d x}{6 x - x ^{2} - 5}$ .

Show answer

Answer: $arcsin (\frac{x - 3}{2}) + C$

Show working and answer

Complete the square: $6 x - x^{2} - 5 = - (x - 3)^{2} + 9 - 5 = 4 - (x - 3)^{2}$ .
The integral is $\int \frac{d x}{2 ^{2} - ( x - 3 ) ^{2}}$ . With $u = x - 3$ and the form $\int \frac{d u}{a ^{2} - u ^{2}} = arcsin \frac{u}{a}$ with $a = 2$ , we obtain the result.

Answer: $arcsin (\frac{x - 3}{2}) + C$

Exercise 12 BasicIB · P1 May 1998

Calculate $\int arctan x d x$ .

Show answer

Answer: $x arctan x - \frac{1}{2} ln (1 + x^{2}) + C$

Show working and answer

Integration by parts with $u = arctan x$ and $d v = d x$ : $d u = \frac{1}{1 + x ^{2}} d x$ , $v = x$ .
$\int arctan x d x = x arctan x - \int \frac{x}{1 + x ^{2}} d x$ .
With $w = 1 + x^{2}$ , $d w = 2 x d x$ : $\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (1 + x^{2})$ .

Answer: $x arctan x - \frac{1}{2} ln (1 + x^{2}) + C$

Exercise 13 BasicIB · P1 November 1999

Calculate the real number $k > 1$ for which $\int_{1}^{k} (1 + \frac{1}{x ^{2}}) d x = \frac{3}{2}$ .

Show answer

Answer: $k = 2$

Show working and answer

An antiderivative of $1 + \frac{1}{x ^{2}} = 1 + x^{- 2}$ is $x - \frac{1}{x}$ .
$\int_{1}^{k} (1 + \frac{1}{x ^{2}}) d x = (k - \frac{1}{k}) - (1 - 1) = k - \frac{1}{k}$ .
Equate: $k - \frac{1}{k} = \frac{3}{2}$ . Multiplying by $2 k$ : $2 k^{2} - 3 k - 2 = 0$ .
$(2 k + 1) (k - 2) = 0$ , so $k = - \frac{1}{2}$ or $k = 2$ . As $k > 1$ , the solution is $k = 2$ .

Answer: $k = 2$

Exercise 14 BasicIB · P1 TZ1 May 2004

Calculate $\int x^{2} e^{x} d x$ .

Show answer

Answer: $x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C$

Show working and answer

Integration by parts with $u = x^{2}$ and $d v = e^{x} d x$ : $d u = 2 x d x$ , $v = e^{x}$ .
$\int x^{2} e^{x} d x = x^{2} e^{x} - 2 \int x e^{x} d x$ .
Again by parts with $u = x$ and $d v = e^{x} d x$ : $\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x}$ .
Substituting: $x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C$ .

Answer: $x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C$

Exercise 15 BasicIB · P2 TZ1 May 2008

Using an appropriate substitution, calculate $\int \frac{tan ( ln y )}{y} d y$ , $y > 0$ .

Show answer

Answer: $- ln ∣ cos (ln y) ∣ + C$

Show working and answer

Substitution $u = ln y$ , $d u = \frac{1}{y} d y$ .
$\int tan u d u = \int \frac{sin u}{cos u} d u = - ln ∣ cos u ∣ + C$ (with $w = cos u$ , $d w = - sin u d u$ ).
Undo the substitution.

Answer: $- ln ∣ cos (ln y) ∣ + C$

Exercise 16 BasicIB · P1 TZ1 May 2015

a) Calculate $\int (1 + tan^{2} x) d x$ .

b) Calculate $\int sin^{2} x d x$ .

Show answer

Answer: a) $tan x + C$ ; b) $\frac{1}{2} (x - \frac{sin 2 x}{2}) + C$

Show working and answer

a) 1. By the identity $1 + tan^{2} x = sec^{2} x$ , the integral is $\int sec^{2} x d x$ . 2. As $\frac{d}{d x} tan x = sec^{2} x$ , the result is $tan x + C$ .

b) 1. We use $cos 2 x = 1 - 2 sin^{2} x$ , that is, $sin^{2} x = \frac{1 - cos 2 x}{2}$ . 2. $\int \frac{1 - cos 2 x}{2} d x = \frac{1}{2} (x - \frac{sin 2 x}{2}) + C$ .

Answer: a) $tan x + C$ ; b) $\frac{1}{2} (x - \frac{sin 2 x}{2}) + C$

Exercise 17 BasicIB · P1 TZ0 November 2015

Use the substitution $u = ln x$ to find the value of $\int_{e}^{e^{2}} \frac{d x}{x ln x}$ .

Show answer

Answer: $ln 2$

Show working and answer

Substitution $u = ln x$ , so that $d u = \frac{1}{x} d x$ .
Change the limits: if $x = e$ , $u = ln e = 1$ ; if $x = e^{2}$ , $u = ln e^{2} = 2$ .
The integral is $\int_{1}^{2} \frac{d u}{u} = [ln ∣ u ∣]_{1}^{2}$ .
$= ln 2 - ln 1 = ln 2$ .

Answer: $ln 2$

Exercise 18 BasicIB · P1 TZ1 May 2004

Given that $\frac{d y}{d x} = 2 x - sin x$ and $y = 2$ when $x = 0$ , find an expression for $y$ in terms of $x$ .

Show answer

Answer: $y = x^{2} + cos x + 1$

Show working and answer

We integrate with respect to $x$ : $y = \int (2 x - sin x) d x = x^{2} + cos x + C$ .
Initial condition $y (0) = 2$ : $0 + 1 + C = 2$ , so $C = 1$ .

Answer: $y = x^{2} + cos x + 1$

Exercise 19 BasicIB · P1 TZ2 May 2004

Given that $\frac{d y}{d x} = e^{x} - 2 x$ , and that when $x = 0$ , $y = 3$ , find the expression of $y$ in terms of $x$ .

Show answer

Answer: $y = e^{x} - x^{2} + 2$

Show working and answer

We integrate: $y = \int (e^{x} - 2 x) d x = e^{x} - x^{2} + C$ .
Initial condition $y (0) = 3$ : $1 - 0 + C = 3$ , so $C = 2$ .

Answer: $y = e^{x} - x^{2} + 2$

Exercise 20 BasicIB · P2 TZ2 May 2022

The derivative of the function $g$ is given by $g^{'} (x) = 3 x^{2} + 5 e^{x}$ , where $x \in R$ . The graph of $g$ passes through the point $(0, 4)$ . Find $g (x)$ .

Show answer

Answer: $g (x) = x^{3} + 5 e^{x} - 1$

Show working and answer

We integrate: $g (x) = \int (3 x^{2} + 5 e^{x}) d x = x^{3} + 5 e^{x} + C$ .
It passes through $(0, 4)$ : $0 + 5 + C = 4$ , so $C = - 1$ .

Answer: $g (x) = x^{3} + 5 e^{x} - 1$

Exercise 21 BasicIB · P1 TZ0 November 2022

The derivative of the function $f$ is given by $f^{'} (x) = \frac{6 x}{x ^{2} + 1}$ . The graph of $y = f (x)$ passes through the point $(1, 5)$ . Find an expression for $f (x)$ .

Show answer

Answer: $f (x) = 3 ln (x^{2} + 1) + 5 - 3 ln 2$

Show working and answer

We integrate: $f (x) = \int \frac{6 x}{x ^{2} + 1} d x$ . Technique: the numerator is $3$ times the derivative of the denominator ( $(x^{2} + 1)^{'} = 2 x$ ), or the substitution $u = x^{2} + 1$ .
$f (x) = 3 ln (x^{2} + 1) + C$ (no absolute value is needed, since $x^{2} + 1 > 0$ ).
It passes through $(1, 5)$ : $3 ln 2 + C = 5$ , so $C = 5 - 3 ln 2$ .

Answer: $f (x) = 3 ln (x^{2} + 1) + 5 - 3 ln 2$

Exercise 22 MediumIB · P1 TZ0 November 2014

Using the substitution $u = 1 + x$ , find $\int \frac{x}{1 + x} d x$ .

Show answer

Answer: $2 [\frac{( 1 + x ) ^{2}}{2} + ln (1 + x) - 2 (1 + x)] + C$

Show working and answer

Substitute $u = 1 + x$ . Then $x = u - 1$ , that is $x = (u - 1)^{2}$ , and $d x = 2 (u - 1) d u$ .
The integral becomes $\int \frac{u - 1}{u} \cdot 2 (u - 1) d u = 2 \int \frac{( u - 1 ) ^{2}}{u} d u$ .
Expanding: $\frac{( u - 1 ) ^{2}}{u} = \frac{u ^{2} - 2 u + 1}{u} = u - 2 + \frac{1}{u}$ .
Integrating: $2 [\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣] + C$ .
Undo the substitution ( $u = 1 + x > 0$ ).

Answer: $2 [\frac{( 1 + x ) ^{2}}{2} + ln (1 + x) - 2 (1 + x)] + C$

Exercise 23 MediumIB · P1 TZ2 May 2005

a) Express $\frac{2 x + 4}{( x ^{2} + 4 ) ( x - 2 )}$ as a sum of partial fractions.

b) Hence, or otherwise, find $\int \frac{2 x + 4}{( x ^{2} + 4 ) ( x - 2 )} d x$ .

Show answer

Answer: a) $\frac{1}{x - 2} - \frac{x}{x ^{2} + 4}$ ; b) $ln ∣ x - 2∣ - \frac{1}{2} ln (x^{2} + 4) + C$

Show working and answer

a) 1. The decomposition is $\frac{2 x + 4}{( x ^{2} + 4 ) ( x - 2 )} = \frac{A}{x - 2} + \frac{B x + C}{x ^{2} + 4}$ (the factor $x^{2} + 4$ has no real roots, which is why it has a linear numerator). 2. Multiplying by the denominator: $2 x + 4 = A (x^{2} + 4) + (B x + C) (x - 2)$ . 3. With $x = 2$ : $8 = 8 A$ , so $A = 1$ . 4. Coefficient of $x^{2}$ : $0 = A + B$ , so $B = - 1$ . 5. Constant term: $4 = 4 A - 2 C = 4 - 2 C$ , so $C = 0$ . 6. Therefore $\frac{2 x + 4}{( x ^{2} + 4 ) ( x - 2 )} = \frac{1}{x - 2} - \frac{x}{x ^{2} + 4}$ .

b) 1. We integrate the decomposition from part a) term by term. 2. $\int \frac{1}{x - 2} d x = ln ∣ x - 2∣$ . 3. $\int \frac{x}{x ^{2} + 4} d x = \frac{1}{2} ln (x^{2} + 4)$ , because the numerator is half the derivative of the denominator. 4. We subtract the two results.

Answer: a) $\frac{1}{x - 2} - \frac{x}{x ^{2} + 4}$ ; b) $ln ∣ x - 2∣ - \frac{1}{2} ln (x^{2} + 4) + C$

Exercise 24 MediumIB · P1 TZ0 November 2009

a) Find $\int tan^{3} x d x$ .

b) Calculate $\int_{π /4}^{π /3} \frac{sec ^{2} x}{3 tan x} d x$ .

Show answer

Answer: a) $\frac{1}{2 cos ^{2} x} + ln ∣ cos x ∣ + C$ ; b) $\frac{3}{2} (33 - 1)$

Show working and answer

a) 1. We write $tan^{3} x = tan x tan^{2} x = tan x (sec^{2} x - 1)$ . 2. Therefore $\int tan^{3} x d x = \int tan x sec^{2} x d x - \int tan x d x$ . 3. For the first, substitute $u = tan x$ , $d u = sec^{2} x d x$ : $\int u d u = \frac{tan ^{2} x}{2}$ . 4. The second is $\int tan x d x = - ln ∣ cos x ∣$ . 5. The result is $\frac{tan ^{2} x}{2} + ln ∣ cos x ∣ + C$ . Since $tan^{2} x = sec^{2} x - 1$ , this is equivalent to $\frac{1}{2 cos ^{2} x} + ln ∣ cos x ∣ + C$ (the difference of $\frac{1}{2}$ is absorbed into the constant).

b) 1. Substitute $u = tan x$ , $d u = sec^{2} x d x$ . 2. Limits: if $x = \frac{π}{4}$ , $u = 1$ ; if $x = \frac{π}{3}$ , $u = 3$ . 3. $\int_{1}^{3} u^{- \frac{1}{3}} d u = [\frac{3}{2} u^{\frac{2}{3}}]_{1}^{3} = \frac{3}{2} ((3^{\frac{1}{2}})^{\frac{2}{3}} - 1) = \frac{3}{2} (3^{\frac{1}{3}} - 1)$ .

Answer: a) $\frac{1}{2 cos ^{2} x} + ln ∣ cos x ∣ + C$ ; b) $\frac{3}{2} (33 - 1)$

Exercise 25 MediumIB · P2 TZ2 May 2009

Using the substitution $x = 2 sin θ$ , verify that $\int 4 - x^{2} d x = A x 4 - x^{2} + B arcsin (\frac{x}{2}) + Constant$ , where $A$ and $B$ are constants whose values you must find.

Show answer

Answer: $A = \frac{1}{2}$ , $B = 2$

Show working and answer

Substitute $x = 2 sin θ$ : $d x = 2 cos θ d θ$ and $4 - x^{2} = 4 - 4 sin^{2} θ = 2 cos θ$ (with $cos θ \geq 0$ , that is $θ \in [- \frac{π}{2}, \frac{π}{2}]$ ).
The integral is $\int 2 cos θ \cdot 2 cos θ d θ = 4 \int cos^{2} θ d θ$ .
Using $cos^{2} θ = \frac{1 + cos 2 θ}{2}$ : $4 \int \frac{1 + cos 2 θ}{2} d θ = 2 θ + sin 2 θ + C$ .
Since $sin 2 θ = 2 sin θ cos θ$ , this becomes $2 θ + 2 sin θ cos θ$ .
Undo the substitution: $θ = arcsin \frac{x}{2}$ , $sin θ = \frac{x}{2}$ , $cos θ = \frac{4 - x ^{2}}{2}$ .
Then $2 θ + 2 sin θ cos θ = 2 arcsin \frac{x}{2} + 2 \cdot \frac{x}{2} \cdot \frac{4 - x ^{2}}{2} = 2 arcsin \frac{x}{2} + \frac{1}{2} x 4 - x^{2}$ .
Comparing with the required form, $A = \frac{1}{2}$ and $B = 2$ .

Answer: $A = \frac{1}{2}$ , $B = 2$

Exercise 26 MediumIB · P1 TZ2 May 2008

Verify that $\int_{0}^{π /6} x sin 2 x d x = \frac{3}{8} - \frac{π}{24}$ .

Show answer

Answer: verified: $\int_{0}^{π /6} x sin 2 x d x = \frac{3}{8} - \frac{π}{24}$ .

Show working and answer

Integration by parts with $u = x$ and $d v = sin 2 x d x$ . Then $d u = d x$ and $v = - \frac{cos 2 x}{2}$ .
$\int_{0}^{π /6} x sin 2 x d x = [- \frac{x cos 2 x}{2}]_{0}^{π /6} + \frac{1}{2} \int_{0}^{π /6} cos 2 x d x$ .
The remaining integral: $\frac{1}{2} [\frac{sin 2 x}{2}]_{0}^{π /6} = \frac{1}{4} sin \frac{π}{3} - 0 = \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8}$ .
The first term: $- \frac{π}{6} \cdot \frac{cos \frac{π}{3}}{2} - 0 = - \frac{π}{6} \cdot \frac{1}{4} = - \frac{π}{24}$ .
Adding: $\frac{3}{8} - \frac{π}{24}$ , which is what we had to verify.

Answer: verified: $\int_{0}^{π /6} x sin 2 x d x = \frac{3}{8} - \frac{π}{24}$ .

Exercise 27 MediumIB · P1 TZ0 November 2008

Calculate the exact value of $\int_{1}^{e} x^{2} ln x d x$ .

Show answer

Answer: $\frac{2 e ^{3} + 1}{9}$

Show working and answer

Integration by parts with $u = ln x$ and $d v = x^{2} d x$ . Then $d u = \frac{1}{x} d x$ and $v = \frac{x ^{3}}{3}$ .
$\int x^{2} ln x d x = \frac{x ^{3}}{3} ln x - \int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9}$ .
Evaluate between $1$ and $e$ : $(\frac{e ^{3}}{3} - \frac{e ^{3}}{9}) - (0 - \frac{1}{9})$ .
$= \frac{2 e ^{3}}{9} + \frac{1}{9}$ .

Answer: $\frac{2 e ^{3} + 1}{9}$

Exercise 28 MediumIB · P1 TZ2 May 2009

Find the value of $k$ such that $\int_{0}^{2} \frac{5 x + 11}{x ^{2} + 4 x + 3} d x = ln k$ .

Show answer

Answer: $k = 75$

Show working and answer

Factorise the denominator: $x^{2} + 4 x + 3 = (x + 1) (x + 3)$ .
Partial fractions: $\frac{5 x + 11}{( x + 1 ) ( x + 3 )} = \frac{A}{x + 1} + \frac{B}{x + 3}$ , so $5 x + 11 = A (x + 3) + B (x + 1)$ .
With $x = - 1$ : $6 = 2 A$ , so $A = 3$ . With $x = - 3$ : $- 4 = - 2 B$ , so $B = 2$ .
$\int_{0}^{2} (\frac{3}{x + 1} + \frac{2}{x + 3}) d x = [3 ln ∣ x + 1∣ + 2 ln ∣ x + 3∣]_{0}^{2}$ .
$= 3 ln 3 + 2 ln 5 - (3 ln 1 + 2 ln 3) = ln 3 + 2 ln 5 = ln 3 + ln 25 = ln 75$ .
Equating with $ln k$ , $k = 75$ .

Answer: $k = 75$

Exercise 29 MediumIB · P2 TZ2 May 2013

a) Find $\int x sec^{2} x d x$ .

b) Determine the value of $m$ , given that $\int_{0}^{m} x sec^{2} x d x = 0.5$ , where $m > 0$ .

Show answer

Answer: a) $x tan x + ln ∣ cos x ∣ + C$ ; b) $m \approx 0.822$

Show working and answer

a) 1. Integration by parts with $u = x$ and $d v = sec^{2} x d x$ . Then $d u = d x$ and $v = tan x$ . 2. $\int x sec^{2} x d x = x tan x - \int tan x d x = x tan x + ln ∣ cos x ∣ + C$ .

b) 1. Using the antiderivative above: $\int_{0}^{m} x sec^{2} x d x = m tan m + ln (cos m) - (0 + ln 1) = m tan m + ln (cos m)$ (for $0 < m < \frac{π}{2}$ , where $cos m > 0$ ). 2. We set up $m tan m + ln (cos m) = 0.5$ . 3. It cannot be solved by hand: using the GDC we solve the equation (or find the intersection of the graph of $y = m tan m + ln (cos m)$ with the line $y = 0.5$ ), and obtain $m \approx 0.822$ .

Answer: a) $x tan x + ln ∣ cos x ∣ + C$ ; b) $m \approx 0.822$

Exercise 30 MediumIB · P2 TZ1 May 2019

Use integration by parts to calculate $\int (ln x)^{2} d x$ .

Show answer

Answer: $x (ln x)^{2} - 2 x ln x + 2 x + C$

Show working and answer

Integration by parts with $u = (ln x)^{2}$ and $d v = d x$ . Then $d u = \frac{2 ln x}{x} d x$ and $v = x$ .
$\int (ln x)^{2} d x = x (ln x)^{2} - \int 2 ln x d x$ .
For $\int ln x d x$ we repeat integration by parts with $u = ln x$ and $d v = d x$ : $d u = \frac{1}{x} d x$ , $v = x$ , and $\int ln x d x = x ln x - \int d x = x ln x - x$ .
We substitute: $x (ln x)^{2} - 2 (x ln x - x) + C$ .

Answer: $x (ln x)^{2} - 2 x ln x + 2 x + C$

Exercise 31 MediumIB · P1 May 2003

Using the substitution $y = 2 - x$ , or in some other way, find $\int (\frac{x}{2 - x})^{2} d x$ .

Show answer

Answer: $\frac{4}{2 - x} - 2 + x + 4 ln ∣2 - x ∣ + C$

Show working and answer

Substitute $y = 2 - x$ : $x = 2 - y$ and $d x = - d y$ .
$\int (\frac{x}{2 - x})^{2} d x = - \int \frac{( 2 - y ) ^{2}}{y ^{2}} d y$ .
Expanding: $\frac{( 2 - y ) ^{2}}{y ^{2}} = \frac{4 - 4 y + y ^{2}}{y ^{2}} = \frac{4}{y ^{2}} - \frac{4}{y} + 1$ .
$- \int (4 y^{- 2} - \frac{4}{y} + 1) d y = - (- \frac{4}{y} - 4 ln ∣ y ∣ + y) = \frac{4}{y} + 4 ln ∣ y ∣ - y$ .
Undo the substitution: $\frac{4}{2 - x} + 4 ln ∣2 - x ∣ - (2 - x) = \frac{4}{2 - x} - 2 + x + 4 ln ∣2 - x ∣$ .

Answer: $\frac{4}{2 - x} - 2 + x + 4 ln ∣2 - x ∣ + C$

Exercise 32 MediumIB · P1 TZ2 May 2004

Find $\int \frac{ln x}{x} d x$ .

Show answer

Answer: $2 x ln x - 4 x + C$

Show working and answer

Integration by parts with $u = ln x$ and $d v = x^{- \frac{1}{2}} d x$ . Then $d u = \frac{1}{x} d x$ and $v = 2 x$ .
$\int \frac{ln x}{x} d x = 2 x ln x - \int 2 x \cdot \frac{1}{x} d x = 2 x ln x - 2 \int x^{- \frac{1}{2}} d x$ .
$2 \int x^{- \frac{1}{2}} d x = 4 x$ .

Answer: $2 x ln x - 4 x + C$

Exercise 33 MediumIB · P1 TZ1 May 2005

Let $f (x) = \frac{x ^{2} + 3 x + 12}{x ( x + 2 ) ^{2}}$ . Calculate $\int f (x) d x$ .

Show answer

Answer: $3 ln ∣ x ∣ - 2 ln ∣ x + 2∣ + \frac{5}{x + 2} + C$

Show working and answer

Partial fractions (repeated factor): $\frac{x ^{2} + 3 x + 12}{x ( x + 2 ) ^{2}} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{( x + 2 ) ^{2}}$ .
Multiplying: $x^{2} + 3 x + 12 = A (x + 2)^{2} + B x (x + 2) + C x$ .
With $x = 0$ : $12 = 4 A$ , so $A = 3$ . With $x = - 2$ : $4 - 6 + 12 = 10 = - 2 C$ , so $C = - 5$ .
Coefficient of $x^{2}$ : $1 = A + B$ , so $B = - 2$ .
We integrate: $\int (\frac{3}{x} - \frac{2}{x + 2} - \frac{5}{( x + 2 ) ^{2}}) d x = 3 ln ∣ x ∣ - 2 ln ∣ x + 2∣ + \frac{5}{x + 2} + C$ .

Answer: $3 ln ∣ x ∣ - 2 ln ∣ x + 2∣ + \frac{5}{x + 2} + C$

Exercise 34 MediumIB · P1 TZ1 May 2004

a) Calculate $\int_{0}^{m} \frac{d x}{x ^{2} + 4}$ , giving your answer in terms of $m$ .

b) Given that $\int_{0}^{m} \frac{d x}{x ^{2} + 4} = \frac{1}{3}$ , calculate the value of $m$ .

Show answer

Answer: a) $\frac{1}{2} arctan \frac{m}{2}$ ; b) $m = 2 tan \frac{2}{3}$

Show working and answer

a) 1. We write $\frac{1}{x ^{2} + 4} = \frac{1}{4} \cdot \frac{1}{1 + ( \frac{x}{2} ) ^{2}}$ . 2. An antiderivative is $\frac{1}{4} \cdot 2 arctan \frac{x}{2} = \frac{1}{2} arctan \frac{x}{2}$ . 3. $\int_{0}^{m} \frac{d x}{x ^{2} + 4} = \frac{1}{2} arctan \frac{m}{2} - \frac{1}{2} arctan 0 = \frac{1}{2} arctan \frac{m}{2}$ .

b) 1. Equating: $\frac{1}{2} arctan \frac{m}{2} = \frac{1}{3}$ , so $arctan \frac{m}{2} = \frac{2}{3}$ . 2. $\frac{m}{2} = tan \frac{2}{3}$ , so $m = 2 tan \frac{2}{3}$ .

Answer: a) $\frac{1}{2} arctan \frac{m}{2}$ ; b) $m = 2 tan \frac{2}{3}$

Exercise 35 MediumIB · P1 November 2004

Using the substitution $2 x = sin θ$ , or in any other way, find $\int 1 - 4 x^{2} d x$ .

Show answer

Answer: $\frac{1}{4} (arcsin 2 x + 2 x 1 - 4 x^{2}) + C$

Show working and answer

Substitute $2 x = sin θ$ : $x = \frac{1}{2} sin θ$ , $d x = \frac{1}{2} cos θ d θ$ and $1 - 4 x^{2} = 1 - sin^{2} θ = cos θ$ (with $cos θ \geq 0$ ).
$\int 1 - 4 x^{2} d x = \frac{1}{2} \int cos^{2} θ d θ = \frac{1}{2} \int \frac{1 + cos 2 θ}{2} d θ = \frac{1}{4} (θ + \frac{sin 2 θ}{2})$ .
Since $\frac{sin 2 θ}{2} = sin θ cos θ = 2 x 1 - 4 x^{2}$ and $θ = arcsin 2 x$ :
the result is $\frac{1}{4} (arcsin 2 x + 2 x 1 - 4 x^{2}) + C$ .

Answer: $\frac{1}{4} (arcsin 2 x + 2 x 1 - 4 x^{2}) + C$

Exercise 36 MediumIB · P1 TZ2 May 2007

Find $\int_{0}^{l n 3} \frac{e ^{x}}{e ^{2 x} + 9} d x$ , giving your answer in exact form.

Show answer

Answer: $\frac{1}{3} (\frac{π}{4} - arctan \frac{1}{3})$

Show working and answer

Substitute $u = e^{x}$ , $d u = e^{x} d x$ , and $e^{2 x} = u^{2}$ .
Limits: $x = 0 \Rightarrow u = 1$ ; $x = ln 3 \Rightarrow u = 3$ .
$\int_{1}^{3} \frac{d u}{u ^{2} + 9} = [\frac{1}{3} arctan \frac{u}{3}]_{1}^{3}$ .
$= \frac{1}{3} (arctan 1 - arctan \frac{1}{3}) = \frac{1}{3} (\frac{π}{4} - arctan \frac{1}{3})$ .

Answer: $\frac{1}{3} (\frac{π}{4} - arctan \frac{1}{3})$

Exercise 37 MediumIB · P1 TZ1 May 2007

Let $f (x) = x ln x - x$ , for $x > 0$ .

a) Find $f^{'} (x)$ .

b) Using integration by parts, find $\int (ln x)^{2} d x$ .

Show answer

Answer: a) $f^{'} (x) = ln x$ ; b) $x (ln x)^{2} - 2 x ln x + 2 x + C$

Show working and answer

a) 1. Differentiating with the product rule: $f^{'} (x) = 1 \cdot ln x + x \cdot \frac{1}{x} - 1$ . 2. $f^{'} (x) = ln x + 1 - 1 = ln x$ .

b) 1. Integration by parts with $u = (ln x)^{2}$ and $d v = d x$ : $d u = \frac{2 ln x}{x} d x$ , $v = x$ . 2. $\int (ln x)^{2} d x = x (ln x)^{2} - 2 \int ln x d x$ . 3. From part a), $f^{'} (x) = ln x$ , so $\int ln x d x = f (x) = x ln x - x$ . 4. We substitute: $x (ln x)^{2} - 2 (x ln x - x) + C$ .

Answer: a) $f^{'} (x) = ln x$ ; b) $x (ln x)^{2} - 2 x ln x + 2 x + C$

Exercise 38 MediumIB · P1 TZ0 November 2007

Calculate $\int_{0}^{a} arcsin x d x$ , $0 < a < 1$ .

Show answer

Answer: $a arcsin a + 1 - a^{2} - 1$

Show working and answer

Integration by parts with $u = arcsin x$ and $d v = d x$ . Then $d u = \frac{1}{1 - x ^{2}} d x$ and $v = x$ .
$\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x$ .
For the last integral, substitute $w = 1 - x^{2}$ , $d w = - 2 x d x$ : $\int \frac{x}{1 - x ^{2}} d x = - \frac{1}{2} \int w^{- \frac{1}{2}} d w = - 1 - x^{2}$ .
Antiderivative: $x arcsin x + 1 - x^{2}$ .
Evaluate between $0$ and $a$ : $(a arcsin a + 1 - a^{2}) - (0 + 1)$ .

Answer: $a arcsin a + 1 - a^{2} - 1$

Exercise 39 MediumIB · P2 TZ1 May 2009

a) Integrate $\int \frac{sin θ}{1 - cos θ} d θ$ .

b) Given that $\int_{π /2}^{a} \frac{sin θ}{1 - cos θ} d θ = \frac{1}{2}$ and $\frac{π}{2} < a < π$ , calculate the value of $a$ .

Show answer

Answer: a) $ln ∣1 - cos θ ∣ + C$ ; b) $a = arccos (1 - e) \approx 2.28$

Show working and answer

a) 1. Substitute $u = 1 - cos θ$ , $d u = sin θ d θ$ (the numerator is the derivative of the denominator). 2. $\int \frac{d u}{u} = ln ∣ u ∣ = ln ∣1 - cos θ ∣ + C$ . (As $1 - cos θ \geq 0$ , this can also be written as $ln (1 - cos θ)$ .)

b) 1. $[ln (1 - cos θ)]_{π /2}^{a} = ln (1 - cos a) - ln (1 - 0) = ln (1 - cos a)$ . 2. Equating: $ln (1 - cos a) = \frac{1}{2}$ , so $1 - cos a = e^{1/2}$ and $cos a = 1 - e \approx - 0.6487$ . 3. For $\frac{π}{2} < a < π$ the cosine is negative, and the solution is $a = arccos (1 - e) \approx 2.28$ .

Answer: a) $ln ∣1 - cos θ ∣ + C$ ; b) $a = arccos (1 - e) \approx 2.28$

Exercise 40 MediumIB · P1 November 1996

Calculate $\int e^{x} sin x d x$ .

Show answer

Answer: $\frac{e ^{x} ( sin x - cos x )}{2} + C$

Show working and answer

Let $I = \int e^{x} sin x d x$ . Integration by parts with $u = sin x$ and $d v = e^{x} d x$ : $d u = cos x d x$ , $v = e^{x}$ .
$I = e^{x} sin x - \int e^{x} cos x d x$ .
Second integration by parts with $u = cos x$ and $d v = e^{x} d x$ : $d u = - sin x d x$ , $v = e^{x}$ .
$\int e^{x} cos x d x = e^{x} cos x + \int e^{x} sin x d x = e^{x} cos x + I$ .
So $I = e^{x} sin x - e^{x} cos x - I$ . The integral reappears, so we solve for it: $2 I = e^{x} (sin x - cos x)$ .
$I = \frac{e ^{x} ( sin x - cos x )}{2} + C$ .

Answer: $\frac{e ^{x} ( sin x - cos x )}{2} + C$

Exercise 41 MediumIB · P2 TZ1 May 2018

a) Given that $2 x^{3} - 3 x + 1$ can be written in the form $A x (x^{2} + 1) + B x + C$ , find the values of the constants $A$ , $B$ and $C$ .

b) Hence, calculate $\int \frac{2 x ^{3} - 3 x + 1}{x ^{2} + 1} d x$ .

Show answer

Answer: a) $A = 2$ , $B = - 5$ , $C = 1$ ; b) $x^{2} - \frac{5}{2} ln (x^{2} + 1) + arctan x + C$

Show working and answer

a) 1. Expand: $A x (x^{2} + 1) + B x + C = A x^{3} + (A + B) x + C$ . 2. Equate with $2 x^{3} - 3 x + 1$ : coefficient of $x^{3}$ , $A = 2$ ; coefficient of $x$ , $A + B = - 3$ , so $B = - 5$ ; constant term, $C = 1$ .

b) 1. Using part (a): $\frac{2 x ^{3} - 3 x + 1}{x ^{2} + 1} = \frac{2 x ( x ^{2} + 1 ) - 5 x + 1}{x ^{2} + 1} = 2 x - \frac{5 x}{x ^{2} + 1} + \frac{1}{x ^{2} + 1}$ . 2. $\int 2 x d x = x^{2}$ . 3. $\int \frac{5 x}{x ^{2} + 1} d x = \frac{5}{2} ln (x^{2} + 1)$ (the numerator is proportional to the derivative of the denominator). 4. $\int \frac{1}{x ^{2} + 1} d x = arctan x$ .

Answer: a) $A = 2$ , $B = - 5$ , $C = 1$ ; b) $x^{2} - \frac{5}{2} ln (x^{2} + 1) + arctan x + C$

Exercise 42 MediumIB · P1 TZ1 May 2018

Let $y = arccos (\frac{x}{2})$ .

a) Calculate $\frac{d y}{d x}$ .

b) Calculate $\int_{0}^{1} arccos (\frac{x}{2}) d x$ .

Show answer

Answer: a) $- \frac{1}{4 - x ^{2}}$ ; b) $\frac{π}{3} - 3 + 2$

Show working and answer

a) 1. Derivative of $arccos u$ : $- \frac{u ^{'}}{1 - u ^{2}}$ , with $u = \frac{x}{2}$ and $u^{'} = \frac{1}{2}$ . 2. $\frac{d y}{d x} = - \frac{1/2}{1 - \frac{x ^{2}}{4}} = - \frac{1/2}{\frac{1}{2} 4 - x ^{2}} = - \frac{1}{4 - x ^{2}}$ .

b) 1. Integration by parts with $u = arccos \frac{x}{2}$ and $d v = d x$ . By part (a), $d u = - \frac{1}{4 - x ^{2}} d x$ , and $v = x$ . 2. $\int arccos \frac{x}{2} d x = x arccos \frac{x}{2} + \int \frac{x}{4 - x ^{2}} d x$ . 3. With $w = 4 - x^{2}$ , $d w = - 2 x d x$ : $\int \frac{x}{4 - x ^{2}} d x = - 4 - x^{2}$ . 4. Antiderivative: $x arccos \frac{x}{2} - 4 - x^{2}$ . 5. Evaluate between $0$ and $1$ : $(arccos \frac{1}{2} - 3) - (0 - 2) = \frac{π}{3} - 3 + 2$ .

Answer: a) $- \frac{1}{4 - x ^{2}}$ ; b) $\frac{π}{3} - 3 + 2$

Exercise 43 MediumIB · P1 TZ2 May 2018

a) Use the substitution $u = x^{\frac{1}{2}}$ to find $\int \frac{d x}{x ^{\frac{3}{2}} + x ^{\frac{1}{2}}}$ .

b) Hence, find the value of $\frac{1}{2} \int_{1}^{9} \frac{d x}{x ^{\frac{3}{2}} + x ^{\frac{1}{2}}}$ , in the form $arctan q$ , where $q \in R$ .

Show answer

Answer: a) $2 arctan (x^{\frac{1}{2}}) + C$ ; b) $arctan \frac{1}{2}$ (that is, $q = \frac{1}{2}$ )

Show working and answer

a) 1. Substitution $u = x^{\frac{1}{2}} = x$ : $x = u^{2}$ , $d x = 2 u d u$ . 2. The denominator: $x^{\frac{3}{2}} + x^{\frac{1}{2}} = u^{3} + u = u (u^{2} + 1)$ . 3. $\int \frac{2 u d u}{u ( u ^{2} + 1 )} = 2 \int \frac{d u}{u ^{2} + 1} = 2 arctan u + C$ . 4. Undo the substitution: $2 arctan (x^{\frac{1}{2}}) + C$ .

b) 1. By part (a), $\frac{1}{2} \int_{1}^{9} \frac{d x}{x ^{\frac{3}{2}} + x ^{\frac{1}{2}}} = \frac{1}{2} [2 arctan x]_{1}^{9} = arctan 3 - arctan 1$ . 2. Use $arctan a - arctan b = arctan \frac{a - b}{1 + ab}$ (this is equivalent to the formula for $tan (α - β)$ ): $arctan \frac{3 - 1}{1 + 3 \cdot 1} = arctan \frac{1}{2}$ .

Answer: a) $2 arctan (x^{\frac{1}{2}}) + C$ ; b) $arctan \frac{1}{2}$ (that is, $q = \frac{1}{2}$ )

Exercise 44 MediumIB · P1 TZ2 May 2017

a) Using the substitution $x = tan θ$ , show that $\int_{0}^{1} \frac{1}{( x ^{2} + 1 ) ^{2}} d x = \int_{0}^{\frac{π}{4}} cos^{2} θ d θ$ .

b) Hence, find the value of $\int_{0}^{1} \frac{1}{( x ^{2} + 1 ) ^{2}} d x$ .

Show answer

Answer: a) proved; b) $\frac{π}{8} + \frac{1}{4}$

Show working and answer

a) 1. Substitution $x = tan θ$ : $d x = sec^{2} θ d θ$ and $x^{2} + 1 = tan^{2} θ + 1 = sec^{2} θ$ . 2. Limits: $x = 0 \Rightarrow θ = 0$ ; $x = 1 \Rightarrow θ = \frac{π}{4}$ . 3. $\int_{0}^{1} \frac{d x}{( x ^{2} + 1 ) ^{2}} = \int_{0}^{π /4} \frac{sec ^{2} θ}{sec ^{4} θ} d θ = \int_{0}^{π /4} \frac{1}{sec ^{2} θ} d θ = \int_{0}^{π /4} cos^{2} θ d θ$ .

b) 1. Use $cos^{2} θ = \frac{1 + cos 2 θ}{2}$ . 2. $\int_{0}^{π /4} \frac{1 + cos 2 θ}{2} d θ = [\frac{θ}{2} + \frac{sin 2 θ}{4}]_{0}^{π /4} = \frac{π}{8} + \frac{1}{4} - 0$ .

Answer: a) proved; b) $\frac{π}{8} + \frac{1}{4}$

Exercise 45 MediumIB · P1 May 1997

Calculate the indefinite integral $\int x^{2} e^{- 2 x} d x$ .

Show answer

Answer: $- \frac{1}{2} e^{- 2 x} (x^{2} + x + \frac{1}{2}) + C$

Show working and answer

Integration by parts with $u = x^{2}$ and $d v = e^{- 2 x} d x$ : $d u = 2 x d x$ , $v = - \frac{1}{2} e^{- 2 x}$ .
$\int x^{2} e^{- 2 x} d x = - \frac{1}{2} x^{2} e^{- 2 x} + \int x e^{- 2 x} d x$ .
Integrate by parts again with $u = x$ and $d v = e^{- 2 x} d x$ : $d u = d x$ , $v = - \frac{1}{2} e^{- 2 x}$ . Then $\int x e^{- 2 x} d x = - \frac{1}{2} x e^{- 2 x} + \frac{1}{2} \int e^{- 2 x} d x = - \frac{1}{2} x e^{- 2 x} - \frac{1}{4} e^{- 2 x}$ .
Putting it together: $- \frac{1}{2} e^{- 2 x} (x^{2} + x + \frac{1}{2})$ .

Answer: $- \frac{1}{2} e^{- 2 x} (x^{2} + x + \frac{1}{2}) + C$

Exercise 46 MediumIB · P1 May 1999

Using the substitution $u = \frac{1}{2} x + 1$ , or by any other method, calculate the integral $\int x \frac{1}{2} x + 1 d x$ .

Show answer

Answer: $4 (\frac{2}{5} (\frac{1}{2} x + 1)^{\frac{5}{2}} - \frac{2}{3} (\frac{1}{2} x + 1)^{\frac{3}{2}}) + C$

Show working and answer

Substitution $u = \frac{1}{2} x + 1$ : $x = 2 (u - 1)$ and $d x = 2 d u$ .
$\int x \frac{1}{2} x + 1 d x = \int 2 (u - 1) u \cdot 2 d u = 4 \int (u^{\frac{3}{2}} - u^{\frac{1}{2}}) d u$ .
Integrating: $4 (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}}) + C$ .
Undo the substitution.

Answer: $4 (\frac{2}{5} (\frac{1}{2} x + 1)^{\frac{5}{2}} - \frac{2}{3} (\frac{1}{2} x + 1)^{\frac{3}{2}}) + C$

Exercise 47 MediumIB · P1 May 2000

Calculate the values of $a > 0$ such that $\int_{a}^{a^{2}} \frac{1}{1 + x ^{2}} d x = 0.22$ .

Show answer

Answer: $a \approx 2.04$ and $a \approx 2.62$

Show working and answer

An antiderivative of $\frac{1}{1 + x ^{2}}$ is $arctan x$ .
$\int_{a}^{a^{2}} \frac{d x}{1 + x ^{2}} = arctan (a^{2}) - arctan a$ .
We set up $arctan (a^{2}) - arctan a = 0.22$ . There is no simple algebraic solution: using the GDC, we find the points where the graph of $y = arctan (x^{2}) - arctan x$ meets the line $y = 0.22$ .
The function is positive only for $x > 1$ (for $0 < x < 1$ we have $x^{2} < x$ and the difference is negative). It rises to a maximum and then falls back towards $0$ , so it meets the line at two points: $a \approx 2.04$ and $a \approx 2.62$ .

Answer: $a \approx 2.04$ and $a \approx 2.62$

Exercise 48 MediumIB · P1 May 2002

a) Use integration by parts to find $\int x^{2} ln x d x$ .

b) Evaluate $\int_{1}^{2} x^{2} ln x d x$ .

Show answer

Answer: a) $\frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9} + C$ ; b) $\frac{8}{3} ln 2 - \frac{7}{9}$

Show working and answer

a) 1. Integration by parts with $u = ln x$ and $d v = x^{2} d x$ : $d u = \frac{1}{x} d x$ , $v = \frac{x ^{3}}{3}$ . 2. $\int x^{2} ln x d x = \frac{x ^{3}}{3} ln x - \int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9} + C$ .

b) 1. Evaluate the antiderivative between $1$ and $2$ : $(\frac{8}{3} ln 2 - \frac{8}{9}) - (0 - \frac{1}{9})$ . 2. $= \frac{8}{3} ln 2 - \frac{7}{9}$ .

Answer: a) $\frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9} + C$ ; b) $\frac{8}{3} ln 2 - \frac{7}{9}$

Exercise 49 MediumIB · P1 TZ2 May 2004

a) Find $\int_{0}^{m} \frac{d x}{2 x + 3}$ , giving the answer in terms of $m$ .

b) Assuming that $\int_{0}^{m} \frac{d x}{2 x + 3} = 1$ , calculate the value of $m$ .

Show answer

Answer: a) $\frac{1}{2} ln (\frac{2 m + 3}{3})$ ; b) $m = \frac{3 e ^{2} - 3}{2}$

Show working and answer

a) 1. An antiderivative of $\frac{1}{2 x + 3}$ is $\frac{1}{2} ln ∣2 x + 3∣$ (adjusting for the chain rule). 2. $\int_{0}^{m} \frac{d x}{2 x + 3} = \frac{1}{2} ln (2 m + 3) - \frac{1}{2} ln 3 = \frac{1}{2} ln \frac{2 m + 3}{3}$ (for $m > 0$ ).

b) 1. $\frac{1}{2} ln \frac{2 m + 3}{3} = 1$ , so $ln \frac{2 m + 3}{3} = 2$ . 2. $\frac{2 m + 3}{3} = e^{2}$ , so $2 m = 3 e^{2} - 3$ and $m = \frac{3 e ^{2} - 3}{2}$ .

Answer: a) $\frac{1}{2} ln (\frac{2 m + 3}{3})$ ; b) $m = \frac{3 e ^{2} - 3}{2}$

Exercise 50 MediumIB · P1 TZ0 November 2005

Find $\int e^{x} cos x d x$ .

Show answer

Answer: $\frac{e ^{x} ( sin x + cos x )}{2} + C$

Show working and answer

Let $I = \int e^{x} cos x d x$ . By parts with $u = cos x$ and $d v = e^{x} d x$ : $d u = - sin x d x$ , $v = e^{x}$ .
$I = e^{x} cos x + \int e^{x} sin x d x$ .
Once more by parts with $u = sin x$ : $\int e^{x} sin x d x = e^{x} sin x - \int e^{x} cos x d x = e^{x} sin x - I$ .
So $I = e^{x} cos x + e^{x} sin x - I$ . The integral reappears, so we solve for it: $2 I = e^{x} (sin x + cos x)$ .

Answer: $\frac{e ^{x} ( sin x + cos x )}{2} + C$

Exercise 51 MediumIB · P1 TZ0 May 2006

Find $\int e^{2 x} sin x d x$ .

Show answer

Answer: $\frac{1}{5} e^{2 x} (2 sin x - cos x) + C$

Show working and answer

Let $I = \int e^{2 x} sin x d x$ . By parts with $u = sin x$ and $d v = e^{2 x} d x$ : $d u = cos x d x$ , $v = \frac{1}{2} e^{2 x}$ .
$I = \frac{1}{2} e^{2 x} sin x - \frac{1}{2} \int e^{2 x} cos x d x$ .
Once more by parts with $u = cos x$ : $\int e^{2 x} cos x d x = \frac{1}{2} e^{2 x} cos x + \frac{1}{2} \int e^{2 x} sin x d x = \frac{1}{2} e^{2 x} cos x + \frac{1}{2} I$ .
Substituting: $I = \frac{1}{2} e^{2 x} sin x - \frac{1}{4} e^{2 x} cos x - \frac{1}{4} I$ .
The integral reappears: $\frac{5}{4} I = \frac{1}{4} e^{2 x} (2 sin x - cos x)$ , so $I = \frac{1}{5} e^{2 x} (2 sin x - cos x)$ .

Answer: $\frac{1}{5} e^{2 x} (2 sin x - cos x) + C$

Exercise 52 MediumIB · P1 TZ0 November 2009

a) Calculate $\int_{π /4}^{π /3} \frac{sec ^{2} x}{3 tan x} d x$ .

b) Find $\int tan^{3} x d x$ .

Show answer

Answer: a) $\frac{3}{2} (33 - 1)$ ; b) $\frac{1}{2 cos ^{2} x} + ln ∣ cos x ∣ + C$

Show working and answer

a) 1. Substitution $u = tan x$ , $d u = sec^{2} x d x$ . 2. Limits: $x = \frac{π}{4} \Rightarrow u = 1$ ; $x = \frac{π}{3} \Rightarrow u = 3$ . 3. $\int_{1}^{3} u^{- \frac{1}{3}} d u = [\frac{3}{2} u^{\frac{2}{3}}]_{1}^{3} = \frac{3}{2} ((3^{\frac{1}{2}})^{\frac{2}{3}} - 1) = \frac{3}{2} (3^{\frac{1}{3}} - 1)$ .

b) 1. $tan^{3} x = tan x (sec^{2} x - 1)$ , so $\int tan^{3} x d x = \int tan x sec^{2} x d x - \int tan x d x$ . 2. For the first, with $u = tan x$ : $\frac{tan ^{2} x}{2}$ . For the second: $- ln ∣ cos x ∣$ . 3. The result is $\frac{tan ^{2} x}{2} + ln ∣ cos x ∣ + C$ , which, using $tan^{2} x = sec^{2} x - 1$ , is equivalent to $\frac{1}{2 cos ^{2} x} + ln ∣ cos x ∣ + C$ (the difference $\frac{1}{2}$ goes into the constant).

Answer: a) $\frac{3}{2} (33 - 1)$ ; b) $\frac{1}{2 cos ^{2} x} + ln ∣ cos x ∣ + C$

Exercise 53 MediumIB · P1 TZ2 May 2015

Using the substitution $t = tan x$ , find $\int \frac{d x}{1 + sin ^{2} x}$ .

Give your answer in the form $m arctan (n tan x) + c$ , where $m$ and $n$ are constants that you must determine.

Show answer

Answer: $\frac{1}{2} arctan (2 tan x) + c$ , with $m = \frac{1}{2}$ and $n = 2$

Show working and answer

Substitution $t = tan x$ : $d t = sec^{2} x d x = (1 + t^{2}) d x$ , so $d x = \frac{d t}{1 + t ^{2}}$ .
$sin^{2} x = \frac{tan ^{2} x}{1 + tan ^{2} x} = \frac{t ^{2}}{1 + t ^{2}}$ .
Therefore $1 + sin^{2} x = \frac{1 + 2 t ^{2}}{1 + t ^{2}}$ .
$\int \frac{d x}{1 + sin ^{2} x} = \int \frac{1 + t ^{2}}{1 + 2 t ^{2}} \cdot \frac{d t}{1 + t ^{2}} = \int \frac{d t}{1 + 2 t ^{2}}$ .
With $s = 2 t$ , $d s = 2 d t$ : $\frac{1}{2} \int \frac{d s}{1 + s ^{2}} = \frac{1}{2} arctan s$ .
Undoing the substitutions: $\frac{1}{2} arctan (2 tan x) + c$ , so $m = \frac{1}{2}$ and $n = 2$ .

Answer: $\frac{1}{2} arctan (2 tan x) + c$ , with $m = \frac{1}{2}$ and $n = 2$

Exercise 54 MediumIB · P2 TZ0 November 2012

Using the substitution $x = sin t$ , calculate $\int \frac{x ^{3}}{1 - x ^{2}} d x$ .

Show answer

Answer: $- 1 - x^{2} + \frac{( 1 - x ^{2} ) ^{3}}{3} + C$

Show working and answer

Substitution $x = sin t$ : $d x = cos t d t$ and $1 - x^{2} = cos t$ (with $cos t > 0$ ).
$\int \frac{x ^{3}}{1 - x ^{2}} d x = \int \frac{sin ^{3} t}{cos t} cos t d t = \int sin^{3} t d t$ .
Write $sin^{3} t = (1 - cos^{2} t) sin t$ and, with $w = cos t$ ( $d w = - sin t d t$ ): $- \int (1 - w^{2}) d w = - w + \frac{w ^{3}}{3}$ .
That is, $- cos t + \frac{cos ^{3} t}{3}$ , and since $cos t = 1 - x^{2}$ :

Answer: $- 1 - x^{2} + \frac{( 1 - x ^{2} ) ^{3}}{3} + C$

Exercise 55 MediumIB · P1 TZ1 May 2014

a) Use the identity $cos 2 θ = 2 cos^{2} θ - 1$ to prove that $cos \frac{1}{2} x = \frac{1 + cos x}{2}$ , $0 \leq x \leq π$ .

b) Find a similar expression for $sin \frac{1}{2} x$ , $0 \leq x \leq π$ .

c) Using the above results, find the value of $\int_{0}^{\frac{π}{2}} (1 + cos x + 1 - cos x) d x$ .

Show answer

Answer: a) proved; b) $sin \frac{1}{2} x = \frac{1 - cos x}{2}$ ; c) $22$

Show working and answer

a) 1. Substitute $θ = \frac{x}{2}$ into $cos 2 θ = 2 cos^{2} θ - 1$ : $cos x = 2 cos^{2} \frac{x}{2} - 1$ . 2. Rearranging: $cos^{2} \frac{x}{2} = \frac{1 + cos x}{2}$ . 3. If $0 \leq x \leq π$ , then $0 \leq \frac{x}{2} \leq \frac{π}{2}$ and $cos \frac{x}{2} \geq 0$ , so we take the positive root: $cos \frac{1}{2} x = \frac{1 + cos x}{2}$ .

b) 1. We start from $cos 2 θ = 1 - 2 sin^{2} θ$ (the identity $cos 2 θ = 2 cos^{2} θ - 1$ with $cos^{2} θ = 1 - sin^{2} θ$ ). With $θ = \frac{x}{2}$ : $cos x = 1 - 2 sin^{2} \frac{x}{2}$ . 2. So $sin^{2} \frac{x}{2} = \frac{1 - cos x}{2}$ and, as $sin \frac{x}{2} \geq 0$ on that interval, $sin \frac{1}{2} x = \frac{1 - cos x}{2}$ .

c) 1. From a) and b): $1 + cos x = 2 cos \frac{x}{2}$ and $1 - cos x = 2 sin \frac{x}{2}$ (on $0 \leq x \leq \frac{π}{2}$ ). 2. $\int_{0}^{π /2} 2 (cos \frac{x}{2} + sin \frac{x}{2}) d x = 2 [2 sin \frac{x}{2} - 2 cos \frac{x}{2}]_{0}^{π /2}$ . 3. At $\frac{π}{2}$ : $2 \cdot \frac{2}{2} - 2 \cdot \frac{2}{2} = 0$ . At $0$ : $0 - 2 = - 2$ . 4. The value is $2 (0 - (- 2)) = 22$ .

Answer: a) proved; b) $sin \frac{1}{2} x = \frac{1 - cos x}{2}$ ; c) $22$

Exercise 56 MediumIB · P1 TZ1 May 2015

Using the substitution $u = e^{x} + 3$ , calculate $\int \frac{e ^{x}}{e ^{2 x} + 6 e ^{x} + 13} d x$ .

Show answer

Answer: $\frac{1}{2} arctan (\frac{e ^{x} + 3}{2}) + C$

Show working and answer

Substitution $u = e^{x} + 3$ , so that $d u = e^{x} d x$ .
Complete the square in the denominator: $e^{2 x} + 6 e^{x} + 13 = (e^{x} + 3)^{2} + 4 = u^{2} + 4$ .
The integral becomes $\int \frac{d u}{u ^{2} + 4}$ , which is an arctangent: $\int \frac{d u}{u ^{2} + a ^{2}} = \frac{1}{a} arctan \frac{u}{a}$ , with $a = 2$ .
Therefore $\int \frac{d u}{u ^{2} + 4} = \frac{1}{2} arctan \frac{u}{2} + C$ .
Undo the substitution, $u = e^{x} + 3$ .

Answer: $\frac{1}{2} arctan (\frac{e ^{x} + 3}{2}) + C$

Exercise 57 MediumIB · P1 TZ2 May 2015

Show that $\int_{1}^{2} x^{3} ln x d x = 4 ln 2 - \frac{15}{16}$ .

Show answer

Answer: proved: $\int_{1}^{2} x^{3} ln x d x = 4 ln 2 - \frac{15}{16}$ .

Show working and answer

We integrate by parts, $\int u d v = uv - \int v d u$ , with $u = ln x$ (it simplifies when differentiated) and $d v = x^{3} d x$ .
Then $d u = \frac{1}{x} d x$ and $v = \frac{x ^{4}}{4}$ .
$\int_{1}^{2} x^{3} ln x d x = [\frac{x ^{4}}{4} ln x]_{1}^{2} - \int_{1}^{2} \frac{x ^{4}}{4} \cdot \frac{1}{x} d x = [\frac{x ^{4}}{4} ln x]_{1}^{2} - \int_{1}^{2} \frac{x ^{3}}{4} d x$ .
First term: $\frac{16}{4} ln 2 - \frac{1}{4} ln 1 = 4 ln 2$ .
Second term: $\int_{1}^{2} \frac{x ^{3}}{4} d x = [\frac{x ^{4}}{16}]_{1}^{2} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}$ .
Subtracting: $4 ln 2 - \frac{15}{16}$ , which is what we had to show.

Answer: proved: $\int_{1}^{2} x^{3} ln x d x = 4 ln 2 - \frac{15}{16}$ .

Exercise 58 MediumIB · P2 TZ1 May 2014

Let $f (x) = x (x + 2)^{6}$ .

a) Solve the inequality $f (x) > x$ .

b) Calculate $\int f (x) d x$ .

Show answer

Answer: a) $x \in] - 3, - 1 [\cup] 0, + \infty [$ ; b) $\frac{( x + 2 ) ^{8}}{8} - \frac{2 ( x + 2 ) ^{7}}{7} + C$

Show working and answer

a) 1. We set up $x (x + 2)^{6} > x$ , that is $x [(x + 2)^{6} - 1] > 0$ . 2. $(x + 2)^{6} > 1 ⟺ ∣ x + 2∣ > 1 ⟺ x > - 1$ or $x < - 3$ . And $(x + 2)^{6} < 1 ⟺ - 3 < x < - 1$ . 3. If $x > 0$ we need $(x + 2)^{6} > 1$ , which always holds (since $x + 2 > 2$ ). 4. If $x < 0$ we need $(x + 2)^{6} < 1$ , that is $- 3 < x < - 1$ . 5. $x = 0$ does not work (it would give $0 > 0$ ).

b) 1. Substitution $u = x + 2$ , $d u = d x$ , $x = u - 2$ . 2. $\int x (x + 2)^{6} d x = \int (u - 2) u^{6} d u = \int (u^{7} - 2 u^{6}) d u = \frac{u ^{8}}{8} - \frac{2 u ^{7}}{7} + C$ . 3. Undo the substitution.

Answer: a) $x \in] - 3, - 1 [\cup] 0, + \infty [$ ; b) $\frac{( x + 2 ) ^{8}}{8} - \frac{2 ( x + 2 ) ^{7}}{7} + C$

Exercise 59 MediumIB · P2 TZ0 November 2005

a) Given that $\frac{x ^{2}}{( 1 + x ) ( 1 + x ^{2} )} = \frac{a}{1 + x} + \frac{b x + c}{1 + x ^{2}}$ , calculate the values of $a$ , $b$ and $c$ .

b) i) Hence, find $I = \int \frac{x ^{2}}{( 1 + x ) ( 1 + x ^{2} )} d x$ .

(ii) If $I = \frac{π}{4}$ when $x = 1$ , calculate the value of the constant of integration.

Give your answer in the form $p + q ln r$ , where $p, q, r \in R$ .

Show answer

Answer: a) $a = \frac{1}{2}$ , $b = \frac{1}{2}$ , $c = - \frac{1}{2}$ ; b) (i) $I = \frac{1}{2} ln ∣1 + x ∣ + \frac{1}{4} ln (1 + x^{2}) - \frac{1}{2} arctan x + C$ ; (ii) $C = \frac{3 π}{8} - \frac{3}{4} ln 2$

Show working and answer

a) 1. Clearing denominators: $x^{2} = a (1 + x^{2}) + (b x + c) (1 + x)$ . 2. Expanding: $x^{2} = (a + b) x^{2} + (b + c) x + (a + c)$ . 3. Equating coefficients: $a + b = 1$ , $b + c = 0$ , $a + c = 0$ . 4. From the last two, $c = - a$ and $b = a$ ; then $2 a = 1$ . Therefore $a = \frac{1}{2}$ , $b = \frac{1}{2}$ , $c = - \frac{1}{2}$ .

b) (i) Integrating by partial fractions (part a).

$I = \int [\frac{1}{2 ( 1 + x )} + \frac{\frac{1}{2} x - \frac{1}{2}}{1 + x ^{2}}] d x = \frac{1}{2} \int \frac{d x}{1 + x} + \frac{1}{2} \int \frac{x}{1 + x ^{2}} d x - \frac{1}{2} \int \frac{d x}{1 + x ^{2}}$ .
The second is done with $u = 1 + x^{2}$ ( $d u = 2 x d x$ ): $\frac{1}{2} \cdot \frac{1}{2} ln (1 + x^{2})$ . The third is the arctangent integral.
$I = \frac{1}{2} ln ∣1 + x ∣ + \frac{1}{4} ln (1 + x^{2}) - \frac{1}{2} arctan x + C$ .

(ii) 1. Imposing $I (1) = \frac{π}{4}$ : $\frac{1}{2} ln 2 + \frac{1}{4} ln 2 - \frac{1}{2} \cdot \frac{π}{4} + C = \frac{π}{4}$ . 2. $C = \frac{π}{4} + \frac{π}{8} - \frac{3}{4} ln 2 = \frac{3 π}{8} - \frac{3}{4} ln 2$ , of the form $p + q ln r$ with $p = \frac{3 π}{8}$ , $q = - \frac{3}{4}$ , $r = 2$ .

Exercise 60 MediumIB · P2 November 1996

Let $I_{n} (x) = \int_{0}^{x} t^{n} e^{- t} d t$ , where $n \in N$ .

a) Calculate $I_{0} (x)$ .

b) Show that $I_{n} (x) = - x^{n} e^{- x} + n I_{n - 1} (x)$ for $n \geq 1$ .

c) Evaluate $I_{3} (1)$ .

Show answer

Answer: a) $I_{0} (x) = 1 - e^{- x}$ ; b) verified; c) $I_{3} (1) = 6 - 16 e^{- 1}$

Show working and answer

a) $I_{0} (x) = \int_{0}^{x} e^{- t} d t = [- e^{- t}]_{0}^{x} = 1 - e^{- x}$ .

b) By parts with $u = t^{n}$ and $d v = e^{- t} d t$ (so $d u = n t^{n - 1} d t$ and $v = - e^{- t}$ ):

$I_{n} (x) = [- t^{n} e^{- t}]_{0}^{x} + n \int_{0}^{x} t^{n - 1} e^{- t} d t$ .
For $n \geq 1$ the term at $t = 0$ is $0$ , so $I_{n} (x) = - x^{n} e^{- x} + n I_{n - 1} (x)$ .

c) We apply the formula with $x = 1$ :

$I_{0} (1) = 1 - e^{- 1}$ .
$I_{1} (1) = - e^{- 1} + 1 \cdot I_{0} (1) = 1 - 2 e^{- 1}$ .
$I_{2} (1) = - e^{- 1} + 2 I_{1} (1) = 2 - 5 e^{- 1}$ .
$I_{3} (1) = - e^{- 1} + 3 I_{2} (1) = 6 - 16 e^{- 1}$ .

Answer: a) $I_{0} (x) = 1 - e^{- x}$ ; b) verified; c) $I_{3} (1) = 6 - 16 e^{- 1}$

Exercise 61 MediumIB · P2 TZ0 November 2018

A function $f$ satisfies $f (0) = - 4$ , $f (1) = 0$ , and its second derivative is $f^{''} (x) = 15 x + \frac{1}{( x + 1 ) ^{2}}$ , $x \geq 0$ . Find $f (x)$ .

Show answer

Answer: $f (x) = 4 x^{\frac{5}{2}} - ln (x + 1) + x ln 2 - 4$

Show working and answer

We integrate $f^{''} (x) = 15 x + \frac{1}{( x + 1 ) ^{2}}$ : $f^{'} (x) = 10 x^{3/2} - \frac{1}{x + 1} + A$ .
Integrating again: $f (x) = 4 x^{5/2} - ln (x + 1) + A x + B$ .
Using $f (0) = - 4$ : $B = - 4$ .
Using $f (1) = 0$ : $4 - ln 2 + A - 4 = 0$ , so $A = ln 2$ .
Hence $f (x) = 4 x^{5/2} - ln (x + 1) + x ln 2 - 4$ .

Answer: $f (x) = 4 x^{\frac{5}{2}} - ln (x + 1) + x ln 2 - 4$

Exercise 62 MediumIB · P1 TZ2 May 2018

Consider the functions $f$ , $g$ , defined for $x \in R$ , by $f (x) = e^{- x} sin x$ and $g (x) = e^{- x} cos x$ .

a) Find

(i) $f^{'} (x)$ ;

(ii) $g^{'} (x)$ .

b) Hence, or otherwise, find $\int_{0}^{π} e^{- x} sin x d x$ .

Show answer

Answer: a) (i) $f^{'} (x) = - e^{- x} sin x + e^{- x} cos x$ ; (ii) $g^{'} (x) = - e^{- x} cos x - e^{- x} sin x$ ; b) $\frac{e ^{- π} + 1}{2}$

Show working and answer

a) (i) Product rule: $f^{'} (x) = - e^{- x} sin x + e^{- x} cos x$ .

(ii) Product rule: $g^{'} (x) = - e^{- x} cos x - e^{- x} sin x$ .

b) 1. Adding: $f^{'} (x) + g^{'} (x) = - 2 e^{- x} sin x$ , so $e^{- x} sin x = - \frac{1}{2} (f^{'} + g^{'})$ . 2. Hence $\int_{0}^{π} e^{- x} sin x d x = - \frac{1}{2} [f (x) + g (x)]_{0}^{π} = - \frac{1}{2} [e^{- x} (sin x + cos x)]_{0}^{π}$ . 3. At $π$ : $e^{- π} (0 - 1) = - e^{- π}$ ; at $0$ : $1 \cdot (0 + 1) = 1$ . 4. $= - \frac{1}{2} (- e^{- π} - 1) = \frac{e ^{- π} + 1}{2}$ .

Answer: a) (i) $f^{'} (x) = - e^{- x} sin x + e^{- x} cos x$ ; (ii) $g^{'} (x) = - e^{- x} cos x - e^{- x} sin x$ ; b) $\frac{e ^{- π} + 1}{2}$

Exercise 63 MediumIB · P1 TZ1 May 2015

A function $f$ is defined by $f (x) = \frac{3 x - 2}{2 x - 1}$ , $x \in R$ , $x \neq = \frac{1}{2}$ .

a) Find an expression for $f^{- 1} (x)$ .

b) Given that $f (x)$ can be written in the form $f (x) = A + \frac{B}{2 x - 1}$ , calculate the values of the constants $A$ and $B$ .

c) Hence, calculate $\int \frac{3 x - 2}{2 x - 1} d x$ .

Show answer

Answer: a) $f^{- 1} (x) = \frac{x - 2}{2 x - 3}$ ; b) $A = \frac{3}{2}$ , $B = - \frac{1}{2}$ ; c) $\frac{3}{2} x - \frac{1}{4} ln ∣2 x - 1∣ + C$

Show working and answer

a) 1. We write $y = \frac{3 x - 2}{2 x - 1}$ and solve for $x$ : $y (2 x - 1) = 3 x - 2$ . 2. $2 x y - y = 3 x - 2$ , so $x (2 y - 3) = y - 2$ and $x = \frac{y - 2}{2 y - 3}$ . 3. Interchanging the variables: $f^{- 1} (x) = \frac{x - 2}{2 x - 3}$ .

b) 1. $A + \frac{B}{2 x - 1} = \frac{A ( 2 x - 1 ) + B}{2 x - 1} = \frac{2 A x - A + B}{2 x - 1}$ . 2. Equating with $\frac{3 x - 2}{2 x - 1}$ : $2 A = 3$ and $- A + B = - 2$ . 3. From the first, $A = \frac{3}{2}$ ; from the second, $B = - 2 + \frac{3}{2} = - \frac{1}{2}$ .

c) 1. From the previous part, $\int \frac{3 x - 2}{2 x - 1} d x = \int (\frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2 x - 1}) d x$ . 2. The integral of $\frac{1}{2 x - 1}$ is $\frac{1}{2} ln ∣2 x - 1∣$ (immediate integration by the chain rule). 3. This gives $\frac{3}{2} x - \frac{1}{2} \cdot \frac{1}{2} ln ∣2 x - 1∣ + C = \frac{3}{2} x - \frac{1}{4} ln ∣2 x - 1∣ + C$ .

Answer: a) $f^{- 1} (x) = \frac{x - 2}{2 x - 3}$ ; b) $A = \frac{3}{2}$ , $B = - \frac{1}{2}$ ; c) $\frac{3}{2} x - \frac{1}{4} ln ∣2 x - 1∣ + C$

Exercise 64 MediumIB · P1 TZ2 May 2021

Using the substitution $u = sin x$ , find $\int \frac{sin x cos x}{sin ^{2} x - sin x - 2} d x$ .

Show answer

Answer: $\frac{2}{3} ln (2 - sin x) + \frac{1}{3} ln ∣ sin x + 1∣ + C$

Show working and answer

Technique: substitution $u = sin x$ , with $d u = cos x d x$ .
The integral becomes $\int \frac{u}{u ^{2} - u - 2} d u = \int \frac{u}{( u - 2 ) ( u + 1 )} d u$ .
Partial fractions: $\frac{u}{( u - 2 ) ( u + 1 )} = \frac{A}{u - 2} + \frac{B}{u + 1}$ , that is, $u = A (u + 1) + B (u - 2)$ .
Putting $u = 2$ : $2 = 3 A$ , so $A = \frac{2}{3}$ . Putting $u = - 1$ : $- 1 = - 3 B$ , so $B = \frac{1}{3}$ .
Integrating: $\frac{2}{3} ln ∣ u - 2∣ + \frac{1}{3} ln ∣ u + 1∣ + C$ .
Undo the substitution. As $sin x - 2 < 0$ , $∣ sin x - 2∣ = 2 - sin x$ .

Answer: $\frac{2}{3} ln (2 - sin x) + \frac{1}{3} ln ∣ sin x + 1∣ + C$

Exercise 65 MediumIB · P1 TZ2 May 2022

Using the substitution $u = sec x$ or otherwise, find an expression for $\int_{0}^{\frac{π}{3}} sec^{n} x tan x d x$ in terms of $n$ , where $n$ is a non-zero real number.

Show answer

Answer: $\frac{2 ^{n} - 1}{n}$

Show working and answer

We rewrite the integrand: $sec^{n} x tan x = sec^{n - 1} x \cdot (sec x tan x)$ .
Technique: substitution $u = sec x$ , with $d u = sec x tan x d x$ .
Limits: $x = 0 \Rightarrow u = sec 0 = 1$ ; $x = \frac{π}{3} \Rightarrow u = sec \frac{π}{3} = 2$ .
$\int_{0}^{π /3} sec^{n} x tan x d x = \int_{1}^{2} u^{n - 1} d u = [\frac{u ^{n}}{n}]_{1}^{2}$ (valid because $n \neq = 0$ ).
$= \frac{2 ^{n} - 1}{n}$ .

Answer: $\frac{2 ^{n} - 1}{n}$

Exercise 66 HardIB · P2 TZ0 November 2013

Using the substitution $x = 2 tan u$ , verify that $\int \frac{d x}{x ^{2} x ^{2} + 4} = \frac{- x ^{2} + 4}{4 x} + C$ .

Show answer

Answer: verified: $\int \frac{d x}{x ^{2} x ^{2} + 4} = \frac{- x ^{2} + 4}{4 x} + C$ .

Show working and answer

Substitute $x = 2 tan u$ : $d x = 2 sec^{2} u d u$ and $x^{2} + 4 = 4 tan^{2} u + 4 = 2 sec u$ (with $sec u > 0$ ).
The denominator: $x^{2} x^{2} + 4 = 4 tan^{2} u \cdot 2 sec u = 8 tan^{2} u sec u$ .
$\int \frac{d x}{x ^{2} x ^{2} + 4} = \int \frac{2 sec ^{2} u}{8 tan ^{2} u sec u} d u = \frac{1}{4} \int \frac{sec u}{tan ^{2} u} d u$ .
Since $\frac{sec u}{tan ^{2} u} = \frac{1/ cos u}{sin ^{2} u / cos ^{2} u} = \frac{cos u}{sin ^{2} u}$ , we get $\frac{1}{4} \int \frac{cos u}{sin ^{2} u} d u$ .
With $w = sin u$ ( $d w = cos u d u$ ): $\frac{1}{4} \int w^{- 2} d w = - \frac{1}{4 w} = - \frac{1}{4 sin u} + C$ .
Undo the substitution: $tan u = \frac{x}{2}$ . In a right-angled triangle with legs $x$ and $2$ the hypotenuse is $x^{2} + 4$ , so $sin u = \frac{x}{x ^{2} + 4}$ .
Then $- \frac{1}{4 sin u} = - \frac{x ^{2} + 4}{4 x}$ .

Answer: verified: $\int \frac{d x}{x ^{2} x ^{2} + 4} = \frac{- x ^{2} + 4}{4 x} + C$ .

Exercise 67 HardIB · P2 TZ0 November 2017Proof

Using the substitution $x^{2} = 2 sec θ$ , show that $\int \frac{d x}{x x ^{4} - 4} = \frac{1}{4} arccos (\frac{2}{x ^{2}}) + c$ .

Show answer

Answer: proved: $\int \frac{d x}{x x ^{4} - 4} = \frac{1}{4} arccos (\frac{2}{x ^{2}}) + c$ .

Show working and answer

Substitution $x^{2} = 2 sec θ$ . Differentiating, $2 x d x = 2 sec θ tan θ d θ$ , that is, $d x = \frac{sec θ tan θ}{x} d θ$ .
The radicand: $x^{4} - 4 = 4 sec^{2} θ - 4 = 4 tan^{2} θ$ , so $x^{4} - 4 = 2 tan θ$ (taking $θ \in] 0, \frac{π}{2} [$ , where $tan θ > 0$ ).
$\int \frac{d x}{x x ^{4} - 4} = \int \frac{sec θ tan θ}{x ^{2} \cdot 2 tan θ} d θ = \int \frac{sec θ}{2 \cdot 2 sec θ} d θ = \frac{1}{4} \int d θ = \frac{θ}{4} + c$ .
Undo the substitution: $sec θ = \frac{x ^{2}}{2}$ , so $cos θ = \frac{2}{x ^{2}}$ and $θ = arccos \frac{2}{x ^{2}}$ .
This gives $\frac{1}{4} arccos (\frac{2}{x ^{2}}) + c$ .

Answer: proved: $\int \frac{d x}{x x ^{4} - 4} = \frac{1}{4} arccos (\frac{2}{x ^{2}}) + c$ .

Exercise 68 HardIB · P1 TZ2 May 2017

Consider the function $f$ defined by $f (x) = x^{2} - a^{2}$ , $x \in R$ , where $a$ is a positive constant.

a) Sketch approximately the following curves on separate sets of axes, showing all intercepts with the $x$ - and $y$ -axes, maxima, minima and any asymptotes.

(i) $y = f (x)$

(ii) $y = \frac{1}{f ( x )}$

(iii) $y = \frac{1}{f ( x )}$ .

b) Find $\int f (x) cos x d x$ .

The function $g$ is defined by $g (x) = x f (x)$ , for $∣ x ∣ > a$ .

c) By finding $g^{'} (x)$ , explain why $g$ is an increasing function.

View solution

a) (i) 1. It is a parabola opening upwards, with $x$ -intercepts $(- a, 0)$ and $(a, 0)$ , and a minimum (which is also the $y$ -intercept) at $(0, - a^{2})$ .

Parabola y = x² − a² opening upwards: it cuts the x-axis at (−a, 0) and (a, 0) and has its minimum at (0, −a²)

(ii) 1. The zeros of $f$ become vertical asymptotes: $x = - a$ and $x = a$ . As $x \to \pm \infty$ , $\frac{1}{f} \to 0$ : horizontal asymptote $y = 0$ . 2. Outside $[- a, a]$ , $f > 0$ and the graph lies above the $x$ -axis. Between $- a$ and $a$ , $f < 0$ and the graph lies below it. 3. The minimum of $f$ at $x = 0$ gives a local maximum of $\frac{1}{f}$ at $(0, - \frac{1}{a ^{2}})$ . It does not cut the $x$ -axis.

Graph of y = 1/(x² − a²) with vertical asymptotes x = −a and x = a and horizontal asymptote y = 0: outer branches above the x-axis and central branch below it, with a local maximum at (0, −1/a²)

(iii) 1. The absolute value reflects upwards the part that lies below the $x$ -axis. The asymptotes are the same ( $x = \pm a$ , $y = 0$ ). 2. The local maximum becomes a local minimum at $(0, \frac{1}{a ^{2}})$ . All branches lie above the $x$ -axis.

Graph of y = |1/(x² − a²)| with vertical asymptotes x = −a and x = a and horizontal asymptote y = 0: all branches above the x-axis and a local minimum at (0, 1/a²)

b) 1. We integrate by parts twice. First, with $u = x^{2} - a^{2}$ and $d v = cos x d x$ ( $d u = 2 x d x$ , $v = sin x$ ): $\int (x^{2} - a^{2}) cos x d x = (x^{2} - a^{2}) sin x - 2 \int x sin x d x$ . 2. Second time, with $u = x$ and $d v = sin x d x$ : $\int x sin x d x = - x cos x + sin x$ . 3. Substituting: $(x^{2} - a^{2}) sin x + 2 x cos x - 2 sin x + C = (x^{2} - a^{2} - 2) sin x + 2 x cos x + C$ .

c) 1. $g (x) = x f (x)$ , so $g^{'} (x) = f (x) + x \cdot \frac{f ^{'} ( x )}{2 f ( x )} = f (x) + \frac{x \cdot 2 x}{2 f ( x )} = \frac{f ( x ) + x ^{2}}{f ( x )}$ . 2. If $∣ x ∣ > a$ , $f (x) = x^{2} - a^{2} > 0$ , so the numerator $f (x) + x^{2}$ and the denominator are both positive. 3. Therefore $g^{'} (x) > 0$ and $g$ is increasing (on each of the two branches, $x > a$ and $x < - a$ ).

Answer: a) (i) parabola with intercepts $(\pm a, 0)$ and minimum $(0, - a^{2})$ ; (ii) asymptotes $x = \pm a$ , $y = 0$ , local maximum $(0, - \frac{1}{a ^{2}})$ ; (iii) the same asymptotes and local minimum $(0, \frac{1}{a ^{2}})$ ; b) $(x^{2} - a^{2} - 2) sin x + 2 x cos x + C$ ; c) $g^{'} (x) = \frac{f ( x ) + x ^{2}}{f ( x )} > 0$ , since $f (x) > 0$ for $∣ x ∣ > a$ .

Exercise 69 HardIB · P2 TZ2 May 2016

The functions $f$ and $g$ are defined by

$f (x) = \frac{e ^{x} + e ^{- x}}{2}$ , $x \in R$ $g (x) = \frac{e ^{x} - e ^{- x}}{2}$ , $x \in R$

a) i) Show that $\frac{1}{4 f ( x ) - 2 g ( x )} = \frac{e ^{x}}{e ^{2 x} + 3}$ .

(ii) Use the substitution $u = e^{x}$ to find $\int_{0}^{l n 3} \frac{1}{4 f ( x ) - 2 g ( x )} d x$ .

Give your answer in the form $\frac{π a}{b}$ where $a, b \in Z^{+}$ .

Let $h (x) = n f (x) + g (x)$ where $n \in R$ , $n > 1$ .

b) i) Solve the equation $h (x) = k$ , where $k \in R^{+}$ , by forming a quadratic equation in $e^{x}$ .

(ii) Hence or otherwise, show that the equation $h (x) = k$ has two real solutions whenever $k > n^{2} - 1$ and $k \in R^{+}$ .

Let $t (x) = \frac{g ( x )}{f ( x )}$ .

c) i) Show that $t^{'} (x) = \frac{[ f ( x ) ] ^{2} - [ g ( x ) ] ^{2}}{[ f ( x ) ] ^{2}}$ for $x \in R$ .

(ii) Hence, show that $t^{'} (x) > 0$ for $x \in R$ .

Show answer

Answer: a) (i) proven; (ii) $\frac{3 π}{18}$ ; b) (i) $x = ln (\frac{k \pm k ^{2} - n ^{2} + 1}{n + 1})$ ; (ii) proven; c) (i) and (ii) proven.

Show working and answer

a) (i) 1. $4 f (x) - 2 g (x) = 2 (e^{x} + e^{- x}) - (e^{x} - e^{- x}) = e^{x} + 3 e^{- x}$ . 2. Therefore $\frac{1}{4 f ( x ) - 2 g ( x )} = \frac{1}{e ^{x} + 3 e ^{- x}}$ . Multiply numerator and denominator by $e^{x}$ : $\frac{e ^{x}}{e ^{2 x} + 3}$ .

(ii) 1. Substitution $u = e^{x}$ , $d u = e^{x} d x$ ; the limits change from $x = 0$ to $u = 1$ and from $x = ln 3$ to $u = 3$ . 2. $\int_{1}^{3} \frac{d u}{u ^{2} + 3} = \frac{1}{3} [arctan \frac{u}{3}]_{1}^{3} = \frac{1}{3} (arctan 3 - arctan \frac{1}{3})$ . 3. $= \frac{1}{3} (\frac{π}{3} - \frac{π}{6}) = \frac{π}{6 3} = \frac{3 π}{18}$ , which is the required form with $a = 3$ and $b = 18$ .

b) (i) 1. $h (x) = n \cdot \frac{e ^{x} + e ^{- x}}{2} + \frac{e ^{x} - e ^{- x}}{2} = \frac{( n + 1 ) e ^{x} + ( n - 1 ) e ^{- x}}{2}$ . 2. Set equal to $k$ and multiply by $2 e^{x}$ : $(n + 1) e^{2 x} - 2 k e^{x} + (n - 1) = 0$ , a quadratic equation in $e^{x}$ . 3. Using the quadratic formula: $e^{x} = \frac{2 k \pm 4 k ^{2} - 4 ( n + 1 ) ( n - 1 )}{2 ( n + 1 )} = \frac{k \pm k ^{2} - n ^{2} + 1}{n + 1}$ . 4. Therefore $x = ln (\frac{k \pm k ^{2} - n ^{2} + 1}{n + 1})$ .

(ii) 1. Each positive value of $e^{x}$ gives one real solution $x$ . So we need two distinct positive roots of the quadratic. 2. They are distinct if the discriminant is positive: $k^{2} - n^{2} + 1 > 0$ , i.e. $k^{2} > n^{2} - 1$ , and since $k > 0$ , $k > n^{2} - 1$ . 3. Both roots are positive: their product is $\frac{n - 1}{n + 1} > 0$ (because $n > 1$ ) and their sum is $\frac{2 k}{n + 1} > 0$ . 4. Therefore, if $k > n^{2} - 1$ there are two real solutions.

c) (i) 1. Quotient rule: $t^{'} (x) = \frac{g ^{'} ( x ) f ( x ) - g ( x ) f ^{'} ( x )}{[ f ( x ) ] ^{2}}$ . 2. Since $g^{'} (x) = \frac{e ^{x} + e ^{- x}}{2} = f (x)$ and $f^{'} (x) = \frac{e ^{x} - e ^{- x}}{2} = g (x)$ , the numerator is $[f (x)]^{2} - [g (x)]^{2}$ .

(ii) 1. $[f (x)]^{2} - [g (x)]^{2} = \frac{( e ^{x} + e ^{- x} ) ^{2} - ( e ^{x} - e ^{- x} ) ^{2}}{4} = \frac{4}{4} = 1$ . 2. Therefore $t^{'} (x) = \frac{1}{[ f ( x ) ] ^{2}} > 0$ for all $x \in R$ .

Answer: a) (i) proven; (ii) $\frac{3 π}{18}$ ; b) (i) $x = ln (\frac{k \pm k ^{2} - n ^{2} + 1}{n + 1})$ ; (ii) proven; c) (i) and (ii) proven.

Exercise 70 HardIB · P2 May 1998

(i) Let $f (x) = x^{3} - 3 x^{2} + 4$ .

a) Show that $(x - 2)$ is a factor of $f (x)$ .

b) Using the above, factorise $f (x)$ completely.

c) Express $\frac{3}{x ^{3} - 3 x ^{2} + 4}$ in partial fractions.

d) Using the above, calculate $\int \frac{3}{x ^{3} - 3 x ^{2} + 4} d x$ .

(ii) a) Calculate the coordinates of the points of intersection of the curve $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ with the $x$ -axis (where $a$ and $b$ are positive constants).

b) Write a simplified expression, in terms of $a$ and $b$ , for the volume of the solid formed when the region enclosed by the curve is rotated $36 0^{\circ}$ about the $x$ -axis.

c) Give a full description of the solid formed when $b = a$ .

Show answer

Answer: i) a) $f (2) = 0$ ; b) $x^{3} - 3 x^{2} + 4 = (x - 2)^{2} (x + 1)$ ; c) $\frac{3}{x ^{3} - 3 x ^{2} + 4} = \frac{- \frac{1}{3}}{x - 2} + \frac{1}{( x - 2 ) ^{2}} + \frac{\frac{1}{3}}{x + 1}$ ; d) $- \frac{1}{3} ln ∣ x - 2∣ - \frac{1}{x - 2} + \frac{1}{3} ln ∣ x + 1∣ + C$ ; ii) a) $(a, 0)$ and $(- a, 0)$ ; b) $V = \frac{4 π a b ^{2}}{3}$ ; c) a sphere with centre $(0, 0)$ and radius $a$ .

Show working and answer

(i) a) 1. $f (2) = 8 - 12 + 4 = 0$ , so $x - 2$ is a factor of $f (x)$ (factor theorem).

b) 1. Divide: $x^{3} - 3 x^{2} + 4 = (x - 2) (x^{2} - x - 2)$ . 2. And $x^{2} - x - 2 = (x - 2) (x + 1)$ . 3. Therefore $f (x) = (x - 2)^{2} (x + 1)$ .

c) 1. Set up $\frac{3}{( x - 2 ) ^{2} ( x + 1 )} = \frac{A}{x - 2} + \frac{B}{( x - 2 ) ^{2}} + \frac{C}{x + 1}$ , i.e. $3 = A (x - 2) (x + 1) + B (x + 1) + C (x - 2)^{2}$ . 2. For $x = 2$ : $3 = 3 B$ , so $B = 1$ . 3. For $x = - 1$ : $3 = 9 C$ , so $C = \frac{1}{3}$ . 4. Comparing coefficients of $x^{2}$ : $0 = A + C$ , so $A = - \frac{1}{3}$ . 5. $\frac{3}{x ^{3} - 3 x ^{2} + 4} = \frac{- \frac{1}{3}}{x - 2} + \frac{1}{( x - 2 ) ^{2}} + \frac{\frac{1}{3}}{x + 1}$ .

d) 1. Integrate each fraction: $\int \frac{3}{x ^{3} - 3 x ^{2} + 4} d x = - \frac{1}{3} ln ∣ x - 2∣ - \frac{1}{x - 2} + \frac{1}{3} ln ∣ x + 1∣ + C$ .

(ii) a) 1. Intercepts with the $x$ -axis ( $y = 0$ ): $\frac{x ^{2}}{a ^{2}} = 1$ , so $x = \pm a$ . Points $(a, 0)$ and $(- a, 0)$ .

b) 1. From the equation, $y^{2} = b^{2} (1 - \frac{x ^{2}}{a ^{2}})$ . Rotating about the $x$ -axis: $V = π \int_{- a}^{a} b^{2} (1 - \frac{x ^{2}}{a ^{2}}) d x = π b^{2} [x - \frac{x ^{3}}{3 a ^{2}}]_{- a}^{a}$ . 2. $= π b^{2} \cdot 2 (a - \frac{a}{3}) = \frac{4 π a b ^{2}}{3}$ .

c) 1. If $b = a$ , the equation is $x^{2} + y^{2} = a^{2}$ : a circle with centre $(0, 0)$ and radius $a$ . 2. Rotating it $36 0^{\circ}$ about the $x$ -axis gives a sphere with centre at the origin and radius $a$ , with volume $\frac{4}{3} π a^{3}$ (which agrees with the formula in b) when $b = a$ ).

Exercise 71 HardIB · P2 TZ0 November 2004

Consider the complex number $z = cos θ + i sin θ$ .

a) Using De Moivre's theorem, verify that $z^{n} + \frac{1}{z ^{n}} = 2 cos n θ$ .

b) By expanding $(z + \frac{1}{z})^{4}$ , verify that $cos^{4} θ = \frac{1}{8} (cos 4 θ + 4 cos 2 θ + 3)$ .

c) Let $g (a) = \int_{0}^{a} cos^{4} θ d θ$ .

(i) Find $g (a)$ .

(ii) Solve $g (a) = 1$ .

Show answer

Answer: c) i) $g (a) = \frac{1}{8} (\frac{sin 4 a}{4} + 2 sin 2 a + 3 a)$ ; ii) $a \approx 2.96$

Show working and answer

a) 1. By De Moivre, $z^{n} = cos n θ + i sin n θ$ . 2. Then $\frac{1}{z ^{n}} = z^{- n} = cos (- n θ) + i sin (- n θ) = cos n θ - i sin n θ$ . 3. Adding, $z^{n} + \frac{1}{z ^{n}} = 2 cos n θ$ .

b) 1. As $z + \frac{1}{z} = 2 cos θ$ , we have $(z + \frac{1}{z})^{4} = 16 cos^{4} θ$ . 2. Expanding with the binomial theorem: $z^{4} + 4 z^{2} + 6 + \frac{4}{z ^{2}} + \frac{1}{z ^{4}} = (z^{4} + \frac{1}{z ^{4}}) + 4 (z^{2} + \frac{1}{z ^{2}}) + 6$ . 3. By (a), this is $2 cos 4 θ + 8 cos 2 θ + 6$ . 4. Equating: $16 cos^{4} θ = 2 cos 4 θ + 8 cos 2 θ + 6$ , and dividing by $16$ : $cos^{4} θ = \frac{1}{8} (cos 4 θ + 4 cos 2 θ + 3)$ .

c) i) 1. Using (b): $g (a) = \frac{1}{8} \int_{0}^{a} (cos 4 θ + 4 cos 2 θ + 3) d θ$ . 2. $g (a) = \frac{1}{8} [\frac{sin 4 θ}{4} + 2 sin 2 θ + 3 θ]_{0}^{a} = \frac{1}{8} (\frac{sin 4 a}{4} + 2 sin 2 a + 3 a)$ .

(ii) 1. Using the GDC we solve $\frac{1}{8} (\frac{sin 4 a}{4} + 2 sin 2 a + 3 a) = 1$ (in radians). 2. $g$ is increasing, since $g^{'} (a) = cos^{4} a \geq 0$ , and therefore there is a unique solution: $a \approx 2.96$ .

Answer: c) i) $g (a) = \frac{1}{8} (\frac{sin 4 a}{4} + 2 sin 2 a + 3 a)$ ; ii) $a \approx 2.96$

Exercise 72 HardIB · P2 May 2003

The function $f$ with domain $[0, \frac{π}{2}]$ is defined by $f (x) = cos x + 3 sin x$ .

This function can also be expressed in the form $R cos (x - α)$ where $R > 0$ and $0 < α < \frac{π}{2}$ .

a) Find the exact values of $R$ and $α$ .

b) i) Find the range of the function.

(ii) State whether or not the inverse function of $f$ exists, giving a reason.

c) Find the exact value of $x$ that satisfies the equation $f (x) = 2$ .

d) Using the result that $\int sec x d x = ln ∣ sec x + tan x ∣ + C$ , where $C$ is a constant, prove that $\int_{0}^{\frac{π}{2}} \frac{d x}{f ( x )} = \frac{1}{2} ln (3 + 23)$ .

Show answer

Answer: a) $R = 2$ , $α = \frac{π}{3}$ ; b) i) $[1, 2]$ ; ii) no inverse exists, because $f$ is not one-to-one; c) $x = \frac{π}{12}$ ; d) proved.

Show working and answer

a) 1. We expand $R cos (x - α) = R cos α cos x + R sin α sin x$ and equate it with $cos x + 3 sin x$ : $R cos α = 1$ and $R sin α = 3$ . 2. Squaring and adding, $R^{2} = 1 + 3 = 4$ , so $R = 2$ . 3. Dividing, $tan α = 3$ , and with $0 < α < \frac{π}{2}$ : $α = \frac{π}{3}$ .

b) i) 1. $f (x) = 2 cos (x - \frac{π}{3})$ , with $x \in [0, \frac{π}{2}]$ , and $x - \frac{π}{3} \in [- \frac{π}{3}, \frac{π}{6}]$ . 2. The maximum is $f (\frac{π}{3}) = 2$ . At the ends, $f (0) = 1$ and $f (\frac{π}{2}) = 3$ ; the minimum is $1$ . 3. Range: $[1, 2]$ .

(ii) 1. $f$ increases on $[0, \frac{π}{3} [$ and decreases on $] \frac{π}{3}, \frac{π}{2}]$ . 2. So there are values of $x$ in both intervals with the same image: for example, $f (\frac{π}{6}) = f (\frac{π}{2}) = 3$ . 3. It is not one-to-one, so $f$ has no inverse function.

c) 1. $2 cos (x - \frac{π}{3}) = 2$ gives $cos (x - \frac{π}{3}) = \frac{2}{2}$ , so $x - \frac{π}{3} = \pm \frac{π}{4}$ . 2. $x = \frac{7 π}{12}$ lies outside the domain ( $> \frac{π}{2}$ ). That leaves $x = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$ .

d) 1. $\frac{1}{f ( x )} = \frac{1}{2 cos ( x - \frac{π}{3} )} = \frac{1}{2} sec (x - \frac{π}{3})$ . 2. Substitution $u = x - \frac{π}{3}$ ( $d u = d x$ ; the limits change from $- \frac{π}{3}$ to $\frac{π}{6}$ ): $\int_{0}^{\frac{π}{2}} \frac{d x}{f ( x )} = \frac{1}{2} [ln ∣ sec u + tan u ∣]_{- \frac{π}{3}}^{\frac{π}{6}}$ . 3. At $u = \frac{π}{6}$ : $sec u + tan u = \frac{2}{3} + \frac{1}{3} = 3$ . At $u = - \frac{π}{3}$ : $2 - 3$ . 4. The result is $\frac{1}{2} ln \frac{3}{2 - 3}$ . Multiplying numerator and denominator by $2 + 3$ (and $(2 - 3) (2 + 3) = 1$ ): $\frac{1}{2} ln (3 (2 + 3)) = \frac{1}{2} ln (23 + 3) = \frac{1}{2} ln (3 + 23)$ .

Answer: a) $R = 2$ , $α = \frac{π}{3}$ ; b) i) $[1, 2]$ ; ii) no inverse exists, because $f$ is not one-to-one; c) $x = \frac{π}{12}$ ; d) proved.

Exercise 73 HardIB · P1 TZ1 May 2012

Let $f (x) = \frac{x}{1 - x}$ $0 < x < 1$ .

a) Show that $f^{'} (x) = \frac{1}{2} x^{- \frac{1}{2}} (1 - x)^{- \frac{3}{2}}$ , and deduce that $f$ is an increasing function.

b) Show that the curve $y = f (x)$ has a point of inflexion, and find its coordinates.

c) Use the substitution $x = sin^{2} θ$ to show that $\int f (x) d x = arcsin x - x - x^{2} + c$ .

Show answer

Answer: a) shown; b) point of inflexion $(\frac{1}{4}, \frac{1}{3})$ ; c) shown.

Show working and answer

a) 1. We write $f (x) = x^{1/2} (1 - x)^{- 1/2}$ and apply the product rule: $f^{'} (x) = \frac{1}{2} x^{- 1/2} (1 - x)^{- 1/2} + x^{1/2} \cdot \frac{1}{2} (1 - x)^{- 3/2}$ . 2. Factor out $\frac{1}{2} x^{- 1/2} (1 - x)^{- 3/2}$ : what remains is $(1 - x) + x = 1$ . 3. $f^{'} (x) = \frac{1}{2} x^{- \frac{1}{2}} (1 - x)^{- \frac{3}{2}}$ . 4. For $0 < x < 1$ both powers are positive, so $f^{'} (x) > 0$ and $f$ is increasing.

b) 1. Differentiating again: $f^{''} (x) = \frac{1}{2} [- \frac{1}{2} x^{- 3/2} (1 - x)^{- 3/2} + \frac{3}{2} x^{- 1/2} (1 - x)^{- 5/2}]$ . 2. Factor out $\frac{1}{4} x^{- 3/2} (1 - x)^{- 5/2}$ : what remains is $- (1 - x) + 3 x = 4 x - 1$ . 3. $f^{''} (x) = \frac{1}{4} x^{- 3/2} (1 - x)^{- 5/2} (4 x - 1)$ , which vanishes only at $x = \frac{1}{4}$ and changes sign (negative before, positive after). So there is a point of inflexion. 4. $f (\frac{1}{4}) = \frac{1/4}{3/4} = \frac{1}{3}$ . Coordinates: $(\frac{1}{4}, \frac{1}{3})$ .

c) 1. With $x = sin^{2} θ$ ( $0 < θ < \frac{π}{2}$ ): $d x = 2 sin θ cos θ d θ$ and $f (x) = \frac{sin ^{2} θ}{cos ^{2} θ} = tan θ$ . 2. $\int f (x) d x = \int tan θ \cdot 2 sin θ cos θ d θ = \int 2 sin^{2} θ d θ = \int (1 - cos 2 θ) d θ = θ - sin θ cos θ + c$ . 3. Undoing the substitution: $θ = arcsin x$ , $sin θ = x$ and $cos θ = 1 - x$ , so $sin θ cos θ = x - x^{2}$ . 4. $\int f (x) d x = arcsin x - x - x^{2} + c$ .

Answer: a) shown; b) point of inflexion $(\frac{1}{4}, \frac{1}{3})$ ; c) shown.

Exercise 74 HardIB · P1 TZ1 May 2013

a) Express $4 x^{2} - 4 x + 5$ in the form $a (x - h)^{2} + k$ , where $a, h, k \in Q$ .

b) The graph of $y = x^{2}$ is transformed into the graph of $4 x^{2} - 4 x + 5$ .

Describe a sequence of transformations that does this, stating the order of the transformations clearly.

The function $f$ is defined by $f (x) = \frac{1}{4 x ^{2} - 4 x + 5}$ .

c) Sketch the graph of $y = f (x)$ .

d) Find the range of $f$ .

e) Using a suitable substitution, show that $\int f (x) d x = \frac{1}{4} \int \frac{1}{u ^{2} + 1} d u$ .

f) Prove that $\int_{1}^{3.5} \frac{1}{4 x ^{2} - 4 x + 5} d x = \frac{π}{16}$ .

View solution

a) 1. Completing the square: $4 x^{2} - 4 x + 5 = 4 (x^{2} - x) + 5 = 4 [(x - \frac{1}{2})^{2} - \frac{1}{4}] + 5$ . 2. $= 4 (x - \frac{1}{2})^{2} - 1 + 5 = 4 (x - \frac{1}{2})^{2} + 4$ .

b) We start from $y = x^{2}$ and, in this order:

Translation by the vector $(\frac{1}{2} 0)$ : gives $y = (x - \frac{1}{2})^{2}$ .
Vertical stretch with scale factor $4$ : gives $y = 4 (x - \frac{1}{2})^{2}$ .
Translation by the vector $(04)$ : gives $y = 4 (x - \frac{1}{2})^{2} + 4$ .

c) 1. The denominator $4 (x - \frac{1}{2})^{2} + 4$ has a minimum of $4$ at $x = \frac{1}{2}$ , so $f$ has a maximum of $\frac{1}{4}$ at $x = \frac{1}{2}$ . 2. $f$ is symmetric about $x = \frac{1}{2}$ , always positive, with $f (0) = \frac{1}{5}$ , and tends to $0$ as $x \to \pm \infty$ (the $x$ -axis is a horizontal asymptote). It is a bell shape.

Graph of f(x) = 1 / (4x² − 4x + 5): bell-shaped curve, symmetric about x = 1/2

d) 1. The denominator takes all values $\geq 4$ , so $f (x) \in] 0, \frac{1}{4}]$ (it never reaches $0$ , but attains $\frac{1}{4}$ ). 2. $R_{f} =] 0, \frac{1}{4}]$ .

e) 1. Technique: substitution $u = x - \frac{1}{2}$ , with $d u = d x$ . 2. From part (a), $4 x^{2} - 4 x + 5 = 4 u^{2} + 4 = 4 (u^{2} + 1)$ . 3. $\int f (x) d x = \int \frac{1}{4 ( u ^{2} + 1 )} d u = \frac{1}{4} \int \frac{1}{u ^{2} + 1} d u$ .

f) 1. Changing the limits: $x = 1 \Rightarrow u = \frac{1}{2}$ ; $x = 3.5 \Rightarrow u = 3$ . 2. $\int_{1}^{3.5} f (x) d x = \frac{1}{4} [arctan u]_{1/2}^{3} = \frac{1}{4} (arctan 3 - arctan \frac{1}{2})$ . 3. Using $tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ with $tan A = 3$ and $tan B = \frac{1}{2}$ : $tan (A - B) = \frac{3 - \frac{1}{2}}{1 + \frac{3}{2}} = \frac{5/2}{5/2} = 1$ . 4. As $A - B$ lies between $0$ and $\frac{π}{2}$ , $A - B = \frac{π}{4}$ . 5. Therefore the integral equals $\frac{1}{4} \cdot \frac{π}{4} = \frac{π}{16}$ .

Answer: a) $4 (x - \frac{1}{2})^{2} + 4$ ; d) $R_{f} =] 0, \frac{1}{4}]$ ; f) shown ( $= \frac{π}{16}$ ).

Differential equations

Exercise 75 BasicIB · P1 May 1998

Let $0 < x < \infty$ , with $y (x) > 0$ . Solve the differential equation $\frac{d y}{d x} = y cos x$ , where $y = 1$ when $x = \frac{π}{2}$ .

Show answer

Answer: $y = e^{s i n x - 1}$

Show working and answer

Separate the variables (with $y > 0$ ): $\frac{d y}{y} = cos x d x$ .
Integrate: $ln y = sin x + C$ .
With $y = 1$ when $x = \frac{π}{2}$ : $0 = 1 + C$ , so $C = - 1$ .
$y = e^{s i n x - 1}$ .

Answer: $y = e^{s i n x - 1}$

Exercise 76 BasicIB · P1 May 1999

When air is released from an inflated balloon, it is observed that the rate at which the volume of the balloon decreases is proportional to the volume of the balloon. This can be represented by the differential equation $\frac{d v}{d t} = - k v$ , where $v$ is the volume, $t$ is the time and $k$ is the constant of proportionality.

a) If the initial volume of the balloon is $v_{0}$ , find an expression, in terms of $k$ , for the volume of the balloon at time $t$ .

b) Find an expression, in terms of $k$ , for the time when the volume is $\frac{v _{0}}{2}$ .

Show answer

Answer: a) $v = v_{0} e^{- k t}$ ; b) $t = \frac{ln 2}{k}$

Show working and answer

a) 1. Separate the variables: $\frac{d v}{v} = - k d t$ . 2. Integrate: $ln v = - k t + C$ , i.e. $v = A e^{- k t}$ . 3. At $t = 0$ the volume is $v_{0}$ : $A = v_{0}$ . 4. $v = v_{0} e^{- k t}$ .

b) 1. Set $v = \frac{v _{0}}{2}$ : $\frac{v _{0}}{2} = v_{0} e^{- k t}$ , so $e^{- k t} = \frac{1}{2}$ . 2. Take logarithms: $- k t = - ln 2$ , so $t = \frac{ln 2}{k}$ .

Answer: a) $v = v_{0} e^{- k t}$ ; b) $t = \frac{ln 2}{k}$

Exercise 77 BasicIB · P1 TZ1 May 2007

Solve the differential equation $(x^{2} + 1) \frac{d y}{d x} - x y = 0$ where $x > 0$ , $y > 0$ , given that $y = 1$ when $x = 1$ .

Show answer

Answer: $y = \frac{x ^{2} + 1}{2}$

Show working and answer

Separate the variables: $\frac{d y}{y} = \frac{x}{x ^{2} + 1} d x$ .
Integrate (with $y > 0$ ): $ln y = \frac{1}{2} ln (x^{2} + 1) + C$ .
With $y = 1$ when $x = 1$ : $0 = \frac{1}{2} ln 2 + C$ , so $C = - \frac{1}{2} ln 2$ .
$ln y = \frac{1}{2} ln \frac{x ^{2} + 1}{2}$ , i.e. $y = \frac{x ^{2} + 1}{2}$ .

Answer: $y = \frac{x ^{2} + 1}{2}$

Exercise 78 BasicIB · P1 TZ2 May 2007

Solve the differential equation $\frac{d y}{d x} = 2 x y^{2}$ , given that $y = 1$ when $x = 0$ .

Express your answer in the form $y = f (x)$ .

Show answer

Answer: $y = \frac{- 1}{x ^{2} - 1}$

Show working and answer

Separate the variables: $\frac{d y}{y ^{2}} = 2 x d x$ .
Integrate: $- \frac{1}{y} = x^{2} + C$ .
With $y = 1$ when $x = 0$ : $- 1 = C$ .
$- \frac{1}{y} = x^{2} - 1$ , so $y = \frac{- 1}{x ^{2} - 1}$ .

Answer: $y = \frac{- 1}{x ^{2} - 1}$

Exercise 79 BasicIB · P1 TZ0 November 2021

Given that $\frac{d y}{d x} = cos (x - \frac{π}{4})$ and $y = 2$ when $x = \frac{3 π}{4}$ , find $y$ in terms of $x$ .

Show answer

Answer: $y = sin (x - \frac{π}{4}) + 1$

Show working and answer

We integrate: $y = \int cos (x - \frac{π}{4}) d x = sin (x - \frac{π}{4}) + C$ (a standard integral by the chain rule, since the derivative of $x - \frac{π}{4}$ is $1$ ).
Condition: $y = 2$ when $x = \frac{3 π}{4}$ : $2 = sin \frac{π}{2} + C = 1 + C$ , so $C = 1$ .

Answer: $y = sin (x - \frac{π}{4}) + 1$

Exercise 80 MediumIB · P1 November 2001

A sample of radioactive material decays at a rate proportional to the amount of material present in the sample. Calculate the half-life of the material if it decreases from $50$ grams to $48$ grams in $10$ years.

Show answer

Answer: half-life $\approx 169.8$ years

Show working and answer

Let $N$ be the mass in grams. "Decays at a rate proportional to the amount present" means $\frac{d N}{d t} = - λ N$ with $λ > 0$ .
Separating variables: $\int \frac{d N}{N} = - λ \int d t$ , so $ln N = - λ t + C$ , that is $N = N_{0} e^{- λ t}$ with $N_{0} = 50$ .
With $N (10) = 48$ : $48 = 50 e^{- 10 λ}$ , so $e^{- 10 λ} = \frac{24}{25}$ and $λ = \frac{1}{10} ln \frac{25}{24} \approx 0.004082$ .
The half-life $T$ satisfies $\frac{N _{0}}{2} = N_{0} e^{- λ T}$ , so $T = \frac{ln 2}{λ} = \frac{10 ln 2}{ln \frac{25}{24}}$ .
$T \approx 169.8$ years.

Answer: half-life $\approx 169.8$ years

Exercise 81 MediumIB · P2 TZ2 May 2009

The acceleration, in $m s^{- 2}$ , of a particle moving in a straight line at time $t$ seconds, with $t \geq 0$ , is given by $a = - \frac{1}{2} v$ . When $t = 0$ , the velocity is $40 m s^{- 1}$ . Find an expression for $v$ in terms of $t$ .

Show answer

Answer: $v = 40 e^{- \frac{t}{2}} m s^{- 1}$

Show working and answer

Since $a = \frac{d v}{d t}$ , the equation is $\frac{d v}{d t} = - \frac{1}{2} v$ , which is separable.
$\int \frac{d v}{v} = - \frac{1}{2} \int d t$ , so $ln v = - \frac{t}{2} + C$ (with $v > 0$ ) and $v = A e^{- \frac{t}{2}}$ , where $A = e^{C}$ .
Initial condition: $v (0) = 40$ gives $A = 40$ .
Check: $v^{'} = - 20 e^{- t /2} = - \frac{1}{2} \cdot 40 e^{- t /2}$ ✓ and $v (0) = 40$ ✓.

Answer: $v = 40 e^{- \frac{t}{2}} m s^{- 1}$

Exercise 82 MediumIB · P1 TZ0 November 2009

A certain population can be modelled by the differential equation $\frac{d y}{d t} = k y cos k t$ , where $y$ represents the population at time $t$ hours and $k$ is a positive constant.

a) Given that $y = y_{0}$ when $t = 0$ , express $y$ in terms of $k$ , $t$ and $y_{0}$ .

b) Find the ratio of the minimum size to the maximum size of the population.

Show answer

Answer: a) $y = y_{0} e^{s i n k t}$ ; b) $\frac{minimum}{maximum} = \frac{1}{e ^{2}}$

Show working and answer

a) 1. Separating variables: $\frac{d y}{y} = k cos k t d t$ . 2. Integrating: $ln ∣ y ∣ = sin k t + C$ (the derivative of $sin k t$ is $k cos k t$ ). 3. Hence $y = A e^{s i n k t}$ with $A = \pm e^{C}$ . 4. With $y (0) = y_{0}$ : $y_{0} = A e^{0}$ , so $A = y_{0}$ and $y = y_{0} e^{s i n k t}$ . 5. Check: $y^{'} = y_{0} e^{s i n k t} \cdot k cos k t = k y cos k t$ ✓.

b) 1. Since $- 1 \leq sin k t \leq 1$ , the exponent oscillates between $- 1$ and $1$ . 2. Minimum size: $y_{0} e^{- 1}$ . Maximum size: $y_{0} e^{1}$ . 3. Ratio: $\frac{y _{0} e ^{- 1}}{y _{0} e} = e^{- 2}$ .

Answer: a) $y = y_{0} e^{s i n k t}$ ; b) $\frac{minimum}{maximum} = \frac{1}{e ^{2}}$

Exercise 83 MediumIB · P2 TZ0 November 2008

The mosquito population in a certain area beside a lake is controlled by means of a pesticide. The rate of decrease of the number of mosquitoes is proportional to the number of mosquitoes at any instant $t$ . Given that the population decreases from $500, 000$ to $400, 000$ over a period of five years, find the time, in years, that must elapse for the mosquito population to be reduced to half.

Show answer

Answer: $N (t) = 500000 e^{- 0.0446 t}$ ; the population is halved in $t \approx 15.5$ years

Show working and answer

Let $N$ be the number of mosquitoes. The rate of decrease is proportional to $N$ : $\frac{d N}{d t} = - λ N$ , with $λ > 0$ .
Separating variables, $N (t) = N_{0} e^{- λ t}$ with $N_{0} = 500000$ .
After $5$ years there are $400000$ : $400000 = 500000 e^{- 5 λ}$ , so $e^{- 5 λ} = \frac{4}{5}$ and $λ = \frac{1}{5} ln \frac{5}{4} \approx 0.0446$ .
So $N (t) = 500000 e^{- 0.0446 t}$ .
It is halved when $e^{- λ T} = \frac{1}{2}$ , that is $T = \frac{ln 2}{λ} = \frac{5 ln 2}{ln \frac{5}{4}} \approx 15.5$ years.

Answer: $N (t) = 500000 e^{- 0.0446 t}$ ; the population is halved in $t \approx 15.5$ years

Exercise 84 MediumIB · P2 TZ2 May 2013

The acceleration of a car is equal to $\frac{1}{40} (60 - v) m s^{- 2}$ , where $v$ is its velocity in $m s^{- 1}$ . Given that the car starts from rest, find the velocity of the car after $30$ seconds.

Show answer

Answer: $v (t) = 60 (1 - e^{- \frac{t}{40}})$ and $v (30) \approx 31.7 m s^{- 1}$

Show working and answer

The acceleration is $\frac{d v}{d t} = \frac{1}{40} (60 - v)$ , a separable differential equation.
$\int \frac{d v}{60 - v} = \int \frac{d t}{40}$ , so $- ln (60 - v) = \frac{t}{40} + C$ (with $v < 60$ ).
Starting from rest: $v (0) = 0$ , so $- ln 60 = C$ .
Then $ln \frac{60}{60 - v} = \frac{t}{40}$ , that is $60 - v = 60 e^{- \frac{t}{40}}$ and $v = 60 (1 - e^{- \frac{t}{40}})$ .
At $t = 30$ : $v = 60 (1 - e^{- 0.75}) \approx 31.7 m s^{- 1}$ .

Answer: $v (t) = 60 (1 - e^{- \frac{t}{40}})$ and $v (30) \approx 31.7 m s^{- 1}$

Exercise 85 MediumIB · P3 TZ0 May 2015

a) Show that $y = \frac{1}{x} \int f (x) d x$ is a solution of the differential equation $x \frac{d y}{d x} + y = f (x)$ , $x > 0$ .

b) Hence, solve $x \frac{d y}{d x} + y = x^{- \frac{1}{2}}$ , $x > 0$ , given that $y = 2$ when $x = 4$ .

Show answer

Answer: a) verified; b) $y = \frac{1}{x} (2 x^{\frac{1}{2}} + 4)$

Show working and answer

a) 1. Let $F (x) = \int f (x) d x$ , so that $F^{'} (x) = f (x)$ , and $y = \frac{F ( x )}{x}$ . 2. Differentiating with the quotient rule: $\frac{d y}{d x} = \frac{f ( x )}{x} - \frac{F ( x )}{x ^{2}}$ . 3. Then $x \frac{d y}{d x} + y = f (x) - \frac{F ( x )}{x} + \frac{F ( x )}{x} = f (x)$ . This completes the proof.

b) 1. With $f (x) = x^{- \frac{1}{2}}$ , by a) the solution is $y = \frac{1}{x} \int x^{- \frac{1}{2}} d x = \frac{1}{x} (2 x^{\frac{1}{2}} + C)$ . 2. Condition $x = 4$ , $y = 2$ : $2 = \frac{1}{4} (2 \cdot 2 + C) = \frac{4 + C}{4}$ , so $C = 4$ . 3. $y = \frac{1}{x} (2 x^{\frac{1}{2}} + 4)$ .

Answer: a) verified; b) $y = \frac{1}{x} (2 x^{\frac{1}{2}} + 4)$

Exercise 86 MediumIB · P1 May 1996

Solve the differential equation $\frac{d y}{d x} + 5 y = e^{8 x}$ , given that $y = \frac{3}{2}$ when $x = 0$ .

Show answer

Answer: $y = \frac{1}{13} e^{8 x} + \frac{37}{26} e^{- 5 x}$

Show working and answer

This is a first-order linear equation $y^{'} + 5 y = e^{8 x}$ , so we use an integrating factor.

The integrating factor is $e^{\int 5 d x} = e^{5 x}$ . Multiplying: $(y e^{5 x})^{'} = e^{13 x}$ .
Integrating: $y e^{5 x} = \frac{1}{13} e^{13 x} + C$ , so $y = \frac{1}{13} e^{8 x} + C e^{- 5 x}$ .
Using the condition $y (0) = \frac{3}{2}$ : $\frac{1}{13} + C = \frac{3}{2}$ , so $C = \frac{37}{26}$ .
Check: $y^{'} + 5 y = \frac{8}{13} e^{8 x} - 5 C e^{- 5 x} + \frac{5}{13} e^{8 x} + 5 C e^{- 5 x} = e^{8 x}$ .

Answer: $y = \frac{1}{13} e^{8 x} + \frac{37}{26} e^{- 5 x}$

Exercise 87 MediumIB · P1 November 1996

Solve $\frac{d y}{d x} + y tan x = cos x$ , $- \frac{π}{2} < x < \frac{π}{2}$ .

Show answer

Answer: $y = x cos x + C cos x$

Show working and answer

This is a first-order linear equation $y^{'} + (tan x) y = cos x$ , so we use an integrating factor.

The integrating factor is $e^{\int t a n x d x} = e^{- l n c o s x} = \frac{1}{cos x} = sec x$ (on $] - \frac{π}{2}, \frac{π}{2} [$ we have $cos x > 0$ ).
Multiplying: $(y sec x)^{'} = cos x \cdot sec x = 1$ .
Integrating: $y sec x = x + C$ .
Hence $y = x cos x + C cos x$ .

Answer: $y = x cos x + C cos x$

Exercise 88 MediumIB · P1 TZ0 November 2010

Find $y$ in terms of $x$ , given that $(1 + x^{3}) \frac{d y}{d x} = 2 x^{2} tan y$ and that $y = \frac{π}{2}$ when $x = 0$ .

Show answer

Answer: $y = arcsin (1 + x^{3})^{\frac{2}{3}}$

Show working and answer

This is a separable variables equation.

Separating the variables: $\frac{d y}{tan y} = \frac{2 x ^{2}}{1 + x ^{3}} d x$ , that is, $\int \frac{cos y}{sin y} d y = \int \frac{2 x ^{2}}{1 + x ^{3}} d x$ .
Both integrals are of the form $\int \frac{u ^{'}}{u}$ : $ln ∣ sin y ∣ = \frac{2}{3} ln 1 + x^{3} + C$ (on the right, $u = 1 + x^{3}$ , $d u = 3 x^{2} d x$ ).
Using $x = 0$ , $y = \frac{π}{2}$ : $ln 1 = 0 = \frac{2}{3} ln 1 + C$ , so $C = 0$ .
Near the initial point $sin y > 0$ and $1 + x^{3} > 0$ , so $sin y = (1 + x^{3})^{2/3}$ .
As $y$ is close to $\frac{π}{2}$ , we choose the branch of arcsine that passes through $(0, \frac{π}{2})$ : $y = arcsin (1 + x^{3})^{2/3}$ (valid while $(1 + x^{3})^{2/3} \leq 1$ , that is, for $- 1 < x \leq 0$ ).

Answer: $y = arcsin (1 + x^{3})^{\frac{2}{3}}$

Exercise 89 MediumIB · P3 TZ0 May 2019

A simple model for predicting the world population is set up as follows: at time $t$ years, the world population is $x$ , which can be assumed to be a continuous variable. The rate at which $x$ increases due to births is $0.056 x$ , and the rate at which $x$ decreases due to deaths is $0.035 x$ .

a) Show that $\frac{d x}{d t} = 0.021 x$ .

b) Find the number of years that must elapse for the world population to double.

Show answer

Answer: a) shown; b) $t \approx 33$ years

Show working and answer

a) The net rate of change is births minus deaths: $\frac{d x}{d t} = 0.056 x - 0.035 x = 0.021 x$ .

b) 1. Separating the variables: $\int \frac{d x}{x} = \int 0.021 d t$ , so $ln x = 0.021 t + C$ and $x = x_{0} e^{0.021 t}$ , where $x_{0}$ is the initial population. 2. It doubles when $x = 2 x_{0}$ : $e^{0.021 t} = 2$ , that is, $t = \frac{ln 2}{0.021} \approx 33.0$ . 3. It takes approximately $33$ years.

Answer: a) shown; b) $t \approx 33$ years

Exercise 90 MediumIB · P3 TZ0 May 2012

Find the general solution of the differential equation $t \frac{d y}{d t} = cos t - 2 y$ , for $t > 0$ .

Show answer

Answer: $y = \frac{t sin t + cos t + C}{t ^{2}}$

Show working and answer

This is a first-order linear equation. Moving all the terms to the left and dividing by $t$ ( $t > 0$ ): $\frac{d y}{d t} + \frac{2}{t} y = \frac{cos t}{t}$ .
Integrating factor: $e^{\int \frac{2}{t} d t} = e^{2 l n t} = t^{2}$ .
Multiplying by $t^{2}$ : $\frac{d}{d t} (t^{2} y) = t cos t$ .
Integrating by parts, with $u = t$ and $d v = cos t d t$ (so $d u = d t$ and $v = sin t$ ): $\int t cos t d t = t sin t - \int sin t d t = t sin t + cos t + C$ .
Hence $t^{2} y = t sin t + cos t + C$ , and we make $y$ the subject.

Answer: $y = \frac{t sin t + cos t + C}{t ^{2}}$

Exercise 91 MediumIB · P3 TZ0 May 2013

Consider the differential equation $\frac{d y}{d x} + y tan x = cos^{2} x$ , given that $y = 2$ when $x = 0$ .

a) Use Euler's method with a step of $0.1$ to find an approximate value for $y$ when $x = 0.3$ .

b) i) Verify that the integrating factor for solving the differential equation is $sec x$ .

(ii) Hence, solve the differential equation, expressing your answer in the form $y = f (x)$ .

Show answer

Answer: a) $y (0.3) \approx 2.23$ ; b) (ii) $y = cos x (sin x + 2)$

Show working and answer

a) 1. We write $\frac{d y}{d x} = cos^{2} x - y tan x$ and apply $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ with $h = 0.1$ , starting from $(0, 2)$ (using the GDC). 2. $x_{0} = 0$ , $y_{0} = 2$ : $f = 1 - 0 = 1$ , so $y_{1} = 2 + 0.1 \cdot 1 = 2.1$ . 3. $x_{1} = 0.1$ : $f = cos^{2} 0.1 - 2.1 tan 0.1 \approx 0.7793$ , so $y_{2} \approx 2.1 + 0.07793 = 2.1779$ . 4. $x_{2} = 0.2$ : $f \approx cos^{2} 0.2 - 2.1779 tan 0.2 \approx 0.5190$ , so $y_{3} \approx 2.1779 + 0.0519 = 2.2298$ . 5. For $x = 0.3$ : $y \approx 2.23$ .

b) (i) 1. The equation is already in the form $\frac{d y}{d x} + P (x) y = Q (x)$ with $P (x) = tan x$ . 2. Integrating factor: $e^{\int t a n x d x} = e^{- l n ∣ c o s x ∣} = \frac{1}{cos x} = sec x$ (near $x = 0$ , $cos x > 0$ ).

(ii) 1. Multiply by $sec x$ : $\frac{d}{d x} (y sec x) = cos^{2} x sec x = cos x$ . 2. Integrate: $y sec x = sin x + C$ . 3. With $x = 0$ , $y = 2$ : $2 \cdot 1 = 0 + C$ , so $C = 2$ . 4. Solve for $y$ : $y = cos x (sin x + 2)$ .

Answer: a) $y (0.3) \approx 2.23$ ; b) (ii) $y = cos x (sin x + 2)$

Exercise 92 MediumIB · P1 May 1997

Consider the differential equation $cos^{2} x \frac{d y}{d x} = 1 - y$ , $- \frac{π}{2} < x < \frac{π}{2}$ , where $y = 2$ when $x = 0$ .

a) Find the integrating factor for the differential equation.

b) Find the solution of the differential equation.

Show answer

Answer: a) integrating factor $e^{t a n x}$ ; b) $y = 1 + \frac{1}{e ^{t a n x}}$

Show working and answer

a) 1. Divide by $cos^{2} x$ and move the term in $y$ to the left: $\frac{d y}{d x} + sec^{2} x y = sec^{2} x$ . 2. Integrating factor: $e^{\int s e c^{2} x d x} = e^{t a n x}$ .

b) 1. Multiply by the integrating factor: $\frac{d}{d x} (y e^{t a n x}) = sec^{2} x e^{t a n x}$ . 2. Integrate (the substitution $u = tan x$ gives $e^{t a n x}$ ): $y e^{t a n x} = e^{t a n x} + C$ . 3. With $y (0) = 2$ : $2 = 1 + C$ , so $C = 1$ . 4. $y = 1 + e^{- t a n x} = 1 + \frac{1}{e ^{t a n x}}$ .

Answer: a) integrating factor $e^{t a n x}$ ; b) $y = 1 + \frac{1}{e ^{t a n x}}$

Exercise 93 MediumIB · P3 TZ0 November 2017

Consider the differential equation $\frac{d y}{d x} + \frac{x}{x ^{2} + 1} y = x$ , where $y = 1$ when $x = 0$ .

a) Show that $x^{2} + 1$ is the integrating factor of the differential equation.

b) Solve the differential equation, giving your answer in the form $y = f (x)$ .

Show answer

Answer: b) $y = \frac{( x ^{2} + 1 ) ^{\frac{3}{2}} + 2}{3 x ^{2} + 1}$

Show working and answer

a) 1. The integrating factor is $e^{\int \frac{x}{x ^{2} + 1} d x}$ . 2. $\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln (x^{2} + 1)$ (substitution $u = x^{2} + 1$ ). 3. Hence the integrating factor is $e^{\frac{1}{2} l n (x^{2} + 1)} = x^{2} + 1$ .

b) 1. Multiply by $x^{2} + 1$ : $\frac{d}{d x} (y x^{2} + 1) = x x^{2} + 1$ . 2. Integrate with $u = x^{2} + 1$ : $\int x x^{2} + 1 d x = \frac{1}{3} (x^{2} + 1)^{3/2} + C$ . 3. With $y (0) = 1$ : $1 = \frac{1}{3} + C$ , so $C = \frac{2}{3}$ . 4. $y x^{2} + 1 = \frac{( x ^{2} + 1 ) ^{3/2} + 2}{3}$ , and solving for $y$ : $y = \frac{( x ^{2} + 1 ) ^{3/2} + 2}{3 x ^{2} + 1}$ .

Answer: b) $y = \frac{( x ^{2} + 1 ) ^{\frac{3}{2}} + 2}{3 x ^{2} + 1}$

Exercise 94 MediumIB · P3 TZ0 November 2016

Consider the differential equation $\frac{d y}{d x} + (\frac{2 x}{1 + x ^{2}}) y = x^{2}$ , given that $y = 2$ when $x = 0$ .

a) Show that $1 + x^{2}$ is an integrating factor for this differential equation.

b) Hence, solve this differential equation. Give your answer in the form $y = f (x)$ .

Show answer

Answer: b) $y = \frac{1}{1 + x ^{2}} (\frac{x ^{3}}{3} + \frac{x ^{5}}{5} + 2)$

Show working and answer

a) 1. The integrating factor is $e^{\int \frac{2 x}{1 + x ^{2}} d x}$ . 2. $\int \frac{2 x}{1 + x ^{2}} d x = ln (1 + x^{2})$ (it is of the form $\frac{u ^{'}}{u}$ ). 3. Therefore the integrating factor is $e^{l n (1 + x^{2})} = 1 + x^{2}$ .

b) 1. Multiply by $1 + x^{2}$ : $\frac{d}{d x} ((1 + x^{2}) y) = x^{2} (1 + x^{2}) = x^{2} + x^{4}$ . 2. Integrate: $(1 + x^{2}) y = \frac{x ^{3}}{3} + \frac{x ^{5}}{5} + C$ . 3. With $y (0) = 2$ : $2 = C$ . 4. $y = \frac{1}{1 + x ^{2}} (\frac{x ^{3}}{3} + \frac{x ^{5}}{5} + 2)$ .

Answer: b) $y = \frac{1}{1 + x ^{2}} (\frac{x ^{3}}{3} + \frac{x ^{5}}{5} + 2)$

Exercise 95 MediumIB · P1 November 2000

Solve the differential equation $x y \frac{d y}{d x} = 1 + y^{2}$ , given that $y = 0$ when $x = 2$ .

Show answer

Answer: $y = \frac{x ^{2} - 4}{2}$

Show working and answer

Separate the variables: $\frac{y}{1 + y ^{2}} d y = \frac{d x}{x}$ .
Integrate: $\frac{1}{2} ln (1 + y^{2}) = ln x + C$ (with $x > 0$ from the initial condition).
With $y = 0$ when $x = 2$ : $0 = ln 2 + C$ , so $C = - ln 2$ .
Therefore $\frac{1}{2} ln (1 + y^{2}) = ln \frac{x}{2}$ , i.e. $1 + y^{2} = \frac{x ^{2}}{4}$ .
Solve for $y$ : $y^{2} = \frac{x ^{2} - 4}{4}$ , and taking the positive branch, $y = \frac{x ^{2} - 4}{2}$ .

Answer: $y = \frac{x ^{2} - 4}{2}$

Exercise 96 MediumIB · P1 November 1999

Solve the differential equation $\frac{d y}{d x} = y tan x + 1$ , $0 \leq x < \frac{π}{2}$ if $y = 1$ when $x = 0$ .

Show answer

Answer: $y = tan x + sec x$

Show working and answer

Write the equation in linear form: $\frac{d y}{d x} - (tan x) y = 1$ .
Integrating factor: $e^{- \int t a n x d x} = e^{l n c o s x} = cos x$ (on $[0, \frac{π}{2} [$ , $cos x > 0$ ).
$\frac{d}{d x} (y cos x) = cos x$ , so $y cos x = sin x + C$ .
With $y = 1$ when $x = 0$ : $1 = 0 + C$ , so $C = 1$ .
$y = \frac{sin x + 1}{cos x} = tan x + sec x$ .

Answer: $y = tan x + sec x$

Exercise 97 MediumIB · P2 November 2002

(ii) Consider the differential equation $\frac{d y}{d x} = \frac{3 y ^{2} + x ^{2}}{2 x y}$ , for $x > 0$ .

a) Substitute $y = v x$ to show that $v + x \frac{d v}{d x} = \frac{3 v ^{2} + 1}{2 v}$ .

b) Hence find the solution of the differential equation, given that $y = 2$ when $x = 1$ .

Show answer

Answer: b) $y = x 5 x - 1$

Show working and answer

a) 1. With $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . 2. The right-hand side: $\frac{3 y ^{2} + x ^{2}}{2 x y} = \frac{3 v ^{2} x ^{2} + x ^{2}}{2 x \cdot v x} = \frac{3 v ^{2} + 1}{2 v}$ . 3. Therefore $v + x \frac{d v}{d x} = \frac{3 v ^{2} + 1}{2 v}$ .

b) 1. Rearranging: $x \frac{d v}{d x} = \frac{3 v ^{2} + 1}{2 v} - v = \frac{v ^{2} + 1}{2 v}$ . 2. Separating variables: $\int \frac{2 v}{v ^{2} + 1} d v = \int \frac{d x}{x}$ . 3. Integrating: $ln (v^{2} + 1) = ln x + C$ , that is, $v^{2} + 1 = A x$ . 4. With $x = 1$ , $y = 2$ ( $v = 2$ ): $5 = A$ . 5. $v^{2} = 5 x - 1$ , and since $v = \frac{y}{x} > 0$ (because $y (1) = 2 > 0$ ), $v = 5 x - 1$ . 6. $y = v x = x 5 x - 1$ .

Answer: b) $y = x 5 x - 1$

Exercise 98 MediumIB · P1 May 2001

Calculate the general solution of the differential equation $\frac{d x}{d t} = k x (5 - x)$ , where $0 < x < 5$ and $k$ is a constant.

Show answer

Answer: $t = \frac{1}{5 k} ln (\frac{x}{5 - x}) + C$

Show working and answer

It is a separable equation: $\frac{d x}{x ( 5 - x )} = k d t$ .
We split into partial fractions: $\frac{1}{x ( 5 - x )} = \frac{1}{5} (\frac{1}{x} + \frac{1}{5 - x})$ .
Integrating (with $0 < x < 5$ we do not need absolute values): $\frac{1}{5} (ln x - ln (5 - x)) = k t + C_{1}$ .
That is, $\frac{1}{5} ln (\frac{x}{5 - x}) = k t + C_{1}$ , and solving for $t$ (with $k \neq = 0$ ): $t = \frac{1}{5 k} ln (\frac{x}{5 - x}) + C$ , where $C = - \frac{C _{1}}{k}$ .

Answer: $t = \frac{1}{5 k} ln (\frac{x}{5 - x}) + C$

Exercise 99 MediumIB · P2 TZ1 May 2004

The temperature of an object $T^{\circ} C$ in a room, after $t$ minutes, satisfies the differential equation $\frac{d T}{d t} = k (T - 22)$ , where $k$ is a constant.

a) Solve this equation to show that $T = A e^{k t} + 22$ , where $A$ is a constant.

b) When $t = 0$ , $T = 100$ , and when $t = 15$ , $T = 70$ .

(i) Use this information to find the values of $A$ and $k$ .

(ii) Hence, find the value of $t$ when $T = 40$ .

Show answer

Answer: a) proved; b) i) $A = 78$ , $k \approx - 0.0324$ ; ii) $t \approx 45.3$ min

Show working and answer

a) 1. We separate the variables: $\frac{d T}{T - 22} = k d t$ . 2. Integrating: $ln ∣ T - 22∣ = k t + C_{1}$ . 3. Taking exponentials: $T - 22 = \pm e^{C_{1}} e^{k t} = A e^{k t}$ , with $A = \pm e^{C_{1}}$ constant. So $T = A e^{k t} + 22$ .

b) i) 1. With $t = 0$ and $T = 100$ : $100 = A + 22$ , so $A = 78$ . 2. With $t = 15$ and $T = 70$ : $70 = 78 e^{15 k} + 22$ , so $e^{15 k} = \frac{48}{78} = \frac{8}{13}$ . 3. Then $k = \frac{1}{15} ln \frac{8}{13} \approx - 0.0324$ .

(ii) 1. Setting $T = 40$ : $40 = 78 e^{k t} + 22$ , so $e^{k t} = \frac{18}{78} = \frac{3}{13}$ . 2. $t = \frac{1}{k} ln \frac{3}{13} = \frac{15 ln ( 3/13 )}{ln ( 8/13 )} \approx 45.3$ minutes.

Answer: a) proved; b) i) $A = 78$ , $k \approx - 0.0324$ ; ii) $t \approx 45.3$ min

Exercise 100 MediumIB · P3 TZ0 May 2006

a) Verify that $\int tan x d x = ln ∣ sec x ∣ + C$ , where $C$ is a constant.

b) Hence, find an integrating factor for solving the differential equation $\frac{d y}{d x} + y tan x = sec x$ .

c) Solve this differential equation given that $y = 2$ when $x = 0$ . Give your answer in the form $y = f (x)$ .

Show answer

Answer: b) $sec x$ ; c) $y = sin x + 2 cos x$

Show working and answer

a) 1. $\int tan x d x = \int \frac{sin x}{cos x} d x$ ; substitution $u = cos x$ , $d u = - sin x d x$ . 2. $= - \int \frac{d u}{u} = - ln ∣ cos x ∣ + C = ln \frac{1}{cos x} + C = ln ∣ sec x ∣ + C$ .

b) 1. The equation $\frac{d y}{d x} + y tan x = sec x$ is linear, with $P (x) = tan x$ . 2. Integrating factor: $e^{\int t a n x d x} = e^{l n ∣ s e c x ∣} = sec x$ (taking the interval where $sec x > 0$ , which contains $x = 0$ ).

c) 1. Multiplying by $sec x$ : $\frac{d}{d x} (y sec x) = sec^{2} x$ . 2. Integrating: $y sec x = tan x + C$ , so $y = sin x + C cos x$ . 3. With $y (0) = 2$ : $0 + C = 2$ , so $C = 2$ .

Answer: b) $sec x$ ; c) $y = sin x + 2 cos x$

Exercise 101 MediumIB · P1 TZ0 November 2006

Solve the differential equation $(x + 2)^{2} \frac{d y}{d x} = 4 x y$ $(x > - 2)$ , given that $y = 1$ when $x = - 1$ .

Show answer

Answer: $y = (x + 2)^{4} e^{\frac{- 8 ( x + 1 )}{x + 2}}$

Show working and answer

We separate the variables (with $x > - 2$ , $y > 0$ from the initial condition): $\frac{d y}{y} = \frac{4 x}{( x + 2 ) ^{2}} d x$ .
We rewrite: $\frac{4 x}{( x + 2 ) ^{2}} = \frac{4 ( x + 2 ) - 8}{( x + 2 ) ^{2}} = \frac{4}{x + 2} - \frac{8}{( x + 2 ) ^{2}}$ .
Integrating: $ln y = 4 ln (x + 2) + \frac{8}{x + 2} + C$ .
With $y = 1$ when $x = - 1$ : $0 = 0 + 8 + C$ , so $C = - 8$ .
$ln y = 4 ln (x + 2) + \frac{8}{x + 2} - 8 = 4 ln (x + 2) - \frac{8 ( x + 1 )}{x + 2}$ , because $\frac{8}{x + 2} - 8 = \frac{8 - 8 x - 16}{x + 2} = \frac{- 8 ( x + 1 )}{x + 2}$ .
$y = (x + 2)^{4} e^{- \frac{8 ( x + 1 )}{x + 2}}$ .

Answer: $y = (x + 2)^{4} e^{\frac{- 8 ( x + 1 )}{x + 2}}$

Exercise 102 MediumIB · P3 TZ0 November 2006

a) Show that $\frac{d}{d x} (ln (\frac{1 + x}{1 - x})) = \frac{2}{1 - x ^{2}}$ , $∣ x ∣ < 1$ .

b) Find the solution of the homogeneous differential equation $x^{2} \frac{d y}{d x} = x^{2} + x y - y^{2}$ , given that $y = \frac{1}{2}$ when $x = 1$ . Give your answer in the form $y = g (x)$ .

Show answer

Answer: a) shown; b) $y = \frac{3 x ^{3} - x}{1 + 3 x ^{2}}$

Show working and answer

a) 1. $ln (\frac{1 + x}{1 - x}) = ln (1 + x) - ln (1 - x)$ for $∣ x ∣ < 1$ . 2. Differentiating: $\frac{1}{1 + x} + \frac{1}{1 - x} = \frac{( 1 - x ) + ( 1 + x )}{1 - x ^{2}} = \frac{2}{1 - x ^{2}}$ .

b) 1. With $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ : $x^{2} (v + x \frac{d v}{d x}) = x^{2} + x^{2} v - x^{2} v^{2}$ . 2. Dividing by $x^{2}$ : $v + x \frac{d v}{d x} = 1 + v - v^{2}$ , i.e. $x \frac{d v}{d x} = 1 - v^{2}$ . 3. Separating: $\frac{d v}{1 - v ^{2}} = \frac{d x}{x}$ . 4. By (a), $\frac{1}{1 - v ^{2}} = \frac{1}{2} \frac{d}{d v} ln \frac{1 + v}{1 - v}$ , so $\frac{1}{2} ln \frac{1 + v}{1 - v} = ln x + C$ (with $∣ v ∣ < 1$ , since $v = \frac{1}{2}$ from the initial condition). 5. With $x = 1$ , $v = \frac{1}{2}$ : $\frac{1}{2} ln \frac{3/2}{1/2} = \frac{1}{2} ln 3 = C$ . 6. Then $ln \frac{1 + v}{1 - v} = 2 ln x + ln 3 = ln (3 x^{2})$ , i.e. $\frac{1 + v}{1 - v} = 3 x^{2}$ . 7. Rearranging: $1 + v = 3 x^{2} - 3 x^{2} v$ , $v (1 + 3 x^{2}) = 3 x^{2} - 1$ , so $v = \frac{3 x ^{2} - 1}{1 + 3 x ^{2}}$ . 8. $y = v x = \frac{3 x ^{3} - x}{1 + 3 x ^{2}}$ .

Answer: a) shown; b) $y = \frac{3 x ^{3} - x}{1 + 3 x ^{2}}$

Exercise 103 MediumIB · P1 TZ0 November 2007

Solve the differential equation $\frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}}$ , given that $y = 3$ when $x = \frac{3}{3}$ .

Express the answer in the form $y = \frac{a x + a}{a - x a}$ where $a \in Z^{+}$ .

Show answer

Answer: $y = \frac{3 x + 3}{3 - x 3}$

Show working and answer

Separate the variables: $\frac{d y}{1 + y ^{2}} = \frac{d x}{1 + x ^{2}}$ .
Integrate: $arctan y = arctan x + C$ .
With $y = 3$ and $x = \frac{3}{3}$ : $arctan 3 = \frac{π}{3}$ and $arctan \frac{3}{3} = \frac{π}{6}$ , so $C = \frac{π}{6}$ .
$y = tan (arctan x + \frac{π}{6}) = \frac{x + tan \frac{π}{6}}{1 - x tan \frac{π}{6}} = \frac{x + \frac{1}{3}}{1 - \frac{x}{3}}$ .
Multiply the numerator and denominator by $3$ , and then by $3$ again: $\frac{3 x + 1}{3 - x} = \frac{3 x + 3}{3 - x 3}$ , which is the form requested with $a = 3$ .

Answer: $y = \frac{3 x + 3}{3 - x 3}$

Exercise 104 MediumIB · P1 TZ1 May 2008

A gourmet chef is famous for a spherical soufflé.

Once it is placed in the oven, its volume increases at a rate proportional to its radius.

a) Show that the radius $r$ cm of the soufflé, at time $t$ minutes after it is placed in the oven, satisfies the differential equation $\frac{d r}{d t} = \frac{k}{r}$ , where $k$ is a constant.

b) Given that the radius of the soufflé is $8$ cm when it enters the oven, and $12$ cm when it is cooked $30$ minutes later, find, to the nearest cm, its radius after $15$ minutes in the oven.

Show answer

Answer: a) shown; b) $r \approx 10$ cm

Show working and answer

a) 1. Volume of the sphere: $V = \frac{4}{3} π r^{3}$ , so $\frac{d V}{d r} = 4 π r^{2}$ . 2. The statement says that $\frac{d V}{d t} = cr$ ( $c$ constant). 3. Chain rule: $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t}$ , so $4 π r^{2} \frac{d r}{d t} = cr$ . 4. $\frac{d r}{d t} = \frac{c}{4 π r} = \frac{k}{r}$ , with $k = \frac{c}{4 π}$ constant.

b) 1. Separate the variables: $r d r = k d t$ , so $\frac{r ^{2}}{2} = k t + C$ . 2. With $r = 8$ at $t = 0$ : $C = 32$ . 3. With $r = 12$ at $t = 30$ : $72 = 30 k + 32$ , so $k = \frac{4}{3}$ . 4. $r^{2} = \frac{8}{3} t + 64$ . 5. At $t = 15$ : $r^{2} = 40 + 64 = 104$ , so $r = 104 \approx 10.2$ .

Answer: a) shown; b) $r \approx 10$ cm

Exercise 105 MediumIB · P3 TZ0 November 2008

a) Verify that the solution of the homogeneous differential equation $\frac{d y}{d x} = \frac{y}{x} + 1$ , $x > 0$ , is $y = x (ln x - 1)$ , given that $y = 0$ when $x = e$ .

b) i) Determine the first three derivatives of the function $f (x) = x (ln x - 1)$ .

(ii) Hence, find the first three non-zero terms of the Taylor series for $f (x)$ about $x = 1$ .

Show answer

Answer: a) shown; b) i) $f^{'} (x) = ln x$ , $f^{''} (x) = \frac{1}{x}$ , $f^{'''} (x) = - \frac{1}{x ^{2}}$ ; ii) $f (x) \approx - 1 + \frac{( x - 1 ) ^{2}}{2} - \frac{( x - 1 ) ^{3}}{6}$

Show working and answer

a) 1. With $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Substituting into $\frac{d y}{d x} = \frac{y}{x} + 1$ gives $v + x \frac{d v}{d x} = v + 1$ . 2. Then $x \frac{d v}{d x} = 1$ , i.e. $d v = \frac{d x}{x}$ and $v = ln x + C$ . 3. So $y = x (ln x + C)$ . With $y = 0$ when $x = e$ : $e (1 + C) = 0$ , so $C = - 1$ . 4. $y = x (ln x - 1)$ .

b) i) 1. $f (x) = x ln x - x$ , so $f^{'} (x) = ln x + 1 - 1 = ln x$ . 2. $f^{''} (x) = \frac{1}{x}$ . 3. $f^{'''} (x) = - \frac{1}{x ^{2}}$ .

(ii) 1. At $x = 1$ : $f (1) = - 1$ , $f^{'} (1) = 0$ , $f^{''} (1) = 1$ , $f^{'''} (1) = - 1$ . 2. Taylor series: $f (x) = f (1) + f^{'} (1) (x - 1) + \frac{f ^{''} ( 1 )}{2 !} (x - 1)^{2} + \frac{f ^{'''} ( 1 )}{3 !} (x - 1)^{3} + \dots$ . 3. The first three non-zero terms: $- 1 + \frac{( x - 1 ) ^{2}}{2} - \frac{( x - 1 ) ^{3}}{6}$ .

Answer: a) shown; b) i) $f^{'} (x) = ln x$ , $f^{''} (x) = \frac{1}{x}$ , $f^{'''} (x) = - \frac{1}{x ^{2}}$ ; ii) $f (x) \approx - 1 + \frac{( x - 1 ) ^{2}}{2} - \frac{( x - 1 ) ^{3}}{6}$

Exercise 106 MediumIB · P3 TZ0 November 2012

A differential equation is given by $\frac{d y}{d x} = \frac{y}{x}$ , where $x > 0$ and $y > 0$ .

a) Solve this differential equation by separating the variables, giving your answer in the form $y = f (x)$ .

b) Solve the same differential equation using the standard homogeneous substitution $y = v x$ .

c) Solve the same differential equation using an integrating factor.

d) If $y = 20$ when $x = 2$ , calculate $y$ when $x = 5$ .

Show answer

Answer: a) $y = A x$ ; b) $y = C x$ ; c) $y = C x$ ; d) $y = 50$

Show working and answer

a) 1. Separating variables: $\frac{d y}{y} = \frac{d x}{x}$ . 2. Integrating (with $x, y > 0$ ): $ln y = ln x + C$ . 3. Then $y = e^{C} x = A x$ , with $A$ a constant.

b) 1. With $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ and $\frac{y}{x} = v$ . 2. The equation becomes $v + x \frac{d v}{d x} = v$ , i.e. $x \frac{d v}{d x} = 0$ . 3. Hence $v$ is constant, $v = C$ , and $y = C x$ .

c) 1. In linear form: $\frac{d y}{d x} - \frac{1}{x} y = 0$ . 2. Integrating factor: $e^{\int - \frac{1}{x} d x} = e^{- l n x} = \frac{1}{x}$ . 3. Multiplying: $\frac{d}{d x} (\frac{y}{x}) = 0$ , so $\frac{y}{x} = C$ and $y = C x$ .

d) 1. With $y = 20$ at $x = 2$ : $20 = 2 C$ , so $C = 10$ and $y = 10 x$ . 2. At $x = 5$ : $y = 50$ .

Answer: a) $y = A x$ ; b) $y = C x$ ; c) $y = C x$ ; d) $y = 50$

Exercise 107 MediumIB · P1 TZ0 November 2021

Solve the differential equation $\frac{d y}{d x} = \frac{ln 2 x}{x ^{2}} - \frac{2}{x} y$ , $x > 0$ , given that $y = 4$ when $x = \frac{1}{2}$ .

Give your answer in the form $y = f (x)$ .

Show answer

Answer: $y = \frac{ln 2 x}{x} - \frac{1}{x} + \frac{3}{2 x ^{2}}$

Show working and answer

We rearrange as a linear equation: $\frac{d y}{d x} + \frac{2}{x} y = \frac{ln 2 x}{x ^{2}}$ .
Integrating factor: $e^{\int \frac{2}{x} d x} = e^{2 l n x} = x^{2}$ (with $x > 0$ ).
We multiply through: $\frac{d}{d x} (x^{2} y) = ln 2 x$ .
Integrating by parts ( $u = ln 2 x$ , $d v = d x$ ): $\int ln 2 x d x = x ln 2 x - \int x \cdot \frac{1}{x} d x = x ln 2 x - x + C$ .
Therefore $y = \frac{ln 2 x}{x} - \frac{1}{x} + \frac{C}{x ^{2}}$ .
Condition $y = 4$ at $x = \frac{1}{2}$ : $4 = \frac{ln 1}{1/2} - 2 + 4 C$ , so $4 C = 6$ and $C = \frac{3}{2}$ .
Check: differentiating the solution gives $\frac{ln 2 x}{x ^{2}} - \frac{2}{x} y$ , as the equation requires.

Answer: $y = \frac{ln 2 x}{x} - \frac{1}{x} + \frac{3}{2 x ^{2}}$

Exercise 108 MediumIB · P1 TZ0 November 2022

Consider the homogeneous differential equation $\frac{d y}{d x} = \frac{y ^{2} - 2 x ^{2}}{x y}$ , where $x, y \neq = 0$ .

It is known that $y = 2$ when $x = 1$ .

a) Using the substitution $y = v x$ , solve the differential equation.

Give your answer in the form $y^{2} = f (x)$ .

The points of zero gradient on the curve $y^{2} = f (x)$ lie on two lines of the form $y = m x$ , where $m \in R$ .

b) Find the values of $m$ .

Show answer

Answer: a) $y^{2} = 2 x^{2} (- 2 ln x + 2)$ ; b) $m = \pm 2$

Show working and answer

a) 1. Substitution $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . 2. The right-hand side is $\frac{v ^{2} x ^{2} - 2 x ^{2}}{x \cdot v x} = \frac{v ^{2} - 2}{v}$ . 3. $x \frac{d v}{d x} = \frac{v ^{2} - 2}{v} - v = - \frac{2}{v}$ . 4. We separate the variables: $v d v = - \frac{2}{x} d x$ , and integrate: $\frac{v ^{2}}{2} = - 2 ln x + C$ (with $x > 0$ , from the initial condition $x = 1$ ). 5. Condition: $x = 1$ , $y = 2$ , so $v = 2$ and $\frac{4}{2} = C$ , that is, $C = 2$ . 6. $v^{2} = - 4 ln x + 4$ , so $y^{2} = x^{2} v^{2} = x^{2} (4 - 4 ln x) = 2 x^{2} (- 2 ln x + 2)$ .

b) 1. Zero gradient: $\frac{d y}{d x} = 0 ⟺ y^{2} - 2 x^{2} = 0$ , that is, $y = \pm 2 x$ . 2. We check that these lines meet the curve: $4 x^{2} (1 - ln x) = 2 x^{2}$ gives $ln x = \frac{1}{2}$ , $x = e$ . 3. $m = 2$ and $m = - 2$ .

Answer: a) $y^{2} = 2 x^{2} (- 2 ln x + 2)$ ; b) $m = \pm 2$

Exercise 109 HardIB · P2 TZ2 May 2012

A particle moves in a straight line with velocity $v$ metres per second. At any instant, $t$ seconds, $0 \leq t \leq \frac{3 π}{4}$ , the velocity is given by the differential equation $\frac{d v}{d t} + v^{2} + 1 = 0$ . It is also known that $v = 1$ when $t = 0$ .

a) Find an expression for $v$ in terms of $t$ .

b) Sketch the graph of $v$ against $t$ , clearly showing the coordinates of all points of intersection with the axes and the equations of all asymptotes.

c) Find an expression for the displacement $s$ in terms of $t$ , given that $s = 0$ when $t = 0$ .

d) Hence, or otherwise, verify that $s = \frac{1}{2} ln \frac{2}{1 + v ^{2}}$ .

View solution

a) 1. From $\frac{d v}{d t} + v^{2} + 1 = 0$ we get $\frac{d v}{d t} = - (1 + v^{2})$ , which is separable. 2. $\int \frac{d v}{1 + v ^{2}} = - \int d t$ , so $arctan v = - t + C$ . 3. With $v (0) = 1$ : $arctan 1 = \frac{π}{4} = C$ . 4. $v = tan (\frac{π}{4} - t)$ . Check: $v^{'} = - sec^{2} (\frac{π}{4} - t) = - (1 + v^{2})$ ✓.

b) 1. Intercept with the $v$ -axis ( $t = 0$ ): $v = tan \frac{π}{4} = 1$ , that is $(0, 1)$ . 2. Intercept with the $t$ -axis: $tan (\frac{π}{4} - t) = 0$ for $t = \frac{π}{4}$ , that is $(\frac{π}{4}, 0)$ . 3. As $t \to \frac{3 π}{4}$ , the argument $\frac{π}{4} - t \to - \frac{π}{2}$ and $v \to - \infty$ : vertical asymptote $t = \frac{3 π}{4}$ . 4. The function is decreasing over the whole interval.

Graph of v against t for t between 0 and 3π/4: a decreasing curve starting at the point (0, 1), crossing the t-axis and approaching a vertical asymptote from below

Intercepts with the axes: $(0, 1)$ and $(\frac{π}{4}, 0)$ . Vertical asymptote: $t = \frac{3 π}{4}$ .

c) 1. $s = \int v d t = \int tan (\frac{π}{4} - t) d t$ . Since $\int tan u d u = - ln ∣ cos u ∣$ and $d u = - d t$ with $u = \frac{π}{4} - t$ , we get $s = ln cos (\frac{π}{4} - t) + C$ . 2. For $0 \leq t < \frac{3 π}{4}$ the argument lies in $] - \frac{π}{2}, \frac{π}{4}]$ , where the cosine is positive: $s = ln cos (\frac{π}{4} - t) + C$ . 3. With $s (0) = 0$ : $ln \frac{2}{2} + C = 0$ , so $C = - ln \frac{2}{2} = \frac{ln 2}{2}$ . 4. $s = ln cos (\frac{π}{4} - t) + \frac{ln 2}{2}$ .

d) 1. We start from $tan^{2} α + 1 = \frac{1}{cos ^{2} α}$ with $α = \frac{π}{4} - t$ : $1 + v^{2} = \frac{1}{cos ^{2} α}$ , so $cos^{2} α = \frac{1}{1 + v ^{2}}$ . 2. Since $cos α > 0$ , $cos α = \frac{1}{1 + v ^{2}}$ and $ln cos α = - \frac{1}{2} ln (1 + v^{2})$ . 3. Substituting into (c): $s = \frac{1}{2} ln 2 - \frac{1}{2} ln (1 + v^{2}) = \frac{1}{2} ln \frac{2}{1 + v ^{2}}$ ✓.

Answer: a) $v = tan (\frac{π}{4} - t)$ ; b) intercepts $(0, 1)$ and $(\frac{π}{4}, 0)$ , asymptote $t = \frac{3 π}{4}$ ; c) $s = ln cos (\frac{π}{4} - t) + \frac{ln 2}{2}$ ; d) shown: $s = \frac{1}{2} ln \frac{2}{1 + v ^{2}}$

Exercise 110 HardIB · P1 TZ0 November 2011

The curve $C$ , whose equation is $y = f (x)$ , satisfies the differential equation $\frac{d y}{d x} = \frac{y}{ln y} (x + 2)$ , $y > 1$ ; also $y = e$ when $x = 2$ .

a) Find the equation of the tangent to $C$ at the point $(2, e)$ .

b) Find $f (x)$ .

c) Determine the largest possible domain of $f$ .

d) Verify that the equation $f (x) = f^{'} (x)$ has no solution.

Show answer

Answer: a) $y = 4 e x - 7 e$ ; b) $f (x) = e^{x^{2} + 4 x - 11}$ ; c) $] - \infty, - 2 - 15] \cup [- 2 + 15, + \infty [$ ; d) verified: there is no solution.

Show working and answer

a) 1. The gradient at $(2, e)$ : $\frac{d y}{d x} = \frac{e}{ln e} (2 + 2) = 4 e$ . 2. Tangent: $y - e = 4 e (x - 2)$ , that is $y = 4 e x - 7 e$ .

b) 1. We separate variables: $\frac{ln y}{y} d y = (x + 2) d x$ . 2. We integrate: on the left, substitute $u = ln y$ (with $d u = \frac{d y}{y}$ ) to get $\frac{( ln y ) ^{2}}{2}$ . On the right: $\frac{x ^{2}}{2} + 2 x$ . 3. $\frac{( ln y ) ^{2}}{2} = \frac{x ^{2}}{2} + 2 x + C$ . With $y = e$ at $x = 2$ : $\frac{1}{2} = 2 + 4 + C$ , so $C = - \frac{11}{2}$ . 4. $(ln y)^{2} = x^{2} + 4 x - 11$ . Since $y > 1$ , $ln y > 0$ , so $ln y = x^{2} + 4 x - 11$ . 5. $f (x) = e^{x^{2} + 4 x - 11}$ .

c) 1. The square root requires $x^{2} + 4 x - 11 \geq 0$ . The roots of $x^{2} + 4 x - 11 = 0$ are $x = - 2 \pm 15$ . 2. The parabola is non-negative outside its roots. Domain: $] - \infty, - 2 - 15] \cup [- 2 + 15, + \infty [$ .

d) 1. From the differential equation, $f^{'} (x) = \frac{f ( x )}{ln f ( x )} (x + 2) = \frac{f ( x ) ( x + 2 )}{x ^{2} + 4 x - 11}$ (since $ln f (x) = x^{2} + 4 x - 11$ ). 2. $f (x) = f^{'} (x)$ with $f (x) > 0$ forces $1 = \frac{x + 2}{x ^{2} + 4 x - 11}$ , that is $x^{2} + 4 x - 11 = x + 2$ . 3. Squaring: $x^{2} + 4 x - 11 = x^{2} + 4 x + 4$ , that is $- 11 = 4$ , which is impossible. 4. Therefore the equation $f (x) = f^{'} (x)$ has no solutions.

Answer: a) $y = 4 e x - 7 e$ ; b) $f (x) = e^{x^{2} + 4 x - 11}$ ; c) $] - \infty, - 2 - 15] \cup [- 2 + 15, + \infty [$ ; d) verified: there is no solution.

Exercise 111 HardIB · P2 TZ2 May 2008

A particle moves from a fixed point $O$ along a straight line in a positive direction. The velocity $v$ in $m s^{- 1}$ at time $t$ seconds, where $t \geq 0$ , satisfies the differential equation $\frac{d v}{d t} = \frac{- v ( 1 + v ^{2} )}{50}$ . The particle starts from $O$ with an initial velocity of $10 m s^{- 1}$ .

a) (i) Express, as a definite integral, the time taken for the velocity of the particle to decrease from $10 m s^{- 1}$ to $5 m s^{- 1}$ .

(ii) Hence, calculate the time taken for the velocity of the particle to decrease from $10 m s^{- 1}$ to $5 m s^{- 1}$ .

b) (i) Verify that when $v > 0$ , the motion of this particle can also be described by the differential equation $\frac{d v}{d x} = \frac{- ( 1 + v ^{2} )}{50}$ , where $x$ represents the displacement in metres from $O$ .

(ii) Given that $v = 10$ when $x = 0$ , solve the differential equation, expressing $x$ in terms of $v$ .

Show answer

Answer: a) (i) $t = \int_{10}^{5} \frac{- 50}{v ( 1 + v ^{2} )} d v$ ; (ii) $t \approx 0.732$ s; b) (i) verified; (ii) $x = - 50 (arctan v - arctan 10)$

Show working and answer

a) (i) 1. We rearrange the differential equation to get $\frac{d t}{d v} = \frac{- 50}{v ( 1 + v ^{2} )}$ . 2. The time to go from $v = 10$ to $v = 5$ is $t = \int_{10}^{5} \frac{- 50}{v ( 1 + v ^{2} )} d v$ (equivalent to $50 \int_{5}^{10} \frac{d v}{v ( 1 + v ^{2} )}$ ).

(ii) 1. Partial fractions: $\frac{1}{v ( 1 + v ^{2} )} = \frac{1}{v} - \frac{v}{1 + v ^{2}}$ (check by combining: $\frac{1 + v ^{2} - v ^{2}}{v ( 1 + v ^{2} )}$ ). 2. An antiderivative is $ln v - \frac{1}{2} ln (1 + v^{2}) = ln \frac{v}{1 + v ^{2}}$ (for $v > 0$ ). 3. $t = 50 [ln \frac{v}{1 + v ^{2}}]_{5}^{10} = 50 [ln \frac{10}{101} - ln \frac{5}{26}] = 50 ln \frac{2 26}{101}$ . 4. Numerically, $t = - 50 ln \frac{101}{2 26} \approx 0.732$ s.

b) (i) 1. Chain rule: $\frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = \frac{d v}{d x} \cdot v$ , because $\frac{d x}{d t} = v$ . 2. Hence $v \frac{d v}{d x} = - \frac{v ( 1 + v ^{2} )}{50}$ . 3. As $v > 0$ we can divide by $v$ : $\frac{d v}{d x} = - \frac{1 + v ^{2}}{50}$ ✓.

(ii) 1. Separate the variables: $\frac{d v}{1 + v ^{2}} = - \frac{d x}{50}$ . 2. Integrate: $arctan v = - \frac{x}{50} + C$ . 3. With $v = 10$ at $x = 0$ : $C = arctan 10$ . 4. Solving for $x$ : $x = - 50 (arctan v - arctan 10) = 50 (arctan 10 - arctan v)$ .

Answer: a) (i) $t = \int_{10}^{5} \frac{- 50}{v ( 1 + v ^{2} )} d v$ ; (ii) $t \approx 0.732$ s; b) (i) verified; (ii) $x = - 50 (arctan v - arctan 10)$

Exercise 112 HardIB · P2 TZ2 May 2011

a) Using integration by parts, verify that: $\int e^{2 x} sin x d x = \frac{1}{5} e^{2 x} (2 sin x - cos x) + C$ .

b) Solve the differential equation $\frac{d y}{d x} = 1 - y^{2} e^{2 x} sin x$ , given that $y = 0$ when $x = 0$ .

Write your answer in the form $y = f (x)$ .

c) (i) Sketch the graph of $y = f (x)$ found in part (b), for $0 \leq x \leq 1.5$ . Find the coordinates of the point $P$ , the first point of intersection with the positive $x$ -axis, and mark this point on your sketch.

(ii) The region bounded by the graph of $f$ and the $x$ -axis, between the origin and $P$ , is rotated through $36 0^{\circ}$ about the $x$ -axis, forming a solid of revolution. Calculate the volume of this solid.

View solution

a) 1. Let $I = \int e^{2 x} sin x d x$ . Integrating by parts with $u = sin x$ and $d v = e^{2 x} d x$ : $I = \frac{1}{2} e^{2 x} sin x - \frac{1}{2} \int e^{2 x} cos x d x$ . 2. For $J = \int e^{2 x} cos x d x$ , integrating by parts again with $u = cos x$ and $d v = e^{2 x} d x$ : $J = \frac{1}{2} e^{2 x} cos x + \frac{1}{2} I$ . 3. Substituting: $I = \frac{1}{2} e^{2 x} sin x - \frac{1}{4} e^{2 x} cos x - \frac{1}{4} I$ . 4. Moving $\frac{1}{4} I$ to the other side: $\frac{5}{4} I = \frac{1}{4} e^{2 x} (2 sin x - cos x)$ . 5. Hence $I = \frac{1}{5} e^{2 x} (2 sin x - cos x) + C$ ✓.

b) 1. Separate the variables: $\frac{d y}{1 - y ^{2}} = e^{2 x} sin x d x$ . 2. Integrate: $arcsin y = \frac{1}{5} e^{2 x} (2 sin x - cos x) + C$ , using (a). 3. With $y = 0$ at $x = 0$ : $0 = \frac{1}{5} (- 1) + C$ , so $C = \frac{1}{5}$ . 4. $y = sin [\frac{e ^{2 x} ( 2 sin x - cos x ) + 1}{5}]$ .

c) (i) 1. Let $a (x) = \frac{e ^{2 x} ( 2 sin x - cos x ) + 1}{5}$ , so that $y = sin a (x)$ . Its derivative is $a^{'} (x) = e^{2 x} sin x > 0$ for $0 < x < π$ , so $a$ increases from $a (0) = 0$ . 2. The first crossing of the $x$ -axis occurs when $a (x) = π$ . Using the GDC: $x \approx 1.16$ , i.e. $P (1.16; 0)$ . 3. Before $P$ , $y$ rises from the origin to a maximum of $1$ (when $a = \frac{π}{2}$ , near $x = 0.95$ ) and falls to $0$ at $P$ .

Graph of y = sin((e^(2x)(2 sin x − cos x) + 1)/5) for x between 0 and 1.5: oscillates between 1 and −1, with the first point P where it meets the positive x-axis

$P (1.16; 0)$

(ii) 1. Volume about the $x$ -axis: $V = π \int_{0}^{1.16} y^{2} d x = π \int_{0}^{1.16} sin^{2} [\frac{e ^{2 x} ( 2 sin x - cos x ) + 1}{5}] d x$ . 2. Using the GDC: $V \approx 1.05$ .

Answer: a) proved; b) $y = sin [\frac{e ^{2 x} ( 2 sin x - cos x ) + 1}{5}]$ ; c) (i) $P (1.16; 0)$ ; (ii) $V \approx 1.05$

Exercise 113 HardIB · P3 TZ0 November 2011

The real and imaginary parts of a complex number $x + i y$ are related by the differential equation $(x + y) \frac{d y}{d x} + (x - y) = 0$ . Solving the differential equation, given that $y = 3$ when $x = 1$ , show that the relationship between the modulus $r$ and the argument $θ$ of the complex number is $r = 2 e^{\frac{π}{3} - θ}$ .

Show answer

Answer: proved: $r = 2 e^{\frac{π}{3} - θ}$ .

Show working and answer

Homogeneous equation: it is solved with $y = v x$ .

The equation is $(x + y) \frac{d y}{d x} + (x - y) = 0$ . With $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
We substitute: $x (1 + v) (v + x \frac{d v}{d x}) + x (1 - v) = 0$ . We divide by $x$ : $(1 + v) v + (1 + v) x \frac{d v}{d x} + 1 - v = 0$ .
We simplify $v + v^{2} + 1 - v = v^{2} + 1$ : $(1 + v) x \frac{d v}{d x} = - (1 + v^{2})$ .
Separate the variables: $\int \frac{1 + v}{1 + v ^{2}} d v = - \int \frac{d x}{x}$ .
We evaluate the left-hand side: $\int \frac{1}{1 + v ^{2}} d v + \int \frac{v}{1 + v ^{2}} d v = arctan v + \frac{1}{2} ln (1 + v^{2})$ . Thus $arctan v + \frac{1}{2} ln (1 + v^{2}) = - ln x + C$ .
Initial condition: $x = 1$ , $y = 3$ , so $v = 3$ : $\frac{π}{3} + \frac{1}{2} ln 4 = C$ , i.e. $C = \frac{π}{3} + ln 2$ .
We move to polar coordinates. With $x > 0$ , $v = \frac{y}{x} = tan θ$ and $θ = arctan v$ . Also $r = x^{2} + y^{2} = x 1 + v^{2}$ , so $\frac{1}{2} ln (1 + v^{2}) = ln r - ln x$ .
Then $θ + ln r - ln x = - ln x + \frac{π}{3} + ln 2$ , and the $ln x$ terms cancel: $ln r = \frac{π}{3} - θ + ln 2$ .
Therefore $r = 2 e^{\frac{π}{3} - θ}$ .

Answer: proved: $r = 2 e^{\frac{π}{3} - θ}$ .

Exercise 114 HardIB · P3 TZ0 May 2019

Consider the differential equation $2 x y \frac{d y}{d x} = y^{2} - x^{2}$ , where $x > 0$ .

a) Solve the differential equation and show that a general solution of this equation is $x^{2} + y^{2} = c x$ , where $c$ is a positive constant.

b) Prove that the curve of this general solution has two horizontal tangent lines, and state the equation of each of them in terms of $c$ .

Show answer

Answer: a) $x^{2} + y^{2} = c x$ ; b) $y = \frac{c}{2}$ and $y = - \frac{c}{2}$

Show working and answer

a) This is a homogeneous equation, so we use the substitution $y = v x$ .

With $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . The equation $2 x y \frac{d y}{d x} = y^{2} - x^{2}$ becomes $2 x \cdot v x (v + x \frac{d v}{d x}) = v^{2} x^{2} - x^{2}$ .
Dividing by $x^{2}$ : $2 v^{2} + 2 v x \frac{d v}{d x} = v^{2} - 1$ , that is, $2 v x \frac{d v}{d x} = - v^{2} - 1$ .
Separating the variables: $\int \frac{2 v}{v ^{2} + 1} d v = - \int \frac{d x}{x}$ , so $ln (v^{2} + 1) = - ln x + K$ (with $x > 0$ ).
Then $ln [x (v^{2} + 1)] = K$ , so $x (v^{2} + 1) = c$ , with $c = e^{K} > 0$ .
Undoing the substitution, $v = \frac{y}{x}$ : $x (\frac{y ^{2}}{x ^{2}} + 1) = c$ , that is, $\frac{x ^{2} + y ^{2}}{x} = c$ and $x^{2} + y^{2} = c x$ .

b) 1. A horizontal tangent requires $\frac{d y}{d x} = 0$ . In the differential equation, $2 x y \cdot 0 = y^{2} - x^{2}$ , so $y^{2} = x^{2}$ , that is, $y = \pm x$ . 2. Substituting into the curve: $x^{2} + x^{2} = c x \Rightarrow x = \frac{c}{2}$ (with $x > 0$ ). Then $y = \pm \frac{c}{2}$ . 3. There are exactly two points, $(\frac{c}{2}, \frac{c}{2})$ and $(\frac{c}{2}, - \frac{c}{2})$ , and at them $y \neq = 0$ and $x \neq = 0$ (the gradient formula $y^{'} = \frac{y ^{2} - x ^{2}}{2 x y}$ is well defined), so there are two horizontal tangents. 4. Their equations are $y = \frac{c}{2}$ and $y = - \frac{c}{2}$ .

Answer: a) $x^{2} + y^{2} = c x$ ; b) $y = \frac{c}{2}$ and $y = - \frac{c}{2}$

Exercise 115 HardIB · P3 TZ0 November 2010

Solve the equation $(x - 1) \frac{d y}{d x} + x y = (x - 1) e^{- x}$ , given that when $x = 0$ , $y = 1$ . Give your answer in the form $y = f (x)$ .

Show answer

Answer: $y = \frac{( x - 1 ) ^{2} - 3}{2 e ^{x} ( x - 1 )}$

Show working and answer

This is a first-order linear equation.

Dividing by $x - 1$ (with $x \neq = 1$ ): $\frac{d y}{d x} + \frac{x}{x - 1} y = e^{- x}$ . As $\frac{x}{x - 1} = 1 + \frac{1}{x - 1}$ , the integrating factor is $e^{x + l n ∣ x - 1∣} = ∣ x - 1∣ e^{x}$ ; we use $(x - 1) e^{x}$ (this gives the same equation, apart from a constant sign).
Multiplying: $[(x - 1) e^{x} y]^{'} = (x - 1) e^{x} \cdot e^{- x} = x - 1$ .
Integrating: $(x - 1) e^{x} y = \frac{( x - 1 ) ^{2}}{2} + C$ .
Using $x = 0$ , $y = 1$ : $(- 1) (1) (1) = \frac{1}{2} + C$ , so $C = - \frac{3}{2}$ .
$y = \frac{\frac{( x - 1 ) ^{2}}{2} - \frac{3}{2}}{( x - 1 ) e ^{x}} = \frac{( x - 1 ) ^{2} - 3}{2 e ^{x} ( x - 1 )}$ .

Answer: $y = \frac{( x - 1 ) ^{2} - 3}{2 e ^{x} ( x - 1 )}$

Exercise 116 HardIB · P3 TZ0 May 2011

Solve the differential equation $x^{2} \frac{d y}{d x} = y^{2} + 3 x y + 2 x^{2}$ , given that when $x = 1$ , $y = - 1$ . Give your answer in the form $y = f (x)$ .

Show answer

Answer: $y = x tan (ln x) - x$

Show working and answer

This is a homogeneous equation: we use the substitution $y = v x$ .

With $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , the equation $x^{2} \frac{d y}{d x} = y^{2} + 3 x y + 2 x^{2}$ becomes $x^{2} (v + x \frac{d v}{d x}) = v^{2} x^{2} + 3 v x^{2} + 2 x^{2}$ .
Dividing by $x^{2}$ : $v + x \frac{d v}{d x} = v^{2} + 3 v + 2$ , that is, $x \frac{d v}{d x} = v^{2} + 2 v + 2 = (v + 1)^{2} + 1$ .
Separating the variables: $\int \frac{d v}{( v + 1 ) ^{2} + 1} = \int \frac{d x}{x}$ , so $arctan (v + 1) = ln x + C$ .
Using $x = 1$ , $y = - 1$ : $v = - 1$ and $arctan 0 = 0 = ln 1 + C$ , so $C = 0$ .
$v + 1 = tan (ln x)$ , and with $v = \frac{y}{x}$ : $y = x tan (ln x) - x$ .

Answer: $y = x tan (ln x) - x$

Exercise 117 HardIB · P3 TZ0 November 2009

Solve the differential equation $\frac{d y}{d x} = \frac{y}{x} + \frac{y ^{2}}{x ^{2}}$ (where $x > 0$ ), given that $y = 2$ when $x = 1$ . Give your answer in the form $y = f (x)$ .

Show answer

Answer: $y = \frac{- x}{ln x - \frac{1}{2}}$

Show working and answer

This is a homogeneous equation: we use the substitution $y = v x$ .

With $y = v x$ : $v + x \frac{d v}{d x} = v + v^{2}$ , so $x \frac{d v}{d x} = v^{2}$ .
Separating the variables: $\int \frac{d v}{v ^{2}} = \int \frac{d x}{x}$ , that is, $- \frac{1}{v} = ln x + C$ .
Using $x = 1$ , $y = 2$ : $v = 2$ , so $- \frac{1}{2} = C$ .
$- \frac{1}{v} = ln x - \frac{1}{2}$ , so $v = \frac{- 1}{ln x - \frac{1}{2}}$ and $y = v x = \frac{- x}{ln x - \frac{1}{2}}$ .

Answer: $y = \frac{- x}{ln x - \frac{1}{2}}$

Exercise 118 HardIB · P3 TZ0 May 2010

Solve the differential equation $x^{2} \frac{d y}{d x} = y^{2} + x y + 4 x^{2}$ , given that when $x = 1$ , $y = 2$ . Give your answer in the form $y = f (x)$ .

Show answer

Answer: $y = 2 x tan (ln x^{2} + \frac{π}{4})$

Show working and answer

Dividing by $x^{2}$ : $\frac{d y}{d x} = (\frac{y}{x})^{2} + \frac{y}{x} + 4$ . This is a homogeneous equation, so we make the substitution $y = v x$ , with $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
We get $v + x \frac{d v}{d x} = v^{2} + v + 4$ , that is, $x \frac{d v}{d x} = v^{2} + 4$ .
Separating the variables: $\int \frac{d v}{v ^{2} + 4} = \int \frac{d x}{x}$ .
The antiderivative on the left is $\frac{1}{2} arctan \frac{v}{2}$ , and on the right it is $ln x$ (with $x > 0$ from the initial condition): $\frac{1}{2} arctan \frac{v}{2} = ln x + C$ .
With $x = 1$ , $y = 2$ we have $v = 2$ : $\frac{1}{2} arctan 1 = C$ , so $C = \frac{π}{8}$ .
Hence $arctan \frac{v}{2} = 2 ln x + \frac{π}{4} = ln x^{2} + \frac{π}{4}$ , so $v = 2 tan (ln x^{2} + \frac{π}{4})$ .
Undoing the substitution, $y = v x$ .

Answer: $y = 2 x tan (ln x^{2} + \frac{π}{4})$

Exercise 119 HardIB · P3 TZ0 May 2009

The variables $x$ and $y$ are related by $\frac{d y}{d x} - y tan x = cos x$ .

Solve the differential equation, given that when $x = π$ , $y = 0$ . Give the solution in the form $y = f (x)$ .

Show answer

Answer: $y = \frac{1}{2 cos x} (x + \frac{sin 2 x}{2}) - \frac{π}{2 cos x}$

Show working and answer

This is a first-order linear equation: $\frac{d y}{d x} - (tan x) y = cos x$ .
Integrating factor: $e^{- \int t a n x d x} = e^{l n ∣ c o s x ∣} = cos x$ (we take $cos x$ without the absolute value, which is valid up to a constant factor).
Multiplying: $\frac{d}{d x} (y cos x) = cos^{2} x$ .
We use $cos^{2} x = \frac{1 + cos 2 x}{2}$ : $\int cos^{2} x d x = \frac{x}{2} + \frac{sin 2 x}{4} + C$ .
So $y cos x = \frac{x}{2} + \frac{sin 2 x}{4} + C$ .
With $x = π$ , $y = 0$ : $0 = \frac{π}{2} + 0 + C$ , so $C = - \frac{π}{2}$ .
Making $y$ the subject: $y = \frac{1}{cos x} (\frac{x}{2} + \frac{sin 2 x}{4} - \frac{π}{2})$ .

Answer: $y = \frac{1}{2 cos x} (x + \frac{sin 2 x}{2}) - \frac{π}{2 cos x}$

Exercise 120 HardIB · P1 TZ2 May 2010

Throughout this question, $x$ satisfies $0 \leq x < \frac{π}{2}$ .

a) Solve the differential equation $sec^{2} x \frac{d y}{d x} = - y^{2}$ , given that $y = 1$ when $x = 0$ .

Give your answer in the form $y = f (x)$ .

b) i) Prove that $1 \leq sec x \leq 1 + tan x$ .

(ii) Deduce that $\frac{π}{4} \leq \int_{0}^{π /4} sec x d x \leq \frac{π}{4} + \frac{1}{2} ln 2$ .

Show answer

Answer: a) $y = \frac{1}{\frac{1}{2} ( x + \frac{sin 2 x}{2} ) + 1}$ ; b) proved.

Show working and answer

a) 1. Separating the variables: $\frac{d y}{y ^{2}} = - cos^{2} x d x$ (dividing by $sec^{2} x$ , that is, multiplying by $cos^{2} x$ ). 2. Integrating: $- \frac{1}{y} = - (\frac{x}{2} + \frac{sin 2 x}{4}) + C$ , that is, $\frac{1}{y} = \frac{x}{2} + \frac{sin 2 x}{4} + K$ . 3. With $y = 1$ for $x = 0$ : $1 = 0 + 0 + K$ , so $K = 1$ . 4. Hence $y = \frac{1}{\frac{1}{2} ( x + \frac{sin 2 x}{2} ) + 1}$ .

b) (i) 1. On $0 \leq x < \frac{π}{2}$ we have $0 < cos x \leq 1$ , so $sec x = \frac{1}{cos x} \geq 1$ . 2. For the other inequality, as $cos x > 0$ we can multiply by $cos x$ without changing the direction of the inequality: $sec x \leq 1 + tan x ⟺ 1 \leq cos x + sin x$ . 3. Squaring both sides (they are positive): $(cos x + sin x)^{2} = 1 + sin 2 x \geq 1$ , because $sin 2 x \geq 0$ for $0 \leq x < \frac{π}{2}$ . 4. This proves that $1 \leq sec x \leq 1 + tan x$ .

(ii) 1. Integrating the inequality between $0$ and $\frac{π}{4}$ (the interval lies within $[0, \frac{π}{2} [$ ): $\int_{0}^{π /4} 1 d x \leq \int_{0}^{π /4} sec x d x \leq \int_{0}^{π /4} (1 + tan x) d x$ . 2. On the left: $\frac{π}{4}$ . 3. On the right: $\int_{0}^{π /4} (1 + tan x) d x = [x - ln cos x]_{0}^{π /4} = \frac{π}{4} - ln \frac{2}{2} = \frac{π}{4} + \frac{1}{2} ln 2$ . 4. Hence $\frac{π}{4} \leq \int_{0}^{π /4} sec x d x \leq \frac{π}{4} + \frac{1}{2} ln 2$ .

Answer: a) $y = \frac{1}{\frac{1}{2} ( x + \frac{sin 2 x}{2} ) + 1}$ ; b) proved.

Exercise 121 HardIB · P2 TZ2 May 2013

Consider the differential equation $y \frac{d y}{d x} = cos 2 x$ .

a) i) Verify that the function $y = cos x + sin x$ satisfies the differential equation.

(ii) Find the general solution of the differential equation. Express the solution in the form $y = f (x)$ , including a constant of integration in your answer.

(iii) For what value of the constant of integration does your solution coincide with the function given in part i)?

b) Another solution of the differential equation, for which $y = 2$ when $x = \frac{π}{4}$ , defines a curve $C$ .

(i) Determine the equation of $C$ , expressing your answer in the form $y = g (x)$ , and state the range of the function $g$ .

c) A region $R$ of the $x y$ -plane is bounded by $C$ , the $x$ -axis and the vertical lines $x = 0$ and $x = \frac{π}{2}$ .

(ii) Find the area of $R$ .

(iii) Find the volume generated when the part of $R$ that lies above the line $y = 1$ is rotated through $2 π$ radians about the $x$ -axis.

Show answer

Answer: a) (ii) $y = sin 2 x + C$ ; (iii) $C = 1$ ; b) (i) $y = sin 2 x + 3$ , range $[2, 2]$ ; c) (ii) area $= 2.99$ ; (iii) volume $= π^{2} + π = 13.01$

Show working and answer

a) (i) 1. We differentiate: $\frac{d y}{d x} = cos x - sin x$ . 2. Then $y \frac{d y}{d x} = (cos x + sin x) (cos x - sin x) = cos^{2} x - sin^{2} x = cos 2 x$ , which is the right-hand side. So the function satisfies the equation.

(ii) 1. Separate the variables: $y d y = cos 2 x d x$ . 2. Integrate: $\frac{y ^{2}}{2} = \frac{sin 2 x}{2} + K$ , i.e. $y^{2} = sin 2 x + C$ with $C = 2 K$ . 3. Solve for $y$ (we take the positive root, as in the function from part i)): $y = sin 2 x + C$ .

(iii) 1. The function from part i) satisfies $y^{2} = cos^{2} x + 2 sin x cos x + sin^{2} x = 1 + sin 2 x$ . 2. Comparing with $y^{2} = sin 2 x + C$ , we get $C = 1$ .

b) (i) 1. With $y = 2$ for $x = \frac{π}{4}$ : $4 = sin \frac{π}{2} + C = 1 + C$ , so $C = 3$ . 2. The curve is $y = g (x) = sin 2 x + 3$ . 3. Since $- 1 \leq sin 2 x \leq 1$ , we have $2 \leq sin 2 x + 3 \leq 4$ , so the range is $[2, 2]$ .

c) (ii) 1. Since $g (x) > 0$ , the area of $R$ is $\int_{0}^{π /2} sin 2 x + 3 d x$ . 2. Using the GDC we obtain $2.99$ .

(iii) 1. Since $g (x) \geq 2 > 1$ for all $x$ , the part of $R$ above $y = 1$ lies between $y = 1$ and $y = g (x)$ . Rotating about the $x$ -axis generates a volume with a "hole" (washer method): $V = π \int_{0}^{π /2} (g (x)^{2} - 1^{2}) d x$ . 2. $V = π \int_{0}^{π /2} (sin 2 x + 2) d x = π [- \frac{cos 2 x}{2} + 2 x]_{0}^{π /2}$ . 3. $V = π [(\frac{1}{2} + π) - (- \frac{1}{2})] = π (1 + π) = π^{2} + π \approx 13.01$ .

Answer: a) (ii) $y = sin 2 x + C$ ; (iii) $C = 1$ ; b) (i) $y = sin 2 x + 3$ , range $[2, 2]$ ; c) (ii) area $= 2.99$ ; (iii) volume $= π^{2} + π = 13.01$

Exercise 122 HardIB · P3 TZ0 November 2007

a) Verify that the change of variables $X = x - 1$ , $Y = y - 2$ transforms the equation $\frac{d y}{d x} = \frac{x + 3 y - 7}{3 x - y - 1}$ into a homogeneous differential equation.

b) Solve this equation.

Show answer

Answer: $3 arctan (\frac{y - 2}{x - 1}) = ln (x - 1)^{2} + (y - 2)^{2} + C$

Show working and answer

a) 1. With $X = x - 1$ and $Y = y - 2$ we have $x = X + 1$ , $y = Y + 2$ , and $\frac{d y}{d x} = \frac{d Y}{d X}$ (the changes are translations). 2. Numerator: $x + 3 y - 7 = X + 1 + 3 Y + 6 - 7 = X + 3 Y$ . 3. Denominator: $3 x - y - 1 = 3 X + 3 - Y - 2 - 1 = 3 X - Y$ . 4. The equation becomes $\frac{d Y}{d X} = \frac{X + 3 Y}{3 X - Y}$ . Dividing numerator and denominator by $X$ shows that it depends only on $\frac{Y}{X}$ : $\frac{d Y}{d X} = \frac{1 + 3 \frac{Y}{X}}{3 - \frac{Y}{X}}$ . It is homogeneous.

b) 1. Let $Y = v X$ , with $\frac{d Y}{d X} = v + X \frac{d v}{d X}$ : $v + X \frac{d v}{d X} = \frac{1 + 3 v}{3 - v}$ . 2. Then $X \frac{d v}{d X} = \frac{1 + 3 v}{3 - v} - v = \frac{1 + 3 v - 3 v + v ^{2}}{3 - v} = \frac{1 + v ^{2}}{3 - v}$ . 3. Separate the variables: $\int \frac{3 - v}{1 + v ^{2}} d v = \int \frac{d X}{X}$ . 4. We integrate: $3 arctan v - \frac{1}{2} ln (1 + v^{2}) = ln ∣ X ∣ + C$ . 5. Move the logarithm to the right: $3 arctan v = ln ∣ X ∣ + \frac{1}{2} ln (1 + v^{2}) + C = ln (∣ X ∣ 1 + v^{2}) + C$ . 6. Since $∣ X ∣ 1 + v^{2} = X^{2} + Y^{2}$ (because $Y = v X$ ), we get $3 arctan \frac{Y}{X} = ln X^{2} + Y^{2} + C$ . 7. Undo the substitution using $X = x - 1$ and $Y = y - 2$ .

Answer: $3 arctan (\frac{y - 2}{x - 1}) = ln (x - 1)^{2} + (y - 2)^{2} + C$

Exercise 123 HardIB · P3 TZ0 May 2014

a) Consider the functions $f (x) = (ln x)^{2}$ , $x > 1$ and $g (x) = ln (f (x))$ , $x > 1$ .

(i) Find $f^{'} (x)$ .

(ii) Find $g^{'} (x)$ .

(iii) Hence, show that $g (x)$ is increasing on $] 1, \infty [$ .

b) Consider the differential equation $(ln x) \frac{d y}{d x} + \frac{2}{x} y = \frac{2 x - 1}{ln x}$ , $x > 1$ .

(i) Find the general solution of the differential equation in the form $y = h (x)$ .

(ii) Show that the particular solution passing through the point with coordinates $(e, e^{2})$ is given by $y = \frac{x ^{2} - x + e}{( ln x ) ^{2}}$ .

(iii) Sketch the graph of this solution for $x > 1$ , clearly indicating all asymptotes, maxima and minima.

View solution

a) (i) 1. By the chain rule, $f^{'} (x) = 2 ln x \cdot \frac{1}{x} = \frac{2 ln x}{x}$ .

(ii) 1. $g (x) = ln ((ln x)^{2}) = 2 ln (ln x)$ (with $x > 1$ , $ln x > 0$ ). 2. Differentiate: $g^{'} (x) = 2 \cdot \frac{1}{ln x} \cdot \frac{1}{x} = \frac{2}{x ln x}$ .

(iii) 1. For $x > 1$ we have $x > 0$ and $ln x > 0$ , so $g^{'} (x) = \frac{2}{x ln x} > 0$ . 2. Since the derivative is positive on the whole of $] 1, \infty [$ , $g$ is increasing on that interval.

b) (i) 1. Divide by $ln x$ to write the equation in linear form: $\frac{d y}{d x} + \frac{2}{x ln x} y = \frac{2 x - 1}{( ln x ) ^{2}}$ . 2. Integrating factor: $e^{\int \frac{2}{x l n x} d x} = e^{g (x)} = e^{2 l n (l n x)} = (ln x)^{2}$ (from part a), (ii)). 3. Multiply: $\frac{d}{d x} (y (ln x)^{2}) = 2 x - 1$ . 4. Integrate: $y (ln x)^{2} = x^{2} - x + C$ . 5. Solve for $y$ : $y = h (x) = \frac{x ^{2} - x + C}{( ln x ) ^{2}}$ .

(ii) 1. Impose that the curve passes through $(e, e^{2})$ : $e^{2} = \frac{e ^{2} - e + C}{( ln e ) ^{2}} = e^{2} - e + C$ . 2. Hence $C = e$ and the particular solution is $y = \frac{x ^{2} - x + e}{( ln x ) ^{2}}$ .

(iii) 1. As $x \to 1^{+}$ , the numerator tends to $e > 0$ and the denominator to $0^{+}$ : $y \to + \infty$ . There is a vertical asymptote $x = 1$ . 2. As $x \to \infty$ , $y \approx \frac{x ^{2}}{( ln x ) ^{2}} \to \infty$ : there is no horizontal asymptote. 3. Using the GDC, we look for the minimum: it is at $(3.13; 7.21)$ . There are no maxima. 4. The curve falls from $+ \infty$ next to $x = 1$ , reaches this minimum and then increases again.

Graph of y = (x² − x + e)/(ln x)² for x > 1: vertical asymptote x = 1 (the curve falls from above next to it), minimum at the point (3.13; 7.21) and an increasing branch to its right

Answer: a) (i) $f^{'} (x) = \frac{2 ln x}{x}$ ; (ii) $g^{'} (x) = \frac{2}{x ln x}$ ; (iii) $g^{'} (x) > 0$ for $x > 1$ ; b) (i) $y = \frac{x ^{2} - x + C}{( ln x ) ^{2}}$ ; (ii) $C = e$ ; (iii) vertical asymptote $x = 1$ and minimum at $(3.13; 7.21)$ .

Exercise 124 HardIB · P3 TZ0 November 2008

a) Verify that the solution of the differential equation $\frac{d y}{d x} = cos x cos^{2} y$ is $y = arctan (1 + sin x)$ , given that when $x = π$ , $y = \frac{π}{4}$ .

b) Determine the value of the constant $a$ for which the following limit exists, $x \to \frac{π}{2} lim \frac{arctan ( 1 + sin x ) - a}{( x - \frac{π}{2} ) ^{2}}$ , and evaluate this limit.

Show answer

Answer: b) $a = arctan 2$ and the limit equals $- \frac{1}{10}$

Show working and answer

a) 1. Separate the variables: $\frac{d y}{cos ^{2} y} = cos x d x$ , i.e. $sec^{2} y d y = cos x d x$ . 2. Integrate: $tan y = sin x + C$ . 3. With $x = π$ , $y = \frac{π}{4}$ : $1 = 0 + C$ , so $C = 1$ . 4. Hence $tan y = 1 + sin x$ . Since $y = \frac{π}{4}$ lies in $] - \frac{π}{2}, \frac{π}{2} [$ , we take the principal branch of the arctangent: $y = arctan (1 + sin x)$ . 5. Check: $y^{'} = \frac{cos x}{1 + ( 1 + sin x ) ^{2}}$ , and since $cos^{2} y = \frac{1}{1 + tan ^{2} y} = \frac{1}{1 + ( 1 + sin x ) ^{2}}$ , we have $y^{'} = cos x cos^{2} y$ . Also, $y (π) = arctan 1 = \frac{π}{4}$ .

b) 1. As $x \to \frac{π}{2}$ the denominator $(x - \frac{π}{2})^{2}$ tends to $0$ . For the limit to exist (and be finite) the numerator must also tend to $0$ : $arctan (1 + 1) - a = 0$ , so $a = arctan 2$ . 2. With this value we have $\frac{0}{0}$ . We apply L'Hôpital's rule: the derivative of the numerator is $\frac{cos x}{1 + ( 1 + sin x ) ^{2}}$ and that of the denominator is $2 (x - \frac{π}{2})$ . 3. It is still $\frac{0}{0}$ (because $cos \frac{π}{2} = 0$ ). We rewrite the quotient as $\frac{1}{2} \cdot \frac{1}{1 + ( 1 + sin x ) ^{2}} \cdot \frac{cos x}{x - \frac{π}{2}}$ and apply L'Hôpital's rule only to the last fraction: $x \to \frac{π}{2} lim \frac{cos x}{x - \frac{π}{2}} = x \to \frac{π}{2} lim \frac{- sin x}{1} = - 1$ . 4. The central factor tends to $\frac{1}{1 + 4} = \frac{1}{5}$ . The limit is $\frac{1}{2} \cdot \frac{1}{5} \cdot (- 1) = - \frac{1}{10}$ .

Answer: b) $a = arctan 2$ and the limit equals $- \frac{1}{10}$

Exercise 125 HardIB · P3 TZ0 May 2009

Consider the differential equation $\frac{d y}{d x} = \frac{y ^{2} + x ^{2}}{2 x ^{2}}$ , in which, when $x = 1$ , $y = - 1$ .

a) Use Euler's method with a step length of $0.25$ to find an estimate of the value of $y$ for $x = 2$ .

b) i) Solve the differential equation, giving your answer in the form $y = f (x)$ .

(ii) Find the value of $y$ when $x = 2$ .

Show answer

Answer: a) $y (2) \approx - 0.3035$ ; b) (i) $y = \frac{- x + x ln x}{ln x + 1}$ ; (ii) $y (2) \approx - 0.362$

Show working and answer

a) 1. We apply $y_{n + 1} = y_{n} + 0.25 \frac{y _{n}^{2} + x _{n}^{2}}{2 x _{n}^{2}}$ from $(1, - 1)$ . 2. $f (1, - 1) = \frac{1 + 1}{2} = 1$ , so $y_{1} = - 1 + 0.25 = - 0.75$ . 3. $f (1.25; - 0.75) = \frac{0.5625 + 1.5625}{3.125} = 0.68$ , so $y_{2} = - 0.75 + 0.17 = - 0.58$ . 4. $f (1.5; - 0.58) = \frac{0.3364 + 2.25}{4.5} \approx 0.5748$ , so $y_{3} \approx - 0.58 + 0.1437 = - 0.4363$ . 5. $f (1.75; - 0.4363) \approx 0.5313$ , so $y_{4} \approx - 0.4363 + 0.1328 = - 0.3035$ .

$n$	$x_{n}$	$y_{n}$
0	1	-1
1	1.25	-0.75
2	1.5	-0.58
3	1.75	-0.4363
4	2	-0.3035

b) (i) 1. The equation is homogeneous: $\frac{d y}{d x} = \frac{1}{2} ((\frac{y}{x})^{2} + 1)$ . Substitute $y = v x$ : $v + x \frac{d v}{d x} = \frac{v ^{2} + 1}{2}$ . 2. Hence $x \frac{d v}{d x} = \frac{v ^{2} + 1}{2} - v = \frac{( v - 1 ) ^{2}}{2}$ . 3. Separate the variables: $\int \frac{d v}{( v - 1 ) ^{2}} = \int \frac{d x}{2 x}$ , i.e. $- \frac{1}{v - 1} = \frac{1}{2} ln x + C$ . 4. With $x = 1$ , $y = - 1$ ( $v = - 1$ ): $- \frac{1}{- 2} = C$ , so $C = \frac{1}{2}$ . 5. Then $- \frac{1}{v - 1} = \frac{ln x + 1}{2}$ , so $v - 1 = - \frac{2}{ln x + 1}$ and $v = \frac{ln x - 1}{ln x + 1}$ . 6. Undo the substitution, $y = v x$ : $y = \frac{x ln x - x}{ln x + 1}$ .

(ii) 1. For $x = 2$ : $y = \frac{2 ln 2 - 2}{ln 2 + 1} \approx - 0.362$ .

Answer: a) $y (2) \approx - 0.3035$ ; b) (i) $y = \frac{- x + x ln x}{ln x + 1}$ ; (ii) $y (2) \approx - 0.362$

Exercise 126 HardIB · P3 TZ0 November 2013

Consider the differential equation $\frac{d y}{d x} = \frac{y}{x + x y}$ , for $x, y > 0$ .

a) Use Euler's method starting at the point $(x, y) = (1, 2)$ , with an interval $h = 0.2$ , to find an approximate value of $y$ for $x = 1.6$ .

b) Use the substitution $y = v x$ to verify that $x \frac{d v}{d x} = \frac{v}{1 + v} - v$ .

c) i) Hence, find the solution of the differential equation in the form $f (x, y) = 0$ , given that $y = 2$ when $x = 1$ .

(ii) Find the value of $y$ when $x = 1.6$ .

Show answer

Answer: a) $y (1.6) \approx 2.46$ ; c) (i) $2 \frac{x}{y} - ln y - 2 + ln 2 = 0$ ; (ii) $y (1.6) = 2.45$

Show working and answer

a) 1. We apply $y_{n + 1} = y_{n} + 0.2 \frac{y _{n}}{x _{n} + x _{n} y _{n}}$ from $(1, 2)$ , up to $x = 1.6$ (3 steps), using the GDC. 2. $f (1, 2) = \frac{2}{1 + 2} \approx 0.8284$ , so $y (1.2) \approx 2 + 0.1657 = 2.1657$ . 3. $f (1.2; 2.1657) \approx 0.7701$ , so $y (1.4) \approx 2.1657 + 0.1540 = 2.3197$ . 4. $f (1.4; 2.3197) \approx 0.7244$ , so $y (1.6) \approx 2.3197 + 0.1449 = 2.4646$ . 5. Hence $y (1.6) \approx 2.46$ .

b) 1. With $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ and $\frac{y}{x + x y} = \frac{v x}{x + x v} = \frac{v}{1 + v}$ . 2. Equating: $v + x \frac{d v}{d x} = \frac{v}{1 + v}$ . 3. Moving $v$ to the right-hand side: $x \frac{d v}{d x} = \frac{v}{1 + v} - v$ , which is what we were asked to verify.

c) (i) 1. Simplify the right-hand side: $\frac{v}{1 + v} - v = \frac{v - v - v v}{1 + v} = - \frac{v ^{3/2}}{1 + v}$ . 2. Separate the variables: $\int \frac{1 + v}{v ^{3/2}} d v = - \int \frac{d x}{x}$ . 3. Since $\frac{1 + v}{v ^{3/2}} = v^{- 3/2} + v^{- 1}$ , we get $- \frac{2}{v} + ln v = - ln x + C$ . 4. Substitute $v = \frac{y}{x}$ : $- 2 \frac{x}{y} + ln \frac{y}{x} + ln x = C$ , i.e. $- 2 \frac{x}{y} + ln y = C$ . 5. With $x = 1$ , $y = 2$ : $C = - \frac{2}{2} + ln 2 = - 2 + ln 2$ . 6. Moving everything to one side and changing sign: $2 \frac{x}{y} - ln y - 2 + ln 2 = 0$ .

(ii) 1. Substitute $x = 1.6$ and solve the equation in $y$ using the GDC: $2 \frac{1.6}{y} - ln y - 2 + ln 2 = 0$ . 2. The solution is $y \approx 2.45$ .

Answer: a) $y (1.6) \approx 2.46$ ; c) (i) $2 \frac{x}{y} - ln y - 2 + ln 2 = 0$ ; (ii) $y (1.6) = 2.45$

Exercise 127 HardIB · P3 TZ0 May 2018

Consider the differential equation $x \frac{d y}{d x} - y = x^{p} + 1$ , where $x \in R$ , $x \neq = 0$ and $p$ is an integer, $p > 1$ .

a) Solve the differential equation given that $y = - 1$ when $x = 1$ . Give your answer in the form $y = f (x)$ .

b) i) Show that the $x$ -coordinates of the points on the curve $y = f (x)$ where $\frac{d y}{d x} = 0$ satisfy the equation $x^{p - 1} = \frac{1}{p}$ .

(ii) Deduce the set of values of $p$ for which there are two points on the curve $y = f (x)$ where $\frac{d y}{d x} = 0$ . Give a reason to justify your answer.

Show answer

Answer: a) $y = \frac{x ^{p} - x}{p - 1} - 1$ ; b) (ii) $p$ is an odd integer ( $p = 3, 5, 7, \dots$ ), because $x^{p - 1} = \frac{1}{p}$ has two solutions if $p - 1$ is even and only one if $p - 1$ is odd.

Show working and answer

a) 1. Divide by $x$ to put the equation in linear form: $\frac{d y}{d x} - \frac{1}{x} y = x^{p - 1} + \frac{1}{x}$ . 2. Integrating factor: $e^{- \int \frac{d x}{x}} = \frac{1}{x}$ (we take $x > 0$ , which contains $x = 1$ ). 3. $\frac{d}{d x} (\frac{y}{x}) = x^{p - 2} + x^{- 2}$ . 4. We integrate (with $p > 1$ ): $\frac{y}{x} = \frac{x ^{p - 1}}{p - 1} - \frac{1}{x} + C$ , and multiplying by $x$ : $y = \frac{x ^{p}}{p - 1} - 1 + C x$ . 5. With $y (1) = - 1$ : $- 1 = \frac{1}{p - 1} - 1 + C$ , so $C = - \frac{1}{p - 1}$ . 6. $y = \frac{x ^{p} - x}{p - 1} - 1$ .

b) (i) 1. Differentiate: $\frac{d y}{d x} = \frac{p x ^{p - 1} - 1}{p - 1}$ . 2. Set equal to $0$ : $p x^{p - 1} - 1 = 0$ , i.e. $x^{p - 1} = \frac{1}{p}$ .

(ii) 1. The equation $x^{p - 1} = \frac{1}{p}$ , with $\frac{1}{p} > 0$ , has two real solutions ( $x = \pm (\frac{1}{p})^{\frac{1}{p - 1}}$ ) when the exponent $p - 1$ is even, and only one positive solution when $p - 1$ is odd (an odd power is one-to-one). 2. $p - 1$ is even if and only if $p$ is odd. 3. Therefore there are two points on the curve with a horizontal tangent when $p$ is an odd integer greater than $1$ (i.e. $p = 3, 5, 7, \dots$ ). For even $p$ there is only one.

Exercise 128 HardIB · P3 TZ0 May 2017

a) Consider the differential equation $\frac{d y}{d x} = f (\frac{y}{x})$ , $x > 0$ .

Use the substitution $y = v x$ to show that the general solution of this differential equation is $\int \frac{d v}{f ( v ) - v} = ln x + Constant$ .

b) Hence, or otherwise, solve the differential equation $\frac{d y}{d x} = \frac{x ^{2} + 3 x y + y ^{2}}{x ^{2}}$ , $x > 0$ , given that $y = 1$ when $x = 1$ . Give your answer in the form $y = g (x)$ .

Show answer

Answer: b) $y = \frac{- x}{ln x - \frac{1}{2}} - x$

Show working and answer

a) 1. With $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . 2. The equation $\frac{d y}{d x} = f (\frac{y}{x})$ becomes $v + x \frac{d v}{d x} = f (v)$ , i.e. $x \frac{d v}{d x} = f (v) - v$ . 3. Separate the variables: $\frac{d v}{f ( v ) - v} = \frac{d x}{x}$ . 4. Integrate (with $x > 0$ ): $\int \frac{d v}{f ( v ) - v} = ln x + Constant$ .

b) 1. We write $\frac{x ^{2} + 3 x y + y ^{2}}{x ^{2}} = 1 + 3 \frac{y}{x} + (\frac{y}{x})^{2}$ , so $f (v) = 1 + 3 v + v^{2}$ . 2. Hence $f (v) - v = v^{2} + 2 v + 1 = (v + 1)^{2}$ . 3. We apply the result from a): $\int \frac{d v}{( v + 1 ) ^{2}} = - \frac{1}{v + 1} = ln x + C$ . 4. With $x = 1$ , $y = 1$ ( $v = 1$ ): $- \frac{1}{2} = C$ . 5. Then $v + 1 = - \frac{1}{ln x - \frac{1}{2}}$ , and $v = - \frac{1}{ln x - \frac{1}{2}} - 1$ . 6. Multiply by $x$ : $y = v x = \frac{- x}{ln x - \frac{1}{2}} - x$ .

Answer: b) $y = \frac{- x}{ln x - \frac{1}{2}} - x$

Exercise 129 HardIB · P3 TZ0 May 2016

Consider the differential equation $\frac{d y}{d x} = \frac{x}{y} - x y$ where $y > 0$ and $y = 2$ when $x = 0$ .

a) Show that the substitution $z = y^{2}$ transforms the differential equation into $\frac{d z}{d x} + 2 x z = 2 x$ .

b) Solve this differential equation in $z$ and obtain an expression for $y$ in terms of $x$ .

Show answer

Answer: b) $y = 1 + \frac{3}{e ^{x^{2}}}$

Show working and answer

a) 1. With $z = y^{2}$ , $\frac{d z}{d x} = 2 y \frac{d y}{d x}$ . 2. Multiply the equation by $2 y$ : $2 y \frac{d y}{d x} = 2 x - 2 x y^{2}$ . 3. That is, $\frac{d z}{d x} = 2 x - 2 x z$ , and moving $2 x z$ to the left: $\frac{d z}{d x} + 2 x z = 2 x$ .

b) 1. This is a linear equation in $z$ . Integrating factor: $e^{\int 2 x d x} = e^{x^{2}}$ . 2. $\frac{d}{d x} (z e^{x^{2}}) = 2 x e^{x^{2}}$ , so $z e^{x^{2}} = e^{x^{2}} + C$ . 3. With $y (0) = 2$ we have $z (0) = 4$ : $4 = 1 + C$ , so $C = 3$ . 4. $z = 1 + 3 e^{- x^{2}}$ . 5. Since $y > 0$ , $y = z = 1 + \frac{3}{e ^{x^{2}}}$ .

Answer: b) $y = 1 + \frac{3}{e ^{x^{2}}}$

Exercise 130 HardIB · P2 November 2003

(i) Verify, by substituting $y = v x$ , that the general solution of the differential equation $(x^{2} + y^{2}) + 2 x y \frac{d y}{d x} = 0$ , $x > 0$ is $x^{3} + 3 x y^{2} = k$ , where $k$ is a constant.

(ii) The equation of a curve is $f (x) = \frac{a}{b + e ^{- c x}}$ , $a \neq = 0$ , $b > 0$ , $c > 0$ .

a) Verify that $f^{''} (x) = \frac{a c ^{2} e ^{- c x} ( e ^{- c x} - b )}{( b + e ^{- c x} ) ^{3}}$ .

b) Find the coordinates of the point on the curve where $f^{''} (x) = 0$ .

c) Verify that the point in part (b) is a point of inflexion.

Show answer

Answer: i) proved; ii) a) proved; b) $(- \frac{ln b}{c}, \frac{a}{2 b})$ ; c) proved.

Show working and answer

(i) 1. With $y = v x$ we have $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . Substituting: $x^{2} (1 + v^{2}) + 2 x \cdot v x (v + x \frac{d v}{d x}) = 0$ . 2. Dividing by $x^{2}$ ( $x > 0$ ): $1 + v^{2} + 2 v^{2} + 2 v x \frac{d v}{d x} = 0$ , that is, $2 v x \frac{d v}{d x} = - (1 + 3 v^{2})$ . 3. Separating variables: $\frac{2 v}{1 + 3 v ^{2}} d v = - \frac{d x}{x}$ . 4. Integrating: $\frac{1}{3} ln (1 + 3 v^{2}) = - ln x + C_{1}$ , so $ln (1 + 3 v^{2}) = - 3 ln x + 3 C_{1}$ and $x^{3} (1 + 3 v^{2}) = K$ with $K = e^{3 C_{1}}$ . 5. Since $v = \frac{y}{x}$ : $x^{3} (1 + \frac{3 y ^{2}}{x ^{2}}) = x^{3} + 3 x y^{2} = k$ .

(ii) a) 1. We write $f (x) = a (b + e^{- c x})^{- 1}$ and differentiate using the chain rule: $f^{'} (x) = a c e^{- c x} (b + e^{- c x})^{- 2}$ . 2. We differentiate again (product rule): $f^{''} (x) = a c [- c e^{- c x} (b + e^{- c x})^{- 2} + e^{- c x} \cdot (- 2) (b + e^{- c x})^{- 3} \cdot (- c e^{- c x})]$ . 3. Taking out the common factor $a c^{2} e^{- c x} (b + e^{- c x})^{- 3}$ , what remains is $[- (b + e^{- c x}) + 2 e^{- c x}] = e^{- c x} - b$ . 4. Therefore $f^{''} (x) = \frac{a c ^{2} e ^{- c x} ( e ^{- c x} - b )}{( b + e ^{- c x} ) ^{3}}$ .

b) 1. As $a \neq = 0$ , $c > 0$ and $e^{- c x} > 0$ , $f^{''} (x) = 0$ requires $e^{- c x} = b$ . 2. Then $- c x = ln b$ , that is, $x = - \frac{ln b}{c}$ . 3. The ordinate: $f = \frac{a}{b + b} = \frac{a}{2 b}$ . 4. The point is $(- \frac{ln b}{c}, \frac{a}{2 b})$ .

c) 1. In $f^{''}$ the only factor that can change sign is $e^{- c x} - b$ , since $a c^{2} e^{- c x}$ has the sign of $a$ and the denominator is positive. 2. $e^{- c x}$ is strictly decreasing: for $x < - \frac{ln b}{c}$ it is greater than $b$ , and for $x > - \frac{ln b}{c}$ it is less than $b$ . 3. So $f^{''}$ changes sign at that point, which is therefore a point of inflexion.

Answer: i) proved; ii) a) proved; b) $(- \frac{ln b}{c}, \frac{a}{2 b})$ ; c) proved.

Exercise 131 HardIB · P2 May 2002

The function $y = f (x)$ satisfies the differential equation $2 x^{2} \frac{d y}{d x} = x^{2} + y^{2}$ $(x > 0)$ .

a) i) By substituting $y = v x$ , show that $2 x \frac{d v}{d x} = (v - 1)^{2}$ .

(ii) Hence show that the solution of the original differential equation is $y = x - \frac{2 x}{( ln x + c )}$ , where $c$ is an arbitrary constant.

(iii) Find the value of $c$ , given that $y = 2$ when $x = 1$ .

b) Below is the graph of $y = f (x)$ . The graph cuts the $x$ -axis at $A$ .

Graph of y = f(x): one branch leaves the origin and rises towards a vertical asymptote; the region under that branch between x = 1 and x = 5 is shaded; the other branch, to the right of the asymptote, comes from below and cuts the x-axis at point A (schematic drawing, not to scale on the x-axis)

(i) Write down the equation of the vertical asymptote.

(ii) Find the exact value of the $x$ -coordinate of the point $A$ .

(iii) Find the area of the shaded region.

Show answer

Answer: a) iii) $c = - 2$ ; b) i) $x = e^{2}$ ; ii) $x = e^{4}$ ; iii) $46.7$

Show working and answer

a) i) 1. With $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . 2. Substituting into $2 x^{2} \frac{d y}{d x} = x^{2} + y^{2}$ : $2 x^{2} (v + x \frac{d v}{d x}) = x^{2} + v^{2} x^{2}$ . 3. Dividing by $x^{2}$ : $2 v + 2 x \frac{d v}{d x} = 1 + v^{2}$ . 4. Hence $2 x \frac{d v}{d x} = v^{2} - 2 v + 1 = (v - 1)^{2}$ .

(ii) 1. Separating variables: $\frac{2 d v}{( v - 1 ) ^{2}} = \frac{d x}{x}$ . 2. Integrating: $- \frac{2}{v - 1} = ln x + c$ ( $x > 0$ ). 3. Rearranging: $v - 1 = - \frac{2}{ln x + c}$ , that is, $v = 1 - \frac{2}{ln x + c}$ . 4. Since $y = v x$ : $y = x - \frac{2 x}{ln x + c}$ .

(iii) 1. With $y = 2$ at $x = 1$ : $2 = 1 - \frac{2}{0 + c}$ . 2. Then $- \frac{2}{c} = 1$ and $c = - 2$ .

b) i) 1. With $c = - 2$ , $y = x - \frac{2 x}{ln x - 2}$ is undefined when $ln x = 2$ . 2. The vertical asymptote is $x = e^{2}$ .

(ii) 1. Point $A$ is where $y = 0$ on the right-hand branch: $x (1 - \frac{2}{ln x - 2}) = 0$ , and since $x > 0$ , $ln x - 2 = 2$ . 2. So $ln x = 4$ , and therefore $x = e^{4}$ .

(iii) 1. The shaded region lies under the left-hand branch, between $x = 1$ and $x = 5$ ( $5 < e^{2} \approx 7.39$ , so it does not cross the asymptote). 2. Using the GDC: $\int_{1}^{5} (x - \frac{2 x}{ln x - 2}) d x \approx 46.7$ .

Answer: a) iii) $c = - 2$ ; b) i) $x = e^{2}$ ; ii) $x = e^{4}$ ; iii) $46.7$

Exercise 132 HardIB · P2 TZ0 November 2004

Consider the differential equation $\frac{d y}{d θ} = \frac{y}{e ^{2 θ} + 1}$ .

a) Use the substitution $x = e^{θ}$ to verify that $\int \frac{d y}{y} = \int \frac{d x}{x ( x ^{2} + 1 )}$ .

b) Find $\int \frac{d x}{x ( x ^{2} + 1 )}$ .

c) Hence, find $y$ in terms of $θ$ , given that $y = 2$ when $θ = 0$ .

Show answer

Answer: b) $ln x - \frac{1}{2} ln (x^{2} + 1) + C$ ; c) $y = \frac{2 e ^{θ}}{e ^{2 θ} + 1}$

Show working and answer

a) 1. Separating variables: $\frac{d y}{y} = \frac{d θ}{e ^{2 θ} + 1}$ . 2. Substitution $x = e^{θ}$ : $d x = e^{θ} d θ = x d θ$ , so $d θ = \frac{d x}{x}$ , and $e^{2 θ} = x^{2}$ . 3. Therefore $\int \frac{d y}{y} = \int \frac{1}{x ^{2} + 1} \cdot \frac{d x}{x} = \int \frac{d x}{x ( x ^{2} + 1 )}$ .

b) 1. Partial fractions: $\frac{1}{x ( x ^{2} + 1 )} = \frac{A}{x} + \frac{B x + C}{x ^{2} + 1}$ , with $1 = A (x^{2} + 1) + (B x + C) x$ . 2. Equating coefficients: $A = 1$ , $B = - 1$ , $C = 0$ . 3. $\int \frac{d x}{x ( x ^{2} + 1 )} = \int (\frac{1}{x} - \frac{x}{x ^{2} + 1}) d x = ln x - \frac{1}{2} ln (x^{2} + 1) + C$ (with $x = e^{θ} > 0$ ).

c) 1. By (a) and (b): $ln y = ln x - \frac{1}{2} ln (x^{2} + 1) + C$ , with $y > 0$ from the initial condition. 2. Returning to $θ$ : $ln y = θ - \frac{1}{2} ln (e^{2 θ} + 1) + C$ . 3. With $y = 2$ and $θ = 0$ : $\frac{1}{2} ln 2 = 0 - \frac{1}{2} ln 2 + C$ , so $C = ln 2$ . 4. $ln y = ln \frac{2 e ^{θ}}{e ^{2 θ} + 1}$ , so $y = \frac{2 e ^{θ}}{e ^{2 θ} + 1}$ .

Answer: b) $ln x - \frac{1}{2} ln (x^{2} + 1) + C$ ; c) $y = \frac{2 e ^{θ}}{e ^{2 θ} + 1}$

Exercise 133 HardIB · P2 TZ0 November 2006

Let $y = cos θ + i sin θ$ .

a) Verify that $\frac{d y}{d θ} = i y$ .

[You may assume, when differentiating and integrating, that $i$ can be treated in the same way as a real constant.]

b) Hence, and using integration, verify that $y = e^{i θ}$ .

c) Use this result to deduce De Moivre's theorem.

d) i) Given that $\frac{sin 6 θ}{sin θ} = a cos^{5} θ + b cos^{3} θ + c cos θ$ , where $sin θ \neq = 0$ , use De Moivre's theorem, with $n = 6$ , to find the values of the constants $a$ , $b$ and $c$ .

(ii) Hence, deduce the value of $θ \to 0 lim \frac{sin 6 θ}{sin θ}$ .

Show answer

Answer: a) shown; b) shown; c) shown; d) i) $a = 32$ , $b = - 32$ , $c = 6$ ; ii) $6$

Show working and answer

a) 1. We differentiate: $\frac{d y}{d θ} = - sin θ + i cos θ$ . 2. Since $i \cdot i = - 1$ , $i y = i cos θ + i^{2} sin θ = i cos θ - sin θ$ . 3. They are equal: $\frac{d y}{d θ} = i y$ .

b) 1. Separate the variables: $\frac{d y}{y} = i d θ$ . 2. Integrate: $ln y = i θ + C$ , i.e. $y = A e^{i θ}$ . 3. With $θ = 0$ : $y = cos 0 + i sin 0 = 1$ , so $A = 1$ . 4. Therefore $cos θ + i sin θ = e^{i θ}$ .

c) 1. Raise to the power $n$ : $(cos θ + i sin θ)^{n} = (e^{i θ})^{n} = e^{i n θ}$ . 2. Applying (b) with $n θ$ : $e^{i n θ} = cos n θ + i sin n θ$ . 3. Thus $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ .

d) i) 1. With $n = 6$ , $cos 6 θ + i sin 6 θ = (c + i s)^{6}$ , with $c = cos θ$ and $s = sin θ$ . 2. Imaginary part of the binomial expansion: $sin 6 θ = 6 c^{5} s - 20 c^{3} s^{3} + 6 c s^{5}$ . 3. Divide by $s \neq = 0$ : $\frac{sin 6 θ}{s} = 6 c^{5} - 20 c^{3} s^{2} + 6 c s^{4}$ . 4. With $s^{2} = 1 - c^{2}$ : $- 20 c^{3} (1 - c^{2}) = - 20 c^{3} + 20 c^{5}$ and $6 c (1 - c^{2})^{2} = 6 c - 12 c^{3} + 6 c^{5}$ . 5. Adding: $32 c^{5} - 32 c^{3} + 6 c$ . 6. $a = 32$ , $b = - 32$ , $c = 6$ .

(ii) 1. $θ \to 0 lim \frac{sin 6 θ}{sin θ} = θ \to 0 lim (32 cos^{5} θ - 32 cos^{3} θ + 6 cos θ)$ . 2. As $cos θ \to 1$ : $32 - 32 + 6 = 6$ .

Answer: a) shown; b) shown; c) shown; d) i) $a = 32$ , $b = - 32$ , $c = 6$ ; ii) $6$

Exercise 134 HardIB · P3 TZ1 May 2007

a) Use integration by parts to show that $\int sin x cos x e^{- s i n x} d x = - e^{- s i n x} (1 + sin x) + C$ .

Consider the differential equation $\frac{d y}{d x} - y cos x = sin x cos x$ .

b) Find an integrating factor for this differential equation.

c) Solve the differential equation, given that $y = - 2$ when $x = 0$ . Give your answer in the form $y = f (x)$ .

Show answer

Answer: b) $e^{- s i n x}$ ; c) $y = - 1 - sin x - e^{s i n x}$

Show working and answer

a) 1. Substitution $w = sin x$ , $d w = cos x d x$ : $\int sin x cos x e^{- s i n x} d x = \int w e^{- w} d w$ . 2. By parts with $u = w$ and $d v = e^{- w} d w$ ( $d u = d w$ , $v = - e^{- w}$ ): $- w e^{- w} + \int e^{- w} d w = - w e^{- w} - e^{- w} + C$ . 3. Undoing the substitution: $- e^{- s i n x} (1 + sin x) + C$ .

b) 1. The equation $\frac{d y}{d x} - y cos x = sin x cos x$ is linear with $P (x) = - cos x$ . 2. Integrating factor: $e^{\int - c o s x d x} = e^{- s i n x}$ .

c) 1. Multiply by the integrating factor: $\frac{d}{d x} (y e^{- s i n x}) = sin x cos x e^{- s i n x}$ . 2. Integrating with (a): $y e^{- s i n x} = - e^{- s i n x} (1 + sin x) + C$ . 3. Rearranging: $y = - (1 + sin x) + C e^{s i n x}$ . 4. With $y (0) = - 2$ : $- 1 + C = - 2$ , so $C = - 1$ . 5. $y = - 1 - sin x - e^{s i n x}$ .

Answer: b) $e^{- s i n x}$ ; c) $y = - 1 - sin x - e^{s i n x}$

Exercise 135 HardIB · P2 TZ1 May 2009

a) Solve the differential equation $\frac{cos ^{2} x}{e ^{e^{y}}} - e^{y} \frac{d y}{d x} = 0$ , given that $y = 0$ when $x = π$ .

b) Calculate the value of $y$ when $x = \frac{π}{2}$ .

Show answer

Answer: a) $y = ln ln [\frac{1}{2} (x + \frac{sin 2 x}{2}) + e - \frac{π}{2}]$ ; b) $y = ln ln (e - \frac{π}{4})$

Show working and answer

a) 1. The equation, in separable form, is $e^{y} e^{e^{y}} \frac{d y}{d x} = cos^{2} x$ (which is equivalent to $\frac{cos ^{2} x}{e ^{e^{y}}} - e^{y} \frac{d y}{d x} = 0$ ). 2. Separate: $e^{y} e^{e^{y}} d y = cos^{2} x d x$ . 3. Antiderivative of the left side with $w = e^{y}$ , $d w = e^{y} d y$ : $\int e^{w} d w = e^{e^{y}}$ . 4. Antiderivative of the right side: $\int cos^{2} x d x = \frac{1}{2} (x + \frac{sin 2 x}{2})$ . 5. So $e^{e^{y}} = \frac{1}{2} (x + \frac{sin 2 x}{2}) + C$ . 6. With $y = 0$ when $x = π$ : $e = \frac{π}{2} + C$ , so $C = e - \frac{π}{2}$ . 7. Taking logarithms twice: $y = ln ln [\frac{1}{2} (x + \frac{sin 2 x}{2}) + e - \frac{π}{2}]$ .

b) 1. When $x = \frac{π}{2}$ : $sin π = 0$ , so the bracket equals $\frac{π}{4} + e - \frac{π}{2} = e - \frac{π}{4}$ . 2. $y = ln ln (e - \frac{π}{4})$ ( $\approx - 0.08$ ).

Answer: a) $y = ln ln [\frac{1}{2} (x + \frac{sin 2 x}{2}) + e - \frac{π}{2}]$ ; b) $y = ln ln (e - \frac{π}{4})$

Exercise 136 HardIB · P3 TZ1 May 2008

Consider the differential equation $x \frac{d y}{d x} - 2 y = \frac{x ^{3}}{x ^{2} + 1}$ .

a) Calculate the integrating factor for this differential equation.

b) Solve the differential equation given that $y = 1$ when $x = 1$ , giving your answer in the form $y = f (x)$ .

Show answer

Answer: a) $\frac{1}{x ^{2}}$ ; b) $y = (arctan x + 1 - \frac{π}{4}) x^{2}$

Show working and answer

a) 1. Divide by $x$ to put it in linear form: $\frac{d y}{d x} - \frac{2}{x} y = \frac{x ^{2}}{x ^{2} + 1}$ . 2. Integrating factor: $e^{\int - \frac{2}{x} d x} = e^{- 2 l n x} = \frac{1}{x ^{2}}$ (with $x > 0$ ).

b) 1. Multiply by $\frac{1}{x ^{2}}$ : $\frac{d}{d x} (\frac{y}{x ^{2}}) = \frac{1}{x ^{2} + 1}$ . 2. Integrate: $\frac{y}{x ^{2}} = arctan x + C$ . 3. With $y = 1$ when $x = 1$ : $1 = \frac{π}{4} + C$ , so $C = 1 - \frac{π}{4}$ . 4. $y = x^{2} (arctan x + 1 - \frac{π}{4})$ .

Answer: a) $\frac{1}{x ^{2}}$ ; b) $y = (arctan x + 1 - \frac{π}{4}) x^{2}$

Exercise 137 HardIB · P1 TZ0 May 2009

Let $f$ be a function with domain $R$ that satisfies $f (x + y) = f (x) f (y)$ for all $x$ and $y$ , and $f (0) \neq = 0$ .

a) Show that $f (0) = 1$ .

b) Prove that $f (x) \neq = 0$ for all $x \in R$ .

c) Assuming that $f^{'} (x)$ exists for all $x \in R$ , use the definition of the derivative to show that $f (x)$ satisfies the differential equation $f^{'} (x) = k f (x)$ , where $k = f^{'} (0)$ .

d) Solve the differential equation to find an expression for $f (x)$ .

Show answer

Answer: a) shown; b) shown; c) shown; d) $f (x) = e^{k x}$

Show working and answer

a) 1. With $y = 0$ : $f (x) = f (x) f (0)$ for all $x$ . 2. In particular, $x = 0$ gives $f (0) = f (0)^{2}$ , i.e. $f (0) (f (0) - 1) = 0$ . 3. As $f (0) \neq = 0$ , $f (0) = 1$ .

b) 1. With $y = - x$ : $f (0) = f (x) f (- x)$ , i.e. $f (x) f (- x) = 1$ . 2. A product equal to $1$ cannot have a zero factor, so $f (x) \neq = 0$ for all $x \in R$ .

c) 1. By definition, $f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h} = h \to 0 lim \frac{f ( x ) f ( h ) - f ( x )}{h}$ . 2. Factor out $f (x)$ and use $f (0) = 1$ : $f^{'} (x) = f (x) h \to 0 lim \frac{f ( h ) - f ( 0 )}{h} = f (x) f^{'} (0)$ . 3. With $k = f^{'} (0)$ : $f^{'} (x) = k f (x)$ .

d) 1. Also $f (x) = f (\frac{x}{2})^{2} > 0$ (it is a non-zero square), so $f > 0$ . 2. Separate: $\frac{f ^{'}}{f} = k$ , and integrate: $ln f (x) = k x + C$ . 3. With $f (0) = 1$ : $C = 0$ . 4. $f (x) = e^{k x}$ .

Answer: a) shown; b) shown; c) shown; d) $f (x) = e^{k x}$

Exercise 138 HardIB · P3 TZ0 November 2014

a) Use an integrating factor to show that the general solution of $\frac{d x}{d t} - \frac{x}{t} = - \frac{2}{t}$ , $t > 0$ is $x = 2 + c t$ , where $c$ is a constant.

The weight in kilograms of a dog, $t$ weeks after it was bought from a pet shop, can be modelled by the following function: $w (t) = ⎩ ⎨ ⎧ 2 + c t 16 - \frac{35}{t} 0 \leq t \leq 5 t > 5$ .

b) Given that $w (t)$ is continuous, find the value of $c$ .

c) Write down

(i) the weight of the dog at the moment it was bought from the pet shop;

(ii) an upper bound for the weight of the dog.

d) Prove, from the definition of the derivative, that $w (t)$ is differentiable at $t = 5$ .

Show answer

Answer: a) integrating factor $\frac{1}{t}$ ; b) $c = \frac{7}{5}$ ; c) (i) $2$ kg; (ii) $16$ kg; d) $w^{'} (5) = \frac{7}{5}$

Show working and answer

a) 1. This is a linear equation $\frac{d x}{d t} + P (t) x = Q (t)$ with $P (t) = - \frac{1}{t}$ . The integrating factor is $e^{\int P d t} = e^{- l n t} = \frac{1}{t}$ (with $t > 0$ ). 2. We multiply the equation by $\frac{1}{t}$ : $\frac{1}{t} \frac{d x}{d t} - \frac{x}{t ^{2}} = - \frac{2}{t ^{2}}$ , and the left-hand side is $\frac{d}{d t} (\frac{x}{t})$ . 3. Integrating: $\frac{x}{t} = \int - \frac{2}{t ^{2}} d t = \frac{2}{t} + c$ . 4. Multiplying by $t$ : $x = 2 + c t$ .

b) 1. Continuity at $t = 5$ : the two pieces must give the same value. 2. From the left, $2 + 5 c$ ; from the right, $16 - \frac{35}{5} = 9$ . 3. $2 + 5 c = 9$ , so $c = \frac{7}{5}$ .

c) (i) 1. On purchase, $t = 0$ : $w (0) = 2 + 0 = 2$ kg.

(ii) 1. For $0 \leq t \leq 5$ the weight increases to $9$ kg. For $t > 5$ , $w = 16 - \frac{35}{t}$ is increasing and tends to $16$ as $t \to \infty$ without reaching it. 2. An upper bound for the weight is $16$ kg.

d) 1. Differentiability at $t = 5$ : the one-sided limits of the difference quotient must exist and coincide. We know $w (5) = 9$ . 2. From the left ( $h < 0$ , first piece): $\frac{w ( 5 + h ) - w ( 5 )}{h} = \frac{2 + \frac{7}{5} ( 5 + h ) - 9}{h} = \frac{\frac{7}{5} h}{h} = \frac{7}{5}$ , and the limit is $\frac{7}{5}$ . 3. From the right ( $h > 0$ , second piece): $\frac{16 - \frac{35}{5 + h} - 9}{h} = \frac{7 - \frac{35}{5 + h}}{h} = \frac{7 ( 5 + h ) - 35}{h ( 5 + h )} = \frac{7}{5 + h} \to \frac{7}{5}$ as $h \to 0^{+}$ . 4. Both one-sided limits equal $\frac{7}{5}$ , so $w$ is differentiable at $t = 5$ and $w^{'} (5) = \frac{7}{5}$ .

Answer: a) integrating factor $\frac{1}{t}$ ; b) $c = \frac{7}{5}$ ; c) (i) $2$ kg; (ii) $16$ kg; d) $w^{'} (5) = \frac{7}{5}$

Exercise 139 HardIB · P1 TZ2 May 2021

The acceleration ( $a$ $m s^{- 2}$ ) of a particle moving along a horizontal straight line at time $t$ seconds, $t \geq 0$ , is given by $a = - (1 + v)$ , where $v$ $m s^{- 1}$ is the velocity of the particle, and $v > - 1$ . At $t = 0$ , the particle is at a fixed origin $O$ and has an initial velocity of $v_{0}$ $m s^{- 1}$ .

a) By solving a suitable differential equation, show that the velocity of the particle at time $t$ is given by $v (t) = (1 + v_{0}) e^{- t} - 1$ .

b) The particle, which is initially at $O$ , moves in the positive direction until it reaches its maximum displacement from $O$ . The particle then returns to $O$ . Let $s$ (in metres) be the displacement of the particle from $O$ , and let $S_{max}$ be its maximum displacement from $O$ .

(i) Show that the time $T$ taken by the particle to reach $S_{max}$ satisfies the equation $e^{T} = 1 + v_{0}$ .

(ii) By solving a suitable differential equation and using the result obtained in part (b)(i), find an expression for $S_{max}$ in terms of $v_{0}$ .

The expression $v (T - k)$ represents the velocity of the particle $k$ seconds before it reaches $S_{max}$ , where $v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$ .

c) Using the result obtained in part (b)(i), show that $v (T - k) = e^{k} - 1$ .

Similarly, the expression $v (T + k)$ represents the velocity of the particle $k$ seconds after it has reached $S_{max}$ .

d) Deduce a similar expression that gives $v (T + k)$ in terms of $k$ .

e) Hence, show that $v (T - k) + v (T + k) \geq 0$ .

Show answer

Answer: b) (ii) $S_{max} = v_{0} - ln (1 + v_{0})$ ; d) $v (T + k) = e^{- k} - 1$ ; parts a), b)(i), c) and e) are shown.

Show working and answer

a) 1. With $a = \frac{d v}{d t}$ , the equation is $\frac{d v}{d t} = - (1 + v)$ , which has separable variables ( $v > - 1$ ). 2. $\int \frac{d v}{1 + v} = - \int d t$ , so $ln (1 + v) = - t + C$ . 3. At $t = 0$ , $v = v_{0}$ : $C = ln (1 + v_{0})$ . 4. $1 + v = (1 + v_{0}) e^{- t}$ , that is, $v (t) = (1 + v_{0}) e^{- t} - 1$ .

b) (i) 1. The maximum displacement occurs when the particle stops, $v (T) = 0$ . 2. $(1 + v_{0}) e^{- T} - 1 = 0$ , so $e^{T} = 1 + v_{0}$ .

(ii) 1. Since $\frac{d s}{d t} = v$ and $s (0) = 0$ : $S_{max} = s (T) = \int_{0}^{T} [(1 + v_{0}) e^{- t} - 1] d t = [- (1 + v_{0}) e^{- t} - t]_{0}^{T}$ . 2. $= - (1 + v_{0}) e^{- T} - T + (1 + v_{0})$ . Using $e^{- T} = \frac{1}{1 + v _{0}}$ and $T = ln (1 + v_{0})$ : $S_{max} = - 1 - ln (1 + v_{0}) + 1 + v_{0}$ . 3. $S_{max} = v_{0} - ln (1 + v_{0})$ .

c) 1. $v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1 = (1 + v_{0}) e^{- T} e^{k} - 1$ . 2. Since $(1 + v_{0}) e^{- T} = 1$ : $v (T - k) = e^{k} - 1$ .

d) 1. Similarly: $v (T + k) = (1 + v_{0}) e^{- T} e^{- k} - 1 = e^{- k} - 1$ .

e) 1. $v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ . 2. This equals $(e^{k /2} - e^{- k /2})^{2} \geq 0$ , since it is a square.

Answer: b) (ii) $S_{max} = v_{0} - ln (1 + v_{0})$ ; d) $v (T + k) = e^{- k} - 1$ ; parts a), b)(i), c) and e) are shown.

Exercise 140 HardIB · P2 TZ1 May 2021

The function $f$ has derivative $f^{'} (x) = \frac{1}{x ( k - x )}$ , $x \in R$ , $x \neq = 0$ , $x \neq = k$ , where $k$ is a positive constant.

a) The expression for $f^{'} (x)$ can be written in the form $\frac{a}{x} + \frac{b}{k - x}$ , where $a, b \in R$ .

Find $a$ and $b$ in terms of $k$ .

b) Using the above, find an expression for $f (x)$ .

Consider $P$ , the population of an ant colony whose initial value is $1200$ .

The rate of change of the population is modelled by the differential equation $\frac{d P}{d t} = \frac{P ( k - P )}{5 k}$ , where $t$ is the time measured in days, $t \geq 0$ , and $k$ is the upper limit of the population.

c) By solving the differential equation, show that $P = \frac{1200 k}{( k - 1200 ) e ^{- \frac{t}{5}} + 1200}$ .

For $t = 10$ the initial population of the colony has doubled.

d) Find the value of $k$ , giving your answer correct to 4 significant figures.

e) Find the value of $t$ when the rate of change of the population is at its maximum.

Show answer

Answer: a) $a = b = \frac{1}{k}$ ; b) $f (x) = \frac{1}{k} ln \frac{x}{k - x} + C$ ; c) shown; d) $k = 2845$ ; e) $t = 1.58$ days

Show working and answer

a) 1. $\frac{1}{x ( k - x )} = \frac{a}{x} + \frac{b}{k - x}$ implies $1 = a (k - x) + b x$ . 2. Setting $x = 0$ : $1 = ak$ , so $a = \frac{1}{k}$ . Setting $x = k$ : $1 = bk$ , so $b = \frac{1}{k}$ .

b) 1. Technique: partial fractions, from the previous part. $f (x) = \int (\frac{1}{k x} + \frac{1}{k ( k - x )}) d x = \frac{1}{k} ln ∣ x ∣ - \frac{1}{k} ln ∣ k - x ∣ + C$ . 2. $f (x) = \frac{1}{k} ln \frac{x}{k - x} + C$ .

c) 1. We separate the variables: $\int \frac{d P}{P ( k - P )} = \int \frac{d t}{5 k}$ . 2. From part (b), $\frac{1}{k} ln \frac{P}{k - P} = \frac{t}{5 k} + C_{1}$ , and multiplying by $k$ : $ln \frac{P}{k - P} = \frac{t}{5} + C_{2}$ . 3. Then $\frac{P}{k - P} = A e^{t /5}$ . With $P (0) = 1200$ : $A = \frac{1200}{k - 1200}$ . 4. We solve for $P$ : $P (k - 1200) = 1200 e^{t /5} (k - P)$ , so $P [(k - 1200) + 1200 e^{t /5}] = 1200 k e^{t /5}$ . 5. Dividing by $e^{t /5}$ : $P = \frac{1200 k}{( k - 1200 ) e ^{- t /5} + 1200}$ .

d) 1. $P (10) = 2400$ : $2400 = \frac{1200 k}{( k - 1200 ) e ^{- 2} + 1200}$ , that is, $2 [(k - 1200) e^{- 2} + 1200] = k$ . 2. $2 k e^{- 2} - 2400 e^{- 2} + 2400 = k$ , so $k (1 - 2 e^{- 2}) = 2400 (1 - e^{- 2})$ . 3. $k = \frac{2400 ( 1 - e ^{- 2} )}{1 - 2 e ^{- 2}} \approx 2845$ (to 4 significant figures).

e) 1. $\frac{d P}{d t} = \frac{P ( k - P )}{5 k}$ is maximum when $P (k - P)$ is maximum, that is, at $P = \frac{k}{2}$ . 2. Equating: $\frac{1200 k}{( k - 1200 ) e ^{- t /5} + 1200} = \frac{k}{2}$ , so $(k - 1200) e^{- t /5} = 1200$ . 3. $t = 5 ln \frac{k - 1200}{1200} = 5 ln \frac{1645.3}{1200} \approx 1.58$ days.

Answer: a) $a = b = \frac{1}{k}$ ; b) $f (x) = \frac{1}{k} ln \frac{x}{k - x} + C$ ; c) shown; d) $k = 2845$ ; e) $t = 1.58$ days

Exercise 141 HardIB · P2 TZ2 May 2022

The population ( $P$ ) of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ , where $t$ is the time measured in years, and $k$ , $N$ are positive constants. The constant $N$ represents the maximum population of this species of marsupial that the island can sustain indefinitely.

a) In the context of this population model, interpret the meaning of $\frac{d P}{d t}$ .

b) Show that $\frac{d ^{2} P}{d t ^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ .

c) Hence, show that the rate at which the marsupial population increases will be a maximum when $P = \frac{N}{2}$ . Justify your answer.

d) Hence, determine the maximum value of $\frac{d P}{d t}$ in terms of $k$ and $N$ .

Let $P_{0}$ be the initial population of marsupials.

e) By solving the logistic differential equation, show that its solution can be expressed in the form $k t = ln (\frac{P}{P _{0}} (\frac{N - P _{0}}{N - P}))$ .

After $10$ years, the population of marsupials is equal to $3 P_{0}$ . Also, it is known that $N = 4 P_{0}$ .

f) Find the value of $k$ for this population model.

Show answer

Answer: a) the rate of change of the marsupial population; b), c) and e) shown; d) $\frac{k N}{4}$ ; f) $k = \frac{ln 9}{10} = 0.220$

Show working and answer

a) 1. $\frac{d P}{d t}$ is the rate of change of the marsupial population with respect to time, in individuals per year.

b) 1. We differentiate $\frac{d P}{d t} = k P (1 - \frac{P}{N}) = k P - \frac{k P ^{2}}{N}$ with respect to $t$ , using the chain rule. 2. $\frac{d ^{2} P}{d t ^{2}} = k \frac{d P}{d t} - \frac{2 k P}{N} \frac{d P}{d t} = k (1 - \frac{2 P}{N}) \frac{d P}{d t}$ . 3. Substituting $\frac{d P}{d t}$ : $\frac{d ^{2} P}{d t ^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ .

c) 1. The growth rate $\frac{d P}{d t}$ is a maximum when $\frac{d ^{2} P}{d t ^{2}} = 0$ . With $0 < P < N$ the factors $k^{2} P$ and $1 - \frac{P}{N}$ are positive, so it is zero when $1 - \frac{2 P}{N} = 0$ , that is, $P = \frac{N}{2}$ . 2. Justification: if $P < \frac{N}{2}$ , $\frac{d ^{2} P}{d t ^{2}} > 0$ (the rate increases); if $\frac{N}{2} < P < N$ , $\frac{d ^{2} P}{d t ^{2}} < 0$ (the rate decreases). As $P$ increases with time, the rate changes from increasing to decreasing at $P = \frac{N}{2}$ : it is a maximum.

d) 1. $(\frac{d P}{d t})_{max} = k \cdot \frac{N}{2} (1 - \frac{1}{2}) = \frac{k N}{4}$ .

e) 1. We separate the variables: $\int \frac{N d P}{P ( N - P )} = \int k d t$ . 2. Partial fractions: $\frac{N}{P ( N - P )} = \frac{1}{P} + \frac{1}{N - P}$ . 3. Integrating: $ln P - ln (N - P) = k t + C$ (with $0 < P < N$ ). 4. At $t = 0$ : $C = ln \frac{P _{0}}{N - P _{0}}$ . 5. $k t = ln \frac{P}{N - P} - ln \frac{P _{0}}{N - P _{0}} = ln (\frac{P}{P _{0}} \cdot \frac{N - P _{0}}{N - P})$ .

f) 1. With $N = 4 P_{0}$ , $P (10) = 3 P_{0}$ : $\frac{P}{P _{0}} = 3$ and $\frac{N - P _{0}}{N - P} = \frac{3 P _{0}}{P _{0}} = 3$ . 2. $10 k = ln (3 \cdot 3) = ln 9$ , so $k = \frac{ln 9}{10} \approx 0.220$ .

Answer: a) the rate of change of the marsupial population; b), c) and e) shown; d) $\frac{k N}{4}$ ; f) $k = \frac{ln 9}{10} = 0.220$

Euler's method

Exercise 142 BasicIB · P3 TZ0 May 2011

Consider the differential equation $\frac{d y}{d x} = x^{2} + y^{2}$ , given that when $x = 0$ , $y = 1$ .

Use Euler's method with a step of $0.1$ to find an approximate value of $y$ when $x = 0.4$ .

Show answer

Answer: $y (0.4) \approx 1.5735$

Show working and answer

We apply $y_{n + 1} = y_{n} + 0.1 (x_{n}^{2} + y_{n}^{2})$ from $(0, 1)$ .
$n = 0$ : $f = 0 + 1 = 1$ , so $y_{1} = 1.1$ .
$n = 1$ : $f = 0. 1^{2} + 1. 1^{2} = 1.22$ , so $y_{2} = 1.1 + 0.122 = 1.222$ .
$n = 2$ : $f = 0. 2^{2} + 1.22 2^{2} = 1.533284$ , so $y_{3} = 1.222 + 0.1533284 \approx 1.3753$ .
$n = 3$ : $f = 0. 3^{2} + 1.375328 4^{2} \approx 1.98170$ , so $y_{4} \approx 1.3753 + 0.19817 \approx 1.5735$ .

$n$	$x_{n}$	$y_{n}$
0	0	1
1	0.1	1.1
2	0.2	1.222
3	0.3	1.3753
4	0.4	1.5735

Answer: $y (0.4) \approx 1.5735$

Exercise 143 MediumIB · P3 TZ0 May 2010

Given that $\frac{d y}{d x} - 2 y^{2} = e^{x}$ , and that $y = 1$ when $x = 0$ , use Euler's method, with a step of $0.1$ , to find an approximate value of $y$ when $x = 0.4$ . Give all intermediate values as accurately as possible.

Show answer

Answer: $y (0.4) \approx 3.8493$

Show working and answer

We write $\frac{d y}{d x} = 2 y^{2} + e^{x}$ and apply $y_{n + 1} = y_{n} + 0.1 f (x_{n}, y_{n})$ from $(0, 1)$ , without rounding the intermediate values.
$n = 0$ : $f (0, 1) = 2 + 1 = 3$ , so $y_{1} = 1 + 0.3 = 1.3$ .
$n = 1$ : $f (0.1; 1.3) = 2 \cdot 1.69 + e^{0.1} = 3.38 + 1.10517 = 4.48517$ , so $y_{2} = 1.3 + 0.448517 = 1.7485$ .
$n = 2$ : $f (0.2; 1.7485) = 2 \cdot 1.748 5^{2} + e^{0.2} \approx 7.3360$ , so $y_{3} \approx 1.7485 + 0.7336 = 2.4821$ .
$n = 3$ : $f (0.3; 2.4821) \approx 13.6717$ , so $y_{4} \approx 2.4821 + 1.3672 = 3.8493$ .

$n$	$x_{n}$	$y_{n}$
0	0	1
1	0.1	1.3
2	0.2	1.7485
3	0.3	2.4821
4	0.4	3.8493

Answer: $y (0.4) \approx 3.8493$

Exercise 144 MediumIB · P3 TZ2 May 2007

Consider the differential equation $\frac{d y}{d x} + (2 x - 1) y = 0$ , given that $y = 2$ when $x = 0$ .

a) Use Euler's method with a step of $0.1$ to obtain an approximation of $y$ when $x = 0.5$ . Write the solution in table form.

b) Solve the differential equation, expressing your answer in the form $y = f (x)$ .

Show answer

Answer: a) $y (0.5) \approx 2.6717$ ; b) $y = 2 e^{x - x^{2}}$

Show working and answer

a) 1. We write $\frac{d y}{d x} = (1 - 2 x) y$ and apply $y_{n + 1} = y_{n} + 0.1 (1 - 2 x_{n}) y_{n}$ from $(0, 2)$ . 2. $y_{1} = 2 + 0.1 \cdot 1 \cdot 2 = 2.2$ . 3. $y_{2} = 2.2 + 0.1 \cdot 0.8 \cdot 2.2 = 2.376$ . 4. $y_{3} = 2.376 + 0.1 \cdot 0.6 \cdot 2.376 = 2.51856$ . 5. $y_{4} = 2.51856 + 0.1 \cdot 0.4 \cdot 2.51856 \approx 2.6193$ . 6. $y_{5} = 2.6193 + 0.1 \cdot 0.2 \cdot 2.6193 \approx 2.6717$ .

$n$	$x_{n}$	$y_{n}$
0	0	2
1	0.1	2.2
2	0.2	2.376
3	0.3	2.5186
4	0.4	2.6193
5	0.5	2.6717

b) 1. Separate the variables: $\frac{d y}{y} = (1 - 2 x) d x$ (the solution is positive because $y (0) = 2$ ). 2. Integrate: $ln y = x - x^{2} + C$ , i.e. $y = A e^{x - x^{2}}$ . 3. With $y (0) = 2$ : $A = 2$ . 4. Check: $y^{'} = 2 e^{x - x^{2}} (1 - 2 x) = (1 - 2 x) y$ , so $y^{'} + (2 x - 1) y = 0$ .

Answer: a) $y (0.5) \approx 2.6717$ ; b) $y = 2 e^{x - x^{2}}$

Exercise 145 MediumIB · P3 TZ2 May 2008

A curve passing through the point $(1, 2)$ is defined by the differential equation $\frac{d y}{d x} = 2 x (1 + x^{2} - y)$ .

a) Use Euler's method to obtain an approximate value of $y$ when $x = 1.3$ , taking steps of $0.1$ . Show the intermediate steps in a table, giving the values to four decimal places.

b) Solve the differential equation, expressing your answer in the form $y = f (x)$ .

Show answer

Answer: a) $y (1.3) \approx 2.1407$ ; b) $y = x^{2} + e^{1 - x^{2}}$

Show working and answer

a) 1. We apply $y_{n + 1} = y_{n} + 0.1 \cdot 2 x_{n} (1 + x_{n}^{2} - y_{n})$ from $(1, 2)$ . 2. $f (1, 2) = 2 (1 + 1 - 2) = 0$ , so $y_{1} = 2.0000$ . 3. $f (1.1; 2) = 2.2 (1 + 1.21 - 2) = 0.462$ , so $y_{2} = 2 + 0.0462 = 2.0462$ . 4. $f (1.2; 2.0462) = 2.4 (1 + 1.44 - 2.0462) = 0.94512$ , so $y_{3} = 2.0462 + 0.0945 \approx 2.1407$ .

$n$	$x_{n}$	$y_{n}$
0	1.0	2.0000
1	1.1	2.0000
2	1.2	2.0462
3	1.3	2.1407

b) 1. It is linear: $\frac{d y}{d x} + 2 x y = 2 x (1 + x^{2})$ . Integrating factor: $e^{\int 2 x d x} = e^{x^{2}}$ . 2. $\frac{d}{d x} (y e^{x^{2}}) = 2 x (1 + x^{2}) e^{x^{2}}$ . 3. Substitute $u = x^{2}$ ( $d u = 2 x d x$ ): $\int (1 + u) e^{u} d u = u e^{u} + C$ (check by differentiating: $(u e^{u})^{'} = (1 + u) e^{u}$ ). 4. Hence $y e^{x^{2}} = x^{2} e^{x^{2}} + C$ , i.e. $y = x^{2} + C e^{- x^{2}}$ . 5. With $y (1) = 2$ : $2 = 1 + C e^{- 1}$ , so $C = e$ . 6. $y = x^{2} + e \cdot e^{- x^{2}} = x^{2} + e^{1 - x^{2}}$ .

Answer: a) $y (1.3) \approx 2.1407$ ; b) $y = x^{2} + e^{1 - x^{2}}$

Exercise 146 MediumIB · P3 TZ0 November 2018

Consider the differential equation $\frac{d y}{d x} = 1 + \frac{y}{x}$ , where $x \neq = 0$ .

a) Given that $y (1) = 1$ , use Euler's method with a step $h = 0.25$ to find an approximate value of $y (2)$ . Give your answer to two significant figures.

b) Solve the equation $\frac{d y}{d x} = 1 + \frac{y}{x}$ given that $y (1) = 1$ .

c) Find the percentage error made when approximating the value of $y (2)$ with the answer given in part a). Express the result to two significant figures.

Show answer

Answer: a) $y (2) \approx 3.3$ ; b) $y = x (ln ∣ x ∣ + 1)$ ; c) the percentage error is $2.5 %$

Show working and answer

a) 1. We apply $y_{n + 1} = y_{n} + 0.25 (1 + \frac{y _{n}}{x _{n}})$ from $(1, 1)$ , up to $x = 2$ (4 steps). 2. $y (1.25) = 1 + 0.25 \cdot 2 = 1.5$ . 3. $y (1.5) = 1.5 + 0.25 (1 + \frac{1.5}{1.25}) = 1.5 + 0.55 = 2.05$ . 4. $y (1.75) = 2.05 + 0.25 (1 + \frac{2.05}{1.5}) \approx 2.6417$ . 5. $y (2) \approx 2.6417 + 0.25 (1 + \frac{2.6417}{1.75}) \approx 3.2690$ . 6. To two significant figures, $y (2) \approx 3.3$ .

b) 1. It is linear: $\frac{d y}{d x} - \frac{1}{x} y = 1$ . Integrating factor: $e^{- \int \frac{d x}{x}} = \frac{1}{x}$ (with $x > 0$ ). 2. $\frac{d}{d x} (\frac{y}{x}) = \frac{1}{x}$ , so $\frac{y}{x} = ln ∣ x ∣ + C$ . 3. With $y (1) = 1$ : $1 = 0 + C$ , so $C = 1$ . 4. $y = x (ln ∣ x ∣ + 1)$ .

c) 1. The exact value is $y (2) = 2 (ln 2 + 1) \approx 3.3863$ . 2. The approximate value given as the answer to part a) is $3.3$ . 3. Percentage error: $\frac{∣3.3863 - 3.3∣}{3.3863} \times 100 \approx 2.5 %$ .

Answer: a) $y (2) \approx 3.3$ ; b) $y = x (ln ∣ x ∣ + 1)$ ; c) the percentage error is $2.5 %$

Exercise 147 MediumIB · P3 TZ0 May 2006

Given that $\frac{d y}{d x} = \frac{y + 2}{x y + 1}$ and $y = 1$ when $x = 0$ , use Euler's method with step $h = 0.5$ to find an approximate value of $y$ when $x = 1$ .

Show answer

Answer: when $x = 1$ , $y \approx 3.5$

Show working and answer

Euler's method with $h = 0.5$ : $x_{n + 1} = x_{n} + h$ , $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ , with $f (x, y) = \frac{y + 2}{x y + 1}$ .
Step 1, from $(0, 1)$ : $f (0, 1) = \frac{1 + 2}{0 + 1} = 3$ , so $y_{1} = 1 + 0.5 \cdot 3 = 2.5$ at $x_{1} = 0.5$ .
Step 2, from $(0.5; 2.5)$ : $f (0.5; 2.5) = \frac{2.5 + 2}{0.5 \cdot 2.5 + 1} = \frac{4.5}{2.25} = 2$ , so $y_{2} = 2.5 + 0.5 \cdot 2 = 3.5$ at $x_{2} = 1$ .

Answer: when $x = 1$ , $y \approx 3.5$

Exercise 148 HardIB · P3 TZ0 November 2015

The curves $y = f (x)$ and $y = g (x)$ both pass through the point $(1, 0)$ and are defined by the differential equations $\frac{d y}{d x} = x - y^{2}$ and $\frac{d y}{d x} = y - x^{2}$ respectively.

a) Show that the tangent to the curve $y = f (x)$ at the point $(1, 0)$ is normal to the curve $y = g (x)$ at the point $(1, 0)$ .

b) Find $g (x)$ .

c) Use Euler's method with a step of $0.2$ to estimate $f (2)$ correct to $5$ decimal places.

Show answer

Answer: a) the two lines coincide (gradient $1$ through $(1, 0)$ ); b) $g (x) = x^{2} + 2 x + 2 - 5 e^{x - 1}$ ; c) $f (2) \approx 1.10033$

Show working and answer

a) 1. For $y = f (x)$ at $(1, 0)$ : $f^{'} (1) = 1 - 0^{2} = 1$ . The tangent has gradient $1$ . 2. For $y = g (x)$ at $(1, 0)$ : $g^{'} (1) = 0 - 1^{2} = - 1$ . The tangent to $g$ has gradient $- 1$ , so the normal to $g$ has gradient $- \frac{1}{- 1} = 1$ . 3. The tangent to $f$ and the normal to $g$ pass through the same point $(1, 0)$ and have the same gradient $1$ : they are the same line.

b) 1. The equation $g^{'} - g = - x^{2}$ is a first-order linear equation. Integrating factor: $e^{- \int d x} = e^{- x}$ . 2. Multiply: $\frac{d}{d x} (g e^{- x}) = - x^{2} e^{- x}$ . 3. Integrate by parts twice ( $u = x^{2}$ , then $u = x$ , with $d v = e^{- x} d x$ ): $\int - x^{2} e^{- x} d x = e^{- x} (x^{2} + 2 x + 2) + C$ . Check by differentiating: $\frac{d}{d x} [e^{- x} (x^{2} + 2 x + 2)] = e^{- x} (2 x + 2 - x^{2} - 2 x - 2) = - x^{2} e^{- x}$ . 4. Hence $g (x) = x^{2} + 2 x + 2 + C e^{x}$ . 5. With $g (1) = 0$ : $5 + C e = 0$ , so $C = - 5 e^{- 1}$ . 6. $g (x) = x^{2} + 2 x + 2 - 5 e^{x - 1}$ .

c) 1. We use $y_{n + 1} = y_{n} + 0.2 (x_{n} - y_{n}^{2})$ from $(1, 0)$ , up to $x = 2$ (5 steps), using the GDC. 2. Values: $y (1.2) = 0.2$ ; $y (1.4) = 0.432$ ; $y (1.6) = 0.674675$ ; $y (1.8) = 0.903638$ ; $y (2) = 1.100326$ .

$x_{n}$	$y_{n}$
1	0
1.2	0.2
1.4	0.432
1.6	0.674675
1.8	0.903638
2	1.100326

To five decimal places, $f (2) \approx 1.10033$ .

Answer: a) the two lines coincide (gradient $1$ through $(1, 0)$ ); b) $g (x) = x^{2} + 2 x + 2 - 5 e^{x - 1}$ ; c) $f (2) \approx 1.10033$

Exercise 149 HardIB · P3 TZ0 May 2012

Consider the differential equation $\frac{d y}{d x} = \frac{y ^{2}}{1 + x}$ , where $x > - 1$ , and $y = 1$ when $x = 0$ .

a) Use Euler's method with a step of $0.1$ to find the approximate value of $y$ when $x = 0.5$ .

b) i) Verify that $\frac{d ^{2} y}{d x ^{2}} = \frac{2 y ^{3} - y ^{2}}{( 1 + x ) ^{2}}$ .

(ii) Hence, find the Maclaurin series for $y$ up to and including the term in $x^{2}$ .

c) i) Solve the differential equation.

(ii) Find the value of $a$ for which $y \to \infty$ as $x \to a$ .

Show answer

Answer: a) $y (0.5) \approx 1.62$ ; b) (ii) $y \approx 1 + x + \frac{x ^{2}}{2}$ ; c) (i) $y = \frac{1}{1 - ln ( 1 + x )}$ ; (ii) $a = e - 1$

Show working and answer

a) 1. We apply $y_{n + 1} = y_{n} + 0.1 \frac{y _{n}^{2}}{1 + x _{n}}$ from $(0, 1)$ , using the GDC. 2. Values: $y (0.1) = 1.1$ ; $y (0.2) = 1.21$ ; $y (0.3) \approx 1.3320$ ; $y (0.4) \approx 1.4685$ ; $y (0.5) \approx 1.6225$ . 3. For $x = 0.5$ : $y \approx 1.62$ .

b) (i) 1. Differentiate $\frac{d y}{d x} = \frac{y ^{2}}{1 + x}$ with respect to $x$ using the quotient rule: $\frac{d ^{2} y}{d x ^{2}} = \frac{2 y \frac{d y}{d x} ( 1 + x ) - y ^{2}}{( 1 + x ) ^{2}}$ . 2. As $(1 + x) \frac{d y}{d x} = y^{2}$ , the first term of the numerator is $2 y \cdot y^{2} = 2 y^{3}$ . 3. Hence $\frac{d ^{2} y}{d x ^{2}} = \frac{2 y ^{3} - y ^{2}}{( 1 + x ) ^{2}}$ .

(ii) 1. At $x = 0$ : $y (0) = 1$ . 2. $y^{'} (0) = \frac{1 ^{2}}{1} = 1$ . 3. $y^{''} (0) = \frac{2 - 1}{1} = 1$ . 4. Maclaurin series: $y \approx y (0) + y^{'} (0) x + \frac{y ^{''} ( 0 )}{2} x^{2} = 1 + x + \frac{x ^{2}}{2}$ .

c) (i) 1. Separate the variables: $\frac{d y}{y ^{2}} = \frac{d x}{1 + x}$ , so $- \frac{1}{y} = ln (1 + x) + C$ . 2. With $y (0) = 1$ : $- 1 = 0 + C$ , so $C = - 1$ . 3. $\frac{1}{y} = 1 - ln (1 + x)$ and $y = \frac{1}{1 - ln ( 1 + x )}$ .

(ii) 1. $y \to \infty$ when the denominator tends to $0$ : $1 - ln (1 + x) = 0$ , i.e. $ln (1 + x) = 1$ . 2. Then $1 + x = e$ and $a = e - 1$ .

Answer: a) $y (0.5) \approx 1.62$ ; b) (ii) $y \approx 1 + x + \frac{x ^{2}}{2}$ ; c) (i) $y = \frac{1}{1 - ln ( 1 + x )}$ ; (ii) $a = e - 1$

Exercise 150 HardIB · P3 TZ0 November 2012

Consider the differential equation $\frac{d y}{d x} = x + y$ , $(x + y \geq 0)$ with $y = 1$ when $x = 1$ . Also let $y = c$ when $x = 2$ .

a) Use Euler's method to find an approximation for the value of $c$ , using a step length $h = 0.1$ . Give your answer correct to four decimal places.

If Euler's method is used with $h = 0.05$ then $c = 2.7921$ , if it is used with $h = 0.01$ then $c = 2.8099$ , and if it is used with $h = 0.005$ then $c = 2.8121$ .

b) On graph paper, with $h$ on the horizontal axis and the approximation for $c$ on the vertical axis, plot the four points (one of which you have calculated and three of which have been given). Use a scale of $1$ cm $= 0.01$ on both axes.

Take the horizontal axis from $0$ to $0.12$ and the vertical axis from $2.76$ to $2.82$ .

c) Draw, by eye, the straight line which best fits these four points, using a ruler.

d) Use your graph to give the best possible estimate for $c$ , giving your answer correct to three decimal places.

View solution

a) 1. Euler's method: $x_{n + 1} = x_{n} + h$ and $y_{n + 1} = y_{n} + h x_{n} + y_{n}$ , with $x_{0} = 1$ , $y_{0} = 1$ and $h = 0.1$ . We need to reach $x = 2$ : ten steps. Using the GDC:

$x_{0} = 1.0$ , $y_{0} = 1$ , $x_{0} + y_{0} = 1.4142$ , $y_{1} = 1.1414$

$x_{1} = 1.1$ , $y_{1} = 1.1414$ , $x_{1} + y_{1} = 1.4971$ , $y_{2} = 1.2911$

$x_{2} = 1.2$ , $y_{2} = 1.2911$ , $x_{2} + y_{2} = 1.5783$ , $y_{3} = 1.4490$

$x_{3} = 1.3$ , $y_{3} = 1.4490$ , $x_{3} + y_{3} = 1.6580$ , $y_{4} = 1.6148$

$x_{4} = 1.4$ , $y_{4} = 1.6148$ , $x_{4} + y_{4} = 1.7363$ , $y_{5} = 1.7884$

$x_{5} = 1.5$ , $y_{5} = 1.7884$ , $x_{5} + y_{5} = 1.8134$ , $y_{6} = 1.9697$

$x_{6} = 1.6$ , $y_{6} = 1.9697$ , $x_{6} + y_{6} = 1.8894$ , $y_{7} = 2.1587$

$x_{7} = 1.7$ , $y_{7} = 2.1587$ , $x_{7} + y_{7} = 1.9644$ , $y_{8} = 2.3551$

$x_{8} = 1.8$ , $y_{8} = 2.3551$ , $x_{8} + y_{8} = 2.0384$ , $y_{9} = 2.5590$

$x_{9} = 1.9$ , $y_{9} = 2.5590$ , $x_{9} + y_{9} = 2.1116$ , $y_{10} = 2.7701$

Therefore $c \approx 2.7701$ .

b) 1. The four points $(h, c)$ are $(0.1; 2.7701)$ , $(0.05; 2.7921)$ , $(0.01; 2.8099)$ and $(0.005; 2.8121)$ . 2. They are plotted on the paper using the scale indicated ( $1$ cm $= 0.01$ on both axes).

c) 1. The points are almost collinear, with negative gradient (the smaller $h$ is, the larger the approximation). We draw with a ruler the line passing as close as possible to all four.

Points of the approximation of c as a function of the step h, and the line of best fit, which meets the vertical axis near 2.814

d) 1. The exact value corresponds to a step $h \to 0$ , i.e. to the point where the line meets the vertical axis. 2. Reading from the graph, the line meets the $c$ -axis at $c \approx 2.814$ .

Answer: a) $c \approx 2.7701$ ; d) $c \approx 2.814$

Exercise 151 HardIB · P1 TZ1 May 2022

Consider the differential equation $x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2}$ for $x > 0$ and $y > 2 x$ . It is known that $y = 3$ when $x = 1$ .

a) Use Euler's method, with a step length of $0.1$ , to calculate an approximate value of $y$ when $x = 1.5$ .

b) Use the substitution $y = v x$ to show that $x \frac{d v}{d x} = v^{2} - v - 2$ .

c) (i) By solving the differential equation, show that $y = \frac{8 x + x ^{4}}{4 - x ^{3}}$ .

(ii) Calculate the actual value of $y$ when $x = 1.5$ .

(iii) Using the graph of $y = \frac{8 x + x ^{4}}{4 - x ^{3}}$ , suggest a reason why the approximation given by Euler's method in part (a) is not a good estimate of the actual value of $y$ at $x = 1.5$ .

Show answer

Answer: a) $y \approx 10.7$ ; c) (i) shown; (ii) $y \approx 27.3$ ; (iii) there is a vertical asymptote at $x = 34 \approx 1.59$ , close to $x = 1.5$ , and Euler's method is not reliable near it

Show working and answer

a) 1. We write the equation as $\frac{d y}{d x} = \frac{y ^{2} - 2 x ^{2}}{x ^{2}}$ . Euler: $y_{n + 1} = y_{n} + 0.1 \cdot \frac{y _{n}^{2} - 2 x _{n}^{2}}{x _{n}^{2}}$ , $x_{n + 1} = x_{n} + 0.1$ , from $(1, 3)$ to $x = 1.5$ : five steps.

$x_{0} = 1$ , $y_{0} = 3$ , gradient $7$ , $y_{1} = 3.7$

$x_{1} = 1.1$ , $y_{1} = 3.7$ , gradient $9.314$ , $y_{2} = 4.6314$

$x_{2} = 1.2$ , $y_{2} = 4.6314$ , gradient $12.896$ , $y_{3} = 5.921$

$x_{3} = 1.3$ , $y_{3} = 5.921$ , gradient $18.744$ , $y_{4} = 7.7954$

$x_{4} = 1.4$ , $y_{4} = 7.7954$ , gradient $29.004$ , $y_{5} = 10.6959$

Hence $y (1.5) \approx 10.7$ .

b) 1. With $y = v x$ : $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . 2. The equation $\frac{d y}{d x} = \frac{y ^{2}}{x ^{2}} - 2$ becomes $v + x \frac{d v}{d x} = v^{2} - 2$ . 3. Therefore $x \frac{d v}{d x} = v^{2} - v - 2$ .

c) (i) 1. We separate the variables: $\int \frac{d v}{( v - 2 ) ( v + 1 )} = \int \frac{d x}{x}$ . 2. Partial fractions: $\frac{1}{( v - 2 ) ( v + 1 )} = \frac{1}{3} (\frac{1}{v - 2} - \frac{1}{v + 1})$ . 3. Integrating: $\frac{1}{3} ln \frac{v - 2}{v + 1} = ln x + C$ . 4. Initial condition: $x = 1$ , $y = 3$ , so $v = 3$ and $\frac{1}{3} ln \frac{1}{4} = C$ . Then $ln \frac{v - 2}{v + 1} = 3 ln x - ln 4$ (with $y > 2 x$ , $v > 2$ , so the absolute value is not needed). 5. $\frac{v - 2}{v + 1} = \frac{x ^{3}}{4}$ , so $4 v - 8 = x^{3} v + x^{3}$ and $v (4 - x^{3}) = 8 + x^{3}$ . 6. $v = \frac{8 + x ^{3}}{4 - x ^{3}}$ and $y = v x = \frac{8 x + x ^{4}}{4 - x ^{3}}$ .

(ii) 1. $y (1.5) = \frac{12 + 5.0625}{4 - 3.375} = \frac{17.0625}{0.625} = 27.3$ .

(iii) 1. The solution has a vertical asymptote at $x = 34 \approx 1.59$ , very close to $x = 1.5$ : the solution increases very rapidly as we approach it. 2. Euler's method uses the gradient at the left-hand end of each step, which underestimates the growth on a curve whose gradient increases so steeply. That is why $10.7$ is well below the actual value.

Graph of y = (8x + x⁴)/(4 − x³) with a vertical asymptote at x = ∛4, approximately 1.59

Answer: a) $y \approx 10.7$ ; c) (i) shown; (ii) $y \approx 27.3$ ; (iii) there is a vertical asymptote at $x = 34 \approx 1.59$ , close to $x = 1.5$ , and Euler's method is not reliable near it

Definite integral and the fundamental theorem of calculus

Exercise 152 BasicIB · P1 TZ1 May 2018

Given that $\int_{- 2}^{2} f (x) d x = 10$ and $\int_{0}^{2} f (x) d x = 12$ , calculate

a) $\int_{- 2}^{0} (f (x) + 2) d x$ .

b) $\int_{- 2}^{0} f (x + 2) d x$ .

Show answer

Answer: a) $2$ ; b) $12$

Show working and answer

Given: $\int_{- 2}^{2} f = 10$ and $\int_{0}^{2} f = 12$ . Then $\int_{- 2}^{0} f = 10 - 12 = - 2$ .

a) $\int_{- 2}^{0} (f (x) + 2) d x = \int_{- 2}^{0} f (x) d x + \int_{- 2}^{0} 2 d x = - 2 + 2 \cdot 2 = 2$ .

b) Substituting $u = x + 2$ : $d u = d x$ , and the limits $x = - 2$ and $x = 0$ become $u = 0$ and $u = 2$ . Hence $\int_{- 2}^{0} f (x + 2) d x = \int_{0}^{2} f (u) d u = 12$ .

Answer: a) $2$ ; b) $12$

Exercise 153 BasicIB · P1 November 1998

Let $F (x) = \int_{0}^{x} \frac{d t}{1 + e ^{t^{2}}}$ .

a) Calculate $f (x) = F^{'} (x)$ .

b) For which values of $k$ does the equation $f (x) = k$ have real solutions?

Show answer

Answer: a) $f (x) = \frac{1}{1 + e ^{x^{2}}}$ ; b) $0 < k \leq \frac{1}{2}$

Show working and answer

a) 1. By the fundamental theorem of calculus, $F^{'} (x) = \frac{1}{1 + e ^{x^{2}}}$ . 2. That is, $f (x) = \frac{1}{1 + e ^{x^{2}}}$ .

b) 1. Since $x^{2} \geq 0$ , $e^{x^{2}} \geq 1$ , so $1 + e^{x^{2}} \geq 2$ . Also $1 + e^{x^{2}}$ can be as large as we like. 2. Therefore $0 < f (x) \leq \frac{1}{2}$ , and the maximum $\frac{1}{2}$ is reached at $x = 0$ . 3. The equation $f (x) = k$ has a real solution if and only if $0 < k \leq \frac{1}{2}$ .

Answer: a) $f (x) = \frac{1}{1 + e ^{x^{2}}}$ ; b) $0 < k \leq \frac{1}{2}$

Exercise 154 BasicIB · P1 TZ1 May 2022

a) The expression $\frac{3 x - 5}{x}$ can be written as $3 - 5 x^{p}$ . Write down the value of $p$ .

b) Using the above, calculate the value of $\int_{1}^{9} \frac{3 x - 5}{x} d x$ .

Show answer

Answer: a) $p = - \frac{1}{2}$ ; b) $4$

Show working and answer

a) 1. $\frac{3 x - 5}{x} = 3 - \frac{5}{x} = 3 - 5 x^{- 1/2}$ . 2. $p = - \frac{1}{2}$ .

b) 1. Using the above: $\int_{1}^{9} (3 - 5 x^{- 1/2}) d x = [3 x - 10 x^{1/2}]_{1}^{9}$ . 2. $= (27 - 30) - (3 - 10) = - 3 + 7 = 4$ .

Answer: a) $p = - \frac{1}{2}$ ; b) $4$

Exercise 155 MediumIB · P1 TZ1 May 2009

a) Let $a > 0$ . Sketch the graph of $y = x - \frac{a}{2}$ for $- a \leq x \leq a$ .

b) Calculate $k$ such that $\int_{- a}^{0} x - \frac{a}{2} d x = k \int_{0}^{a} x - \frac{a}{2} d x$ .

View solution

a) 1. $y = x - \frac{a}{2}$ is a "V" shape with its vertex at $(\frac{a}{2}, 0)$ . 2. For $x < \frac{a}{2}$ , $y = \frac{a}{2} - x$ (a decreasing line), and for $x \geq \frac{a}{2}$ , $y = x - \frac{a}{2}$ (an increasing line). 3. At the endpoints: $x = - a$ gives $y = \frac{3 a}{2}$ ; $x = a$ gives $y = \frac{a}{2}$ ; and it meets the $y$ -axis at $(0, \frac{a}{2})$ .

Graph of y = |x − a/2| for x between −a and a: a decreasing segment from the point (−a, 3a/2), passing through (0, a/2), down to the vertex (a/2, 0), then an increasing segment up to (a, a/2)

b) 1. In $[- a, 0]$ , $x - \frac{a}{2} < 0$ , so $x - \frac{a}{2} = \frac{a}{2} - x$ : $\int_{- a}^{0} (\frac{a}{2} - x) d x = [\frac{a}{2} x - \frac{x ^{2}}{2}]_{- a}^{0} = 0 - (- \frac{a ^{2}}{2} - \frac{a ^{2}}{2}) = a^{2}$ . 2. On $[0, a]$ the absolute value changes at $x = \frac{a}{2}$ . Each half is a triangle with base $\frac{a}{2}$ and height $\frac{a}{2}$ , with area $\frac{a ^{2}}{8}$ ; in total $\int_{0}^{a} x - \frac{a}{2} d x = \frac{a ^{2}}{4}$ . 3. Then $a^{2} = k \cdot \frac{a ^{2}}{4}$ and, as $a > 0$ , $k = 4$ .

Answer: a) graph; b) $k = 4$

Exercise 156 MediumIB · P1 TZ1 May 2019

a) Show that $(sin x + cos x)^{2} = 1 + sin 2 x$ .

b) Show that $sec 2 x + tan 2 x = \frac{cos x + sin x}{cos x - sin x}$ .

c) Using the above, or otherwise, calculate $\int_{0}^{π /6} (sec 2 x + tan 2 x) d x$ in the form $ln (a + b)$ where $a, b \in Z$ .

Show answer

Answer: a) and b) proved; c) $ln (1 + 3)$

Show working and answer

a) $(sin x + cos x)^{2} = sin^{2} x + 2 sin x cos x + cos^{2} x = 1 + sin 2 x$ .

b) 1. $sec 2 x + tan 2 x = \frac{1}{cos 2 x} + \frac{sin 2 x}{cos 2 x} = \frac{1 + sin 2 x}{cos 2 x}$ . 2. By a) the numerator is $(sin x + cos x)^{2}$ , and $cos 2 x = cos^{2} x - sin^{2} x = (cos x - sin x) (cos x + sin x)$ . 3. Cancelling the factor $cos x + sin x$ : $sec 2 x + tan 2 x = \frac{cos x + sin x}{cos x - sin x}$ .

c) 1. We use the antiderivatives $\int sec 2 x d x = \frac{1}{2} ln ∣ sec 2 x + tan 2 x ∣$ and $\int tan 2 x d x = - \frac{1}{2} ln ∣ cos 2 x ∣$ . 2. Between $0$ and $\frac{π}{6}$ : at $\frac{π}{6}$ , $sec \frac{π}{3} = 2$ and $tan \frac{π}{3} = 3$ , so we get $\frac{1}{2} ln (2 + 3) - \frac{1}{2} ln \frac{1}{2}$ ; at $0$ , both parts are $0$ . 3. Result: $\frac{1}{2} ln [2 (2 + 3)] = \frac{1}{2} ln (4 + 23)$ . 4. As $4 + 23 = (1 + 3)^{2}$ , the value is $ln (1 + 3)$ , with $a = 1$ and $b = 3$ .

Answer: a) and b) proved; c) $ln (1 + 3)$

Exercise 157 MediumIB · P1 TZ1 May 2019

The graph of $y = f^{'} (x)$ , $0 \leq x \leq 5$ is shown in the following diagram. The curve cuts the $x$ -axis at $(1, 0)$ and $(4, 0)$ and has a local minimum at $(3, - 1)$ .

Graph of y = f'(x) for x between 0 and 5: starts at (0, 1), cuts the x-axis at (1, 0), has its local minimum at (3, −1), cuts the axis again at (4, 0) and rises; the region between the curve, the x-axis and the y-axis is shaded

a) Write down the $x$ -coordinate of the point of inflexion on the graph of $y = f (x)$ .

The shaded area bounded by the curve $y = f^{'} (x)$ , the $x$ -axis and the $y$ -axis is $0.5$ . Given that $f (0) = 3$ ,

b) Find the value of $f (1)$ .

The area bounded by the curve $y = f^{'} (x)$ and the $x$ -axis between $x = 1$ and $x = 4$ is $2.5$ .

c) Find the value of $f (4)$ .

d) Sketch the curve $y = f (x)$ , $0 \leq x \leq 5$ , clearly indicating the coordinates of the maximum and minimum points and the points where the curve cuts the axes.

View solution

a) The point of inflexion of $y = f (x)$ is where $f^{''} = 0$ and changes sign, that is, where $f^{'}$ has an extremum. $f^{'}$ has its local minimum at $x = 3$ , so the point of inflexion is at $x = 3$ .

b) 1. On $[0, 1]$ we have $f^{'} \geq 0$ , and the shaded area is $\int_{0}^{1} f^{'} (x) d x = 0.5$ . 2. Hence $f (1) = f (0) + \int_{0}^{1} f^{'} (x) d x = 3 + 0.5 = 3.5$ .

c) 1. On $[1, 4]$ we have $f^{'} \leq 0$ , so $\int_{1}^{4} f^{'} (x) d x = - 2.5$ . 2. $f (4) = f (1) + \int_{1}^{4} f^{'} (x) d x = 3.5 - 2.5 = 1$ .

d) 1. $f (0) = 3$ : the $y$ -intercept is $(0, 3)$ . 2. $f^{'} > 0$ on $[0, 1 [$ : $f$ increases; $f^{'} (1) = 0$ with a change from $+$ to $-$ : maximum at $(1; 3.5)$ . 3. $f^{'} < 0$ on $] 1, 4 [$ : $f$ decreases (with the steepest negative gradient at $x = 3$ , where the point of inflexion is). $f^{'} (4) = 0$ with a change from $-$ to $+$ : minimum at $(4, 1)$ . 4. For $x > 4$ , $f^{'} > 0$ and $f$ increases again up to $x = 5$ . 5. As the minimum is $1 > 0$ , the curve does not cut the $x$ -axis.

Graph of y = f(x) for x between 0 and 5: starts at (0, 3), rises to the maximum (1; 3.5), falls to the minimum (4, 1) and rises again up to x = 5; does not cut the x-axis

Answer: a) $x = 3$ ; b) $f (1) = 3.5$ ; c) $f (4) = 1$ ; d) graph with maximum $(1; 3.5)$ , minimum $(4, 1)$ and $y$ -intercept $(0, 3)$

Exercise 158 MediumIB · P1 TZ2 May 2016

a) Show that $cot α = tan (\frac{π}{2} - α)$ for $0 < α < \frac{π}{2}$ .

b) Hence, find $\int_{t a n α}^{c o t α} \frac{1}{1 + x ^{2}} d x$ , $0 < α < \frac{π}{2}$ .

Show answer

Answer: a) proven; b) $\frac{π}{2} - 2 α$

Show working and answer

a) 1. For $0 < α < \frac{π}{2}$ , $tan (\frac{π}{2} - α) = \frac{sin ( \frac{π}{2} - α )}{cos ( \frac{π}{2} - α )}$ . 2. Since $sin (\frac{π}{2} - α) = cos α$ and $cos (\frac{π}{2} - α) = sin α$ , we get $\frac{cos α}{sin α} = cot α$ .

b) 1. $\int \frac{1}{1 + x ^{2}} d x = arctan x$ , so the integral equals $arctan (cot α) - arctan (tan α)$ . 2. By part a), $arctan (cot α) = arctan (tan (\frac{π}{2} - α)) = \frac{π}{2} - α$ , because $\frac{π}{2} - α$ lies in $] 0, \frac{π}{2} [$ , where arctan undoes tan. 3. In the same way, $arctan (tan α) = α$ . 4. The result is $\frac{π}{2} - α - α = \frac{π}{2} - 2 α$ .

Answer: a) proven; b) $\frac{π}{2} - 2 α$

Exercise 159 MediumIB · P3 TZ0 May 2016

The function $f$ is given by $f (x) = \int_{0}^{x} ln (2 + sin t) d t$ .

a) Write down $f^{'} (x)$ .

b) By differentiating $f (x^{2})$ , obtain an expression for the derivative of $\int_{0}^{x^{2}} ln (2 + sin t) d t$ with respect to $x$ .

c) Hence, obtain an expression for the derivative of $\int_{x}^{x^{2}} ln (2 + sin t) d t$ with respect to $x$ .

Show answer

Answer: a) $f^{'} (x) = ln (2 + sin x)$ ; b) $2 x ln (2 + sin x^{2})$ ; c) $2 x ln (2 + sin x^{2}) - ln (2 + sin x)$

Show working and answer

a) 1. By the fundamental theorem of calculus, $f^{'} (x) = ln (2 + sin x)$ .

b) 1. The integral $\int_{0}^{x^{2}} ln (2 + sin t) d t$ is $f (x^{2})$ . 2. Chain rule: $\frac{d}{d x} f (x^{2}) = f^{'} (x^{2}) \cdot 2 x = 2 x ln (2 + sin x^{2})$ .

c) 1. Split the integral: $\int_{x}^{x^{2}} ln (2 + sin t) d t = \int_{0}^{x^{2}} ln (2 + sin t) d t - \int_{0}^{x} ln (2 + sin t) d t = f (x^{2}) - f (x)$ . 2. Differentiating, using parts a) and b): $2 x ln (2 + sin x^{2}) - ln (2 + sin x)$ .

Answer: a) $f^{'} (x) = ln (2 + sin x)$ ; b) $2 x ln (2 + sin x^{2})$ ; c) $2 x ln (2 + sin x^{2}) - ln (2 + sin x)$

Exercise 160 MediumIB · P1 May 2002

The following diagram shows the graph of $y_{1} = f (x)$ , $0 \leq x \leq 4$ .

Graph of y₁ = f(x) for 0 ≤ x ≤ 4: starts at the origin, rises to a maximum near x = 1, cuts the x-axis at x = 2, falls to a minimum near x = 3.3 and rises again up to x = 4

On the axes below, sketch the graph of $y_{2} = \int_{0}^{x} f (t) d t$ , and clearly mark the points of inflexion.

View solution

Since $\frac{d}{d x} y_{2} = f (x)$ (fundamental theorem), the graph of $y_{2}$ can be read from that of $f$ : its gradient is the value of $f$ .
$y_{2} (0) = 0$ , and since $f (0) = 0$ the tangent is horizontal at the origin.
For $0 < x < 2$ , $f > 0$ : $y_{2}$ increases. At $x = 2$ , $f$ changes from positive to negative: $y_{2}$ has a maximum. For $2 < x < 4$ , $f < 0$ : $y_{2}$ decreases.
The points of inflexion of $y_{2}$ are the extrema of $f$ (where $y_{2}^{''} = f^{'} = 0$ and it changes sign): at $x \approx 1$ (maximum of $f$ , $y_{2}$ changes from concave up to concave down) and at $x \approx 3.3$ (minimum of $f$ , it changes from concave down to concave up).

Graph of y₂ = integral of f(t) from 0 to x: starts at the origin with a horizontal tangent, rises to a maximum at x = 2 and decreases afterwards; with two points of inflexion, one near x = 1 and another near x = 3.3

Answer: graph of $y_{2}$ with a horizontal tangent at the origin, a maximum at $x = 2$ and points of inflexion at $x \approx 1$ and $x \approx 3.3$ .

Exercise 161 MediumIB · P1 TZ1 May 2012

The graphs of $y = ∣ x + 1∣$ and $y = ∣ x - 3∣$ are shown below.

Graphs of y = |x + 1| (vertex at (−1, 0)) and y = |x − 3| (vertex at (3, 0)) on a grid from −6 to 6 on both axes; they intersect at (1, 2)

Let $f (x) = ∣ x + 1∣ - ∣ x - 3∣$ .

a) Sketch the graph of $y = f (x)$ .

b) Hence, state the value of

(i) $f^{'} (- 3)$ ;

(ii) $f^{'} (2.7)$ ;

(iii) $\int_{- 3}^{- 2} f (x) d x$

View solution

a) 1. We study $f (x) = ∣ x + 1∣ - ∣ x - 3∣$ piece by piece, according to the sign of $x + 1$ and $x - 3$ . 2. If $x < - 1$ : $∣ x + 1∣ = - (x + 1)$ and $∣ x - 3∣ = 3 - x$ , so $f (x) = - x - 1 - 3 + x = - 4$ (horizontal line). 3. If $- 1 \leq x \leq 3$ : $∣ x + 1∣ = x + 1$ and $∣ x - 3∣ = 3 - x$ , so $f (x) = x + 1 - 3 + x = 2 x - 2$ (straight line of gradient $2$ , from $(- 1, - 4)$ to $(3, 4)$ ). 4. If $x > 3$ : $∣ x + 1∣ = x + 1$ and $∣ x - 3∣ = x - 3$ , so $f (x) = 4$ (horizontal line). 5. The graph is horizontal at $y = - 4$ up to $x = - 1$ , rises with gradient $2$ to $(3, 4)$ , and continues horizontally at $y = 4$ .

Graph of y = f(x) = |x + 1| − |x − 3|: constant equal to −4 up to x = −1, straight line of gradient 2 between (−1, −4) and (3, 4), and constant equal to 4 from x = 3

b) i) 1. At $x = - 3$ we are on the horizontal section $y = - 4$ , which has zero gradient: $f^{'} (- 3) = 0$ .

(ii) 1. $x = 2.7$ lies on the section of gradient $2$ (between $- 1$ and $3$ ): $f^{'} (2.7) = 2$ .

(iii) 1. On $[- 3, - 2]$ the function equals $- 4$ , so the integral is the signed area of a rectangle of base $1$ and height $- 4$ . 2. $\int_{- 3}^{- 2} f (x) d x = - 4 \cdot 1 = - 4$ .

Answer: b) i) $f^{'} (- 3) = 0$ ; ii) $f^{'} (2.7) = 2$ ; iii) $\int_{- 3}^{- 2} f (x) d x = - 4$

Exercise 162 MediumIB · P1 TZ1 May 2014

The first set of axes shows the graph of $y = f (x)$ for $- 4 \leq x \leq 4$ .

First set of axes with the graph of f: horizontal at y = −1 from x = −4 to x = 0, and a straight line of gradient 1 from (0, −1) to (4, 3); second set of axes, empty, with the same scale

Let $g (x) = \int_{- 4}^{x} f (t) d t$ for $- 4 \leq x \leq 4$ .

a) State the value of $x$ for which $g (x)$ has a minimum.

b) On the second set of axes, sketch the graph of $y = g (x)$ .

View solution

a) 1. By the fundamental theorem of calculus, $g^{'} (x) = f (x)$ . 2. A minimum requires $g^{'} = f = 0$ with a change of sign from negative to positive. $f < 0$ for $x < 1$ and $f > 0$ for $x > 1$ , and $f (1) = 0$ . 3. The minimum of $g$ is at $x = 1$ .

b) 1. For $- 4 \leq x \leq 0$ : $f (t) = - 1$ , so $g (x) = \int_{- 4}^{x} (- 1) d t = - (x + 4) = - x - 4$ (from $g (- 4) = 0$ to $g (0) = - 4$ , a straight line of gradient $- 1$ ). 2. For $0 < x \leq 4$ : $f (t) = t - 1$ , so $g (x) = g (0) + \int_{0}^{x} (t - 1) d t = - 4 + \frac{x ^{2}}{2} - x$ . 3. The minimum value is $g (1) = - 4 + \frac{1}{2} - 1 = - \frac{9}{2}$ , and $g (4) = - 4 + 8 - 4 = 0$ . 4. The graph is a segment from $(- 4, 0)$ to $(0, - 4)$ , followed by an arc of a parabola from $(0, - 4)$ to $(4, 0)$ with minimum at $(1, - \frac{9}{2})$ .

Graph of y = g(x) for −4 ≤ x ≤ 4: segment from (−4, 0) to (0, −4) and parabolic arc from (0, −4) to (4, 0) with minimum at (1, −9/2)

Answer: a) $x = 1$ ; b) $g (x) = - x - 4$ if $- 4 \leq x \leq 0$ and $g (x) = - 4 + \frac{x ^{2}}{2} - x$ if $0 < x \leq 4$

Exercise 163 HardIB · P1 TZ1 May 2010Proof

a) Given that $α > 1$ , use the substitution $u = \frac{1}{x}$ to show that $\int_{1}^{α} \frac{1}{1 + x ^{2}} d x = \int_{\frac{1}{α}}^{1} \frac{1}{1 + u ^{2}} d u$ .

b) Using the above, show that $arctan α + arctan \frac{1}{α} = \frac{π}{2}$ .

Show answer

Answer: proved: $arctan α + arctan \frac{1}{α} = \frac{π}{2}$ .

Show working and answer

a) 1. Substitution $u = \frac{1}{x}$ , that is $x = \frac{1}{u}$ , so $d x = - \frac{1}{u ^{2}} d u$ . 2. Limits: if $x = 1$ , $u = 1$ ; if $x = α$ , $u = \frac{1}{α}$ . 3. The integrand: $\frac{1}{1 + x ^{2}} = \frac{1}{1 + \frac{1}{u ^{2}}} = \frac{u ^{2}}{u ^{2} + 1}$ . 4. Therefore $\int_{1}^{α} \frac{d x}{1 + x ^{2}} = \int_{1}^{\frac{1}{α}} \frac{u ^{2}}{u ^{2} + 1} (- \frac{1}{u ^{2}}) d u = - \int_{1}^{\frac{1}{α}} \frac{d u}{1 + u ^{2}}$ . 5. Swapping the limits changes the sign, so $= \int_{\frac{1}{α}}^{1} \frac{d u}{1 + u ^{2}}$ , as required.

b) 1. We evaluate the left-hand side: $\int_{1}^{α} \frac{d x}{1 + x ^{2}} = [arctan x]_{1}^{α} = arctan α - \frac{π}{4}$ . 2. We evaluate the right-hand side: $\int_{\frac{1}{α}}^{1} \frac{d u}{1 + u ^{2}} = [arctan u]_{\frac{1}{α}}^{1} = \frac{π}{4} - arctan \frac{1}{α}$ . 3. By part (a) they are equal: $arctan α - \frac{π}{4} = \frac{π}{4} - arctan \frac{1}{α}$ . 4. Moving $arctan \frac{1}{α}$ to the left and $\frac{π}{4}$ to the right: $arctan α + arctan \frac{1}{α} = \frac{π}{2}$ .

Answer: proved: $arctan α + arctan \frac{1}{α} = \frac{π}{2}$ .

Exercise 164 HardIB · P3 TZ0 May 2013

a) Figure 1 shows part of the graph of $y = \frac{1}{x}$ , together with line segments parallel to the coordinate axes.

Part of the graph of y = 1/x with two rectangles above and two below the curve over the intervals (a − 1, a) and (a, a + 1)

(i) Starting from the areas of suitable rectangles, verify that: $\frac{2 a + 1}{a ( a + 1 )} < ln (\frac{a + 1}{a - 1}) < \frac{2 a - 1}{a ( a - 1 )}$ .

(ii) Hence, find a lower bound and an upper bound for $ln (1.2)$ .

b) An improved value for the upper bound can be obtained by considering Figure 2, which again shows part of the graph of $y = \frac{1}{x}$ .

Part of the graph of y = 1/x with the vertical lines at x = a − 1 and x = a up to the curve, and the segment joining the two points on the curve

(i) Starting from the areas of suitable regions, verify that: $ln (\frac{a}{a - 1}) < \frac{2 a - 1}{2 a ( a - 1 )}$ .

(ii) Hence, find an upper bound for $ln (1.2)$ .

Show answer

Answer: a) (i) verified; (ii) $0.174 < ln (1.2) < 0.191$ ; b) (i) verified; (ii) $ln (1.2) < 0.183$

Show working and answer

a) (i) 1. Let $a > 1$ . As $y = \frac{1}{x}$ is decreasing, on $[a - 1, a]$ the smallest value of the curve is $\frac{1}{a}$ and the largest is $\frac{1}{a - 1}$ ; on $[a, a + 1]$ the smallest is $\frac{1}{a + 1}$ and the largest is $\frac{1}{a}$ . 2. The rectangles of the smaller height lie below the curve and those of the greater height lie above it. Each rectangle has base $1$ , so its area equals its height. 3. Area below: $\frac{1}{a} + \frac{1}{a + 1}$ . Area above: $\frac{1}{a - 1} + \frac{1}{a}$ . 4. The area under the curve between $a - 1$ and $a + 1$ lies between these two: $\frac{1}{a} + \frac{1}{a + 1} < \int_{a - 1}^{a + 1} \frac{1}{x} d x < \frac{1}{a - 1} + \frac{1}{a}$ . 5. The integral equals $ln (a + 1) - ln (a - 1) = ln \frac{a + 1}{a - 1}$ , and the sums are $\frac{2 a + 1}{a ( a + 1 )}$ and $\frac{2 a - 1}{a ( a - 1 )}$ . Verified.

(ii) 1. We impose $\frac{a + 1}{a - 1} = 1.2$ : $a + 1 = 1.2 a - 1.2$ , so $a = 11$ . 2. Lower bound: $\frac{2 \cdot 11 + 1}{11 \cdot 12} = \frac{23}{132} \approx 0.174$ . Upper bound: $\frac{2 \cdot 11 - 1}{11 \cdot 10} = \frac{21}{110} \approx 0.191$ . 3. Hence $0.174 < ln (1.2) < 0.191$ .

b) (i) 1. On $[a - 1, a]$ the curve $y = \frac{1}{x}$ is convex, so the chord joining $(a - 1, \frac{1}{a - 1})$ and $(a, \frac{1}{a})$ lies above the curve. 2. The area under the chord is that of a trapezium with parallel sides $\frac{1}{a - 1}$ and $\frac{1}{a}$ and width $1$ : $\frac{1}{2} (\frac{1}{a - 1} + \frac{1}{a})$ . 3. Therefore $\int_{a - 1}^{a} \frac{1}{x} d x < \frac{1}{2} (\frac{1}{a - 1} + \frac{1}{a})$ . 4. The left-hand side is $ln \frac{a}{a - 1}$ and the right-hand side is $\frac{1}{2} \cdot \frac{a + a - 1}{a ( a - 1 )} = \frac{2 a - 1}{2 a ( a - 1 )}$ . Verified.

(ii) 1. $\frac{a}{a - 1} = 1.2 \Rightarrow a = 1.2 a - 1.2 \Rightarrow a = 6$ . 2. $ln (1.2) < \frac{2 \cdot 6 - 1}{2 \cdot 6 \cdot 5} = \frac{11}{60} \approx 0.183$ .

Answer: a) (i) verified; (ii) $0.174 < ln (1.2) < 0.191$ ; b) (i) verified; (ii) $ln (1.2) < 0.183$

Exercise 165 HardIB · P1 TZ0 November 2013

Consider the complex number $z = cos θ + i sin θ$ .

a) Use De Moivre's theorem to verify that $z^{n} + z^{- n} = 2 cos n θ$ , $n \in Z^{+}$ .

b) Expand $(z + z^{- 1})^{4}$ .

c) Hence, verify that $cos^{4} θ = p cos 4 θ + q cos 2 θ + r$ , where $p$ , $q$ and $r$ are constants to be determined.

d) Verify that $cos^{6} θ = \frac{1}{32} cos 6 θ + \frac{3}{16} cos 4 θ + \frac{15}{32} cos 2 θ + \frac{5}{16}$ .

e) Hence, find the value of $\int_{0}^{π /2} cos^{6} θ d θ$ .

The region $S$ is enclosed by the curve $y = sin x cos^{2} x$ and the $x$ -axis, between $x = 0$ and $x = \frac{π}{2}$ .

f) $S$ is rotated through $2 π$ radians about the $x$ -axis. Find the volume generated.

g) i) Write down an expression for the constant term in the expansion of $(z + z^{- 1})^{2 k}$ , $k \in Z^{+}$ .

(ii) Hence, determine an expression for $\int_{0}^{π /2} cos^{2 k} θ d θ$ in terms of $k$ .

Show answer

Answer: a) verified; b) $(z + z^{- 1})^{4} = z^{4} + 4 z^{2} + 6 + 4 z^{- 2} + z^{- 4}$ ; c) $p = \frac{1}{8}$ , $q = \frac{1}{2}$ , $r = \frac{3}{8}$ ; d) verified; e) $\frac{5 π}{32}$ ; f) $V = \frac{π ^{2}}{32} \approx 0.308$ ; g) i) $(k 2 k)$ ; ii) $\frac{( k 2 k )}{2 ^{2 k}} \cdot \frac{π}{2}$

Show working and answer

a) 1. By De Moivre's theorem, $z^{n} = cos n θ + i sin n θ$ and $z^{- n} = cos (- n θ) + i sin (- n θ) = cos n θ - i sin n θ$ . 2. Adding: $z^{n} + z^{- n} = 2 cos n θ$ .

b) Using the binomial theorem: $(z + z^{- 1})^{4} = z^{4} + 4 z^{2} + 6 + 4 z^{- 2} + z^{- 4} .$

c) 1. From a) with $n = 1$ , $z + z^{- 1} = 2 cos θ$ , so $(z + z^{- 1})^{4} = 16 cos^{4} θ$ . 2. Grouping the expansion from b): $(z^{4} + z^{- 4}) + 4 (z^{2} + z^{- 2}) + 6 = 2 cos 4 θ + 8 cos 2 θ + 6$ . 3. Equating: $16 cos^{4} θ = 2 cos 4 θ + 8 cos 2 θ + 6$ , that is, $cos^{4} θ = \frac{1}{8} cos 4 θ + \frac{1}{2} cos 2 θ + \frac{3}{8}$ . 4. Therefore $p = \frac{1}{8}$ , $q = \frac{1}{2}$ , $r = \frac{3}{8}$ .

d) 1. $(z + z^{- 1})^{6} = z^{6} + 6 z^{4} + 15 z^{2} + 20 + 15 z^{- 2} + 6 z^{- 4} + z^{- 6} = 2 cos 6 θ + 12 cos 4 θ + 30 cos 2 θ + 20$ . 2. As $(z + z^{- 1})^{6} = 64 cos^{6} θ$ , we divide by $64$ : $cos^{6} θ = \frac{1}{32} cos 6 θ + \frac{3}{16} cos 4 θ + \frac{15}{32} cos 2 θ + \frac{5}{16}$ .

e) 1. Integrating the result from d): $\int_{0}^{π /2} cos 6 θ d θ = [\frac{sin 6 θ}{6}]_{0}^{π /2} = 0$ ; likewise the integrals of $cos 4 θ$ and $cos 2 θ$ vanish (the sines of $2 π$ and $π$ are $0$ ). 2. Only the constant term remains: $\frac{5}{16} \cdot \frac{π}{2} = \frac{5 π}{32}$ .

f) 1. Volume of revolution about the $x$ -axis: $V = π \int_{0}^{π /2} (sin x cos^{2} x)^{2} d x = π \int_{0}^{π /2} sin^{2} x cos^{4} x d x$ . 2. Writing $sin^{2} x = 1 - cos^{2} x$ : $V = π (\int_{0}^{π /2} cos^{4} x d x - \int_{0}^{π /2} cos^{6} x d x)$ . 3. By c), $\int_{0}^{π /2} cos^{4} x d x = \frac{3}{8} \cdot \frac{π}{2} = \frac{3 π}{16}$ (again only the constant term survives). By e), the second integral is $\frac{5 π}{32}$ . 4. $V = π (\frac{6 π}{32} - \frac{5 π}{32}) = \frac{π ^{2}}{32} \approx 0.308$ .

g) i) The constant term of $(z + z^{- 1})^{2 k}$ appears when we choose $z$ $k$ times and $z^{- 1}$ $k$ times: it is $(k 2 k)$ .

(ii) 1. $(2 cos θ)^{2 k} = (z + z^{- 1})^{2 k}$ . The other terms group in pairs $z^{m} + z^{- m} = 2 cos m θ$ (with $m$ even), whose integral between $0$ and $\frac{π}{2}$ is $0$ . 2. Hence $2^{2 k} \int_{0}^{π /2} cos^{2 k} θ d θ = (k 2 k) \cdot \frac{π}{2}$ . 3. Therefore $\int_{0}^{π /2} cos^{2 k} θ d θ = \frac{( k 2 k )}{2 ^{2 k}} \cdot \frac{π}{2}$ .

Exercise 166 HardIB · P1 TZ0 November 2014

a) i) Show that $(1 + i tan θ)^{n} + (1 - i tan θ)^{n} = \frac{2 cos n θ}{cos ^{n} θ}$ , $cos θ \neq = 0$ .

(ii) Hence, verify that $i tan \frac{3 π}{8}$ is a root of the equation $(1 + z)^{4} + (1 - z)^{4} = 0$ , $z \in C$ .

(iii) State another root of the equation $(1 + z)^{4} + (1 - z)^{4} = 0$ , $z \in C$ .

b) i) Use the double angle identity $tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ to show that $tan \frac{π}{8} = 2 - 1$ .

(ii) Show that $cos 4 x = 8 cos^{4} x - 8 cos^{2} x + 1$ .

(iii) Hence, find the value of $\int_{0}^{π /8} \frac{2 cos 4 x}{cos ^{2} x} d x$ .

Show answer

Answer: a) (i) and (ii) verified; (iii) another root $- i tan \frac{3 π}{8}$ ; b) (i) and (ii) verified; (iii) $42 - 2 - π$

Show working and answer

a) (i) 1. We write $1 + i tan θ = \frac{cos θ + i sin θ}{cos θ}$ and $1 - i tan θ = \frac{cos θ - i sin θ}{cos θ}$ . 2. By De Moivre: $(1 + i tan θ)^{n} = \frac{cos n θ + i sin n θ}{cos ^{n} θ}$ and $(1 - i tan θ)^{n} = \frac{cos n θ - i sin n θ}{cos ^{n} θ}$ . 3. Adding: $\frac{2 cos n θ}{cos ^{n} θ}$ .

(ii) 1. With $n = 4$ and $θ = \frac{3 π}{8}$ (where $cos θ \neq = 0$ ): $(1 + i tan θ)^{4} + (1 - i tan θ)^{4} = \frac{2 cos \frac{3 π}{2}}{cos ^{4} θ} = 0$ . 2. With $z = i tan \frac{3 π}{8}$ this is $(1 + z)^{4} + (1 - z)^{4} = 0$ , so $z$ is a root.

(iii) The coefficients of the equation are real, so the complex conjugate is also a root: $z = - i tan \frac{3 π}{8}$ . (This is also clear because the equation contains only even powers of $z$ .)

b) (i) 1. With $θ = \frac{π}{8}$ and $t = tan \frac{π}{8}$ : $tan \frac{π}{4} = 1 = \frac{2 t}{1 - t ^{2}}$ . 2. $1 - t^{2} = 2 t$ , that is, $t^{2} + 2 t - 1 = 0$ , so $t = - 1 \pm 2$ . 3. As $\frac{π}{8}$ lies in the first quadrant, $t > 0$ : $tan \frac{π}{8} = 2 - 1$ .

(ii) $cos 4 x = 2 cos^{2} 2 x - 1 = 2 (2 cos^{2} x - 1)^{2} - 1 = 2 (4 cos^{4} x - 4 cos^{2} x + 1) - 1 = 8 cos^{4} x - 8 cos^{2} x + 1$ .

(iii) 1. By (ii): $\frac{2 cos 4 x}{cos ^{2} x} = \frac{16 cos ^{4} x - 16 cos ^{2} x + 2}{cos ^{2} x} = 16 cos^{2} x - 16 + \frac{2}{cos ^{2} x}$ . 2. Using $cos^{2} x = \frac{1 + cos 2 x}{2}$ : $= 8 cos 2 x - 8 + 2 sec^{2} x$ . 3. $\int_{0}^{π /8} (8 cos 2 x - 8 + 2 sec^{2} x) d x = [4 sin 2 x - 8 x + 2 tan x]_{0}^{π /8}$ . 4. $= 4 \cdot \frac{2}{2} - π + 2 (2 - 1) = 42 - 2 - π$ (using b) (i)).

Answer: a) (i) and (ii) verified; (iii) another root $- i tan \frac{3 π}{8}$ ; b) (i) and (ii) verified; (iii) $42 - 2 - π$

Exercise 167 HardIB · P3 TZ0 November 2018

a) Use L'Hôpital's rule to determine the value of $x \to 0 lim (\frac{e ^{- 3 x^{2}} + 3 cos ( 2 x ) - 4}{3 x ^{2}})$ .

b) Hence, find $x \to 0 lim \frac{\int _{0}^{x} ( e ^{- 3 t^{2}} + 3 cos ( 2 t ) - 4 ) d t}{\int _{0}^{x} 3 t ^{2} d t}$ .

Show answer

Answer: a) $- 3$ ; b) $- 3$

Show working and answer

a) 1. As $x \to 0$ , the numerator tends to $1 + 3 - 4 = 0$ and the denominator tends to $0$ : form $\frac{0}{0}$ . We apply L'Hôpital's rule. 2. Derivatives: numerator $- 6 x e^{- 3 x^{2}} - 6 sin 2 x$ ; denominator $6 x$ . The quotient is still $\frac{0}{0}$ at $x = 0$ . 3. We apply L'Hôpital's rule again. Numerator: $- 6 e^{- 3 x^{2}} + 36 x^{2} e^{- 3 x^{2}} - 12 cos 2 x$ ; denominator: $6$ . 4. At $x = 0$ : $\frac{- 6 + 0 - 12}{6} = - 3$ .

b) 1. The numerator and denominator are integrals with upper limit $x$ , and both tend to $0$ as $x \to 0$ : form $\frac{0}{0}$ . 2. By the fundamental theorem of calculus, the derivative of the numerator is $e^{- 3 x^{2}} + 3 cos 2 x - 4$ and that of the denominator is $3 x^{2}$ . 3. Applying L'Hôpital's rule, the limit is $x \to 0 lim \frac{e ^{- 3 x^{2}} + 3 cos 2 x - 4}{3 x ^{2}}$ , which is the limit from part a). 4. It equals $- 3$ .

Answer: a) $- 3$ ; b) $- 3$

Exercise 168 HardIB · P3 TZ0 November 2006

a) i) Calculate $I_{n} = \int_{- n}^{α n} \frac{x}{1 + x ^{2}} d x$ , where $α$ is a positive constant and $n$ is a positive integer.

(ii) Calculate $n \to \infty lim I_{n}$ .

b) Using l'Hôpital's rule, calculate $x \to 0 lim (\frac{tan β x - β tan x}{sin β x - β sin x})$ , where $β$ is a non-zero constant that is not equal to $\pm 1$ .

Show answer

Answer: a) i) $I_{n} = \frac{1}{2} ln (\frac{1 + α ^{2} n ^{2}}{1 + n ^{2}})$ ; ii) $ln α$ ; b) $- 2$

Show working and answer

a) i) 1. Substitution $u = 1 + x^{2}$ , $d u = 2 x d x$ : $\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (1 + x^{2})$ . 2. $I_{n} = \frac{1}{2} [ln (1 + α^{2} n^{2}) - ln (1 + n^{2})] = \frac{1}{2} ln (\frac{1 + α ^{2} n ^{2}}{1 + n ^{2}})$ .

(ii) 1. Dividing numerator and denominator by $n^{2}$ : $\frac{1 + α ^{2} n ^{2}}{1 + n ^{2}} = \frac{\frac{1}{n ^{2}} + α ^{2}}{\frac{1}{n ^{2}} + 1} \to α^{2}$ . 2. By continuity of the logarithm, $n \to \infty lim I_{n} = \frac{1}{2} ln α^{2} = ln α$ (with $α > 0$ ).

b) 1. It is of the form $\frac{0}{0}$ at $x = 0$ . We apply l'Hôpital's rule; let $N = tan β x - β tan x$ and $D = sin β x - β sin x$ . 2. $N^{'} = β sec^{2} β x - β sec^{2} x$ and $D^{'} = β cos β x - β cos x$ : it is still $\frac{0}{0}$ . 3. $N^{''} = 2 β^{2} sec^{2} β x tan β x - 2 β sec^{2} x tan x$ and $D^{''} = - β^{2} sin β x + β sin x$ : again $\frac{0}{0}$ . 4. $N^{'''} = 2 β^{3} (2 sec^{2} β x tan^{2} β x + sec^{4} β x) - 2 β (2 sec^{2} x tan^{2} x + sec^{4} x)$ , which at $x = 0$ equals $2 β^{3} - 2 β$ . $D^{'''} = - β^{3} cos β x + β cos x$ , which at $x = 0$ equals $- β^{3} + β$ . 5. Limit $= \frac{2 β ^{3} - 2 β}{- β ^{3} + β} = \frac{2 β ( β ^{2} - 1 )}{- β ( β ^{2} - 1 )} = - 2$ (we can simplify because $β \neq = 0, \pm 1$ ).

Answer: a) i) $I_{n} = \frac{1}{2} ln (\frac{1 + α ^{2} n ^{2}}{1 + n ^{2}})$ ; ii) $ln α$ ; b) $- 2$

Exercise 169 HardIB · P1 TZ1 May 2013

a) Find all values of $x$ , $0.1 \leq x \leq 1$ , for which $sin (π x^{- 1}) = 0$ .

b) Calculate $\int_{\frac{1}{n + 1}}^{\frac{1}{n}} π x^{- 2} sin (π x^{- 1}) d x$ , showing that it takes different integer values when $n$ is even and when $n$ is odd.

c) Evaluate $\int_{0.1}^{1} π x^{- 2} sin (π x^{- 1}) d x$ .

Show answer

Answer: a) $x = 1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{10}$ ; b) $2$ if $n$ is even, $- 2$ if $n$ is odd; c) $18$

Show working and answer

a) 1. $sin (\frac{π}{x}) = 0$ when $\frac{π}{x} = mπ$ with $m \in Z$ , i.e. $x = \frac{1}{m}$ . 2. We impose $0.1 \leq \frac{1}{m} \leq 1$ , i.e. $1 \leq m \leq 10$ . 3. The solutions are $x = 1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{10}$ .

b) 1. Method: substitution $u = π x^{- 1}$ , so that $d u = - π x^{- 2} d x$ . 2. Limits: if $x = \frac{1}{n + 1}$ then $u = (n + 1) π$ ; if $x = \frac{1}{n}$ then $u = nπ$ . 3. $\int_{\frac{1}{n + 1}}^{\frac{1}{n}} π x^{- 2} sin (π x^{- 1}) d x = \int_{(n + 1) π}^{nπ} - sin u d u = \int_{nπ}^{(n + 1) π} sin u d u$ . 4. $= [- cos u]_{nπ}^{(n + 1) π} = - cos ((n + 1) π) + cos (nπ) = (- 1)^{n} + (- 1)^{n} = 2 (- 1)^{n}$ . 5. If $n$ is even it equals $2$ , and if $n$ is odd it equals $- 2$ : two different integers.

c) 1. The interval $[0.1, 1]$ is split into the intervals $[\frac{1}{n + 1}, \frac{1}{n}]$ for $n = 1, 2, \dots, 9$ . 2. On each of them the sine does not change sign (its endpoints are consecutive zeros), so $\int ∣ \dots ∣ d x = \int \dots d x = ∣2 (- 1)^{n} ∣ = 2$ . 3. Adding the nine intervals: $9 \cdot 2 = 18$ .

Answer: a) $x = 1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{10}$ ; b) $2$ if $n$ is even, $- 2$ if $n$ is odd; c) $18$

Exercise 170 HardIB · P3 TZ0 November 2015

Consider the function $f (x) = \frac{1}{1 + x ^{2}}$ , $x \in R$ .

a) Illustrate graphically the inequality $\frac{1}{5} r = 1 \sum 5 f (\frac{r}{5}) < \int_{0}^{1} f (x) d x < \frac{1}{5} r = 0 \sum 4 f (\frac{r}{5})$ .

b) Use the inequality in part (a) to find a lower bound and an upper bound for $π$ .

c) Show that $r = 0 \sum n - 1 (- 1)^{r} x^{2 r} = \frac{1 + ( - 1 ) ^{n - 1} x ^{2 n}}{1 + x ^{2}}$ .

d) Hence, show that $π = 4 (r = 0 \sum n - 1 \frac{( - 1 ) ^{r}}{2 r + 1} - (- 1)^{n - 1} \int_{0}^{1} \frac{x ^{2 n}}{1 + x ^{2}} d x)$ .

View solution

a) 1. $f (x) = \frac{1}{1 + x ^{2}}$ is decreasing on $[0, 1]$ . We divide $[0, 1]$ into five subintervals of width $\frac{1}{5}$ and draw rectangles on them. 2. If the height of each rectangle is the value of $f$ at the right-hand endpoint of its subinterval ( $f (\frac{r}{5})$ , $r = 1, \dots, 5$ ), the rectangles lie below the curve: their sum is the lower bound. 3. If the height is the value at the left-hand endpoint ( $f (\frac{r}{5})$ , $r = 0, \dots, 4$ ), the rectangles stick out above the curve: their sum is the upper bound. 4. Therefore, the sum of the five lower rectangles is less than $\int_{0}^{1} f (x) d x$ , and this is less than the sum of the five upper rectangles.

Graph of f(x) = 1/(1 + x²) on (0, 1) with five lower and five upper rectangles of base 1/5

b) 1. $\int_{0}^{1} \frac{1}{1 + x ^{2}} d x = [arctan x]_{0}^{1} = \frac{π}{4}$ . 2. Lower sum: $\frac{1}{5} (0.9615 + 0.8621 + 0.7353 + 0.6098 + 0.5) \approx 0.7337$ . 3. Upper sum: $\frac{1}{5} (1 + 0.9615 + 0.8621 + 0.7353 + 0.6098) \approx 0.8337$ . 4. Hence $0.7337 < \frac{π}{4} < 0.8337$ , and multiplying by $4$ : $2.9349 < π < 3.3349$ .

c) 1. This is the sum of a geometric sequence of $n$ terms, with first term $1$ and common ratio $- x^{2}$ . 2. $r = 0 \sum n - 1 (- x^{2})^{r} = \frac{1 - ( - x ^{2} ) ^{n}}{1 - ( - x ^{2} )} = \frac{1 - ( - 1 ) ^{n} x ^{2 n}}{1 + x ^{2}}$ . 3. As $- (- 1)^{n} = (- 1)^{n - 1}$ , this becomes $\frac{1 + ( - 1 ) ^{n - 1} x ^{2 n}}{1 + x ^{2}}$ . Also $(- x^{2})^{r} = (- 1)^{r} x^{2 r}$ .

d) 1. Integrate the result of (c) between $0$ and $1$ : $r = 0 \sum n - 1 (- 1)^{r} \int_{0}^{1} x^{2 r} d x = \int_{0}^{1} \frac{1}{1 + x ^{2}} d x + (- 1)^{n - 1} \int_{0}^{1} \frac{x ^{2 n}}{1 + x ^{2}} d x$ . 2. Since $\int_{0}^{1} x^{2 r} d x = \frac{1}{2 r + 1}$ and the first integral on the right is $\frac{π}{4}$ : $r = 0 \sum n - 1 \frac{( - 1 ) ^{r}}{2 r + 1} = \frac{π}{4} + (- 1)^{n - 1} \int_{0}^{1} \frac{x ^{2 n}}{1 + x ^{2}} d x$ . 3. Rearranging for $π$ : $π = 4 (r = 0 \sum n - 1 \frac{( - 1 ) ^{r}}{2 r + 1} - (- 1)^{n - 1} \int_{0}^{1} \frac{x ^{2 n}}{1 + x ^{2}} d x)$ .

Answer: b) $2.9349 < π < 3.3349$ ; c) and d) proved.

Exercise 171 HardIB · P3 TZ0 November 2013

Let $g (x) = sin (x^{2})$ , where $x \in R$ .

a) Using the result $t \to 0 lim \frac{sin t}{t} = 1$ , or otherwise, calculate $x \to 0 lim \frac{g ( 2 x ) - g ( 3 x )}{4 x ^{2}}$ .

b) Use the Maclaurin series for $sin x$ to verify that $g (x) = n = 0 \sum \infty (- 1)^{n} \frac{x ^{4 n + 2}}{( 2 n + 1 )!}$ .

c) Hence, determine the minimum number of terms of the expansion of $g (x)$ needed to approximate the value of $\int_{0}^{1} g (x) d x$ correct to four decimal places.

Show answer

Answer: a) $- \frac{5}{4}$ ; c) $3$ terms ( $n = 0, 1, 2$ ), giving $\int_{0}^{1} g (x) d x \approx 0.3103$

Show working and answer

a) 1. $g (2 x) = sin (4 x^{2})$ and $g (3 x) = sin (9 x^{2})$ . 2. We separate: $\frac{sin 4 x ^{2} - sin 9 x ^{2}}{4 x ^{2}} = \frac{sin 4 x ^{2}}{4 x ^{2}} - \frac{9}{4} \cdot \frac{sin 9 x ^{2}}{9 x ^{2}}$ . 3. As $x \to 0$ , $t = 4 x^{2} \to 0$ and $t = 9 x^{2} \to 0$ , so each quotient tends to $1$ . 4. The limit is $1 - \frac{9}{4} = - \frac{5}{4}$ .

b) 1. $sin t = n = 0 \sum \infty (- 1)^{n} \frac{t ^{2 n + 1}}{( 2 n + 1 )!}$ . 2. Substituting $t = x^{2}$ : $sin x^{2} = n = 0 \sum \infty (- 1)^{n} \frac{( x ^{2} ) ^{2 n + 1}}{( 2 n + 1 )!} = n = 0 \sum \infty (- 1)^{n} \frac{x ^{4 n + 2}}{( 2 n + 1 )!}$ .

c) 1. Integrate term by term: $\int_{0}^{1} g (x) d x = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{( 2 n + 1 )! ( 4 n + 3 )}$ . 2. The terms are: $n = 0$ : $\frac{1}{3} \approx 0.333333$ ; $n = 1$ : $- \frac{1}{42} \approx - 0.023810$ ; $n = 2$ : $\frac{1}{1320} \approx 0.000758$ ; $n = 3$ : $- \frac{1}{75600} \approx - 0.0000132$ . 3. This is an alternating series with decreasing terms: the error when stopping is less than the first omitted term. To have four correct decimal places, an error less than $0.00005$ is sufficient. 4. With two terms the error can reach $0.00076$ (not sufficient). With three terms ( $n = 0, 1, 2$ ) the error is less than $0.0000132 < 0.00005$ : $0.333333 - 0.023810 + 0.000758 \approx 0.3103$ . 5. $3$ terms of the expansion are needed.

Answer: a) $- \frac{5}{4}$ ; c) $3$ terms ( $n = 0, 1, 2$ ), giving $\int_{0}^{1} g (x) d x \approx 0.3103$

Exercise 172 HardIB · P3 TZ0 May 2017Proof

Consider the curve $y = \frac{1}{x}$ , $x > 0$ .

a) By drawing an accurate diagram and considering the area under the curve of a suitable region, show that for $r > 0$ , $\frac{1}{r + 1} < ln (\frac{r + 1}{r}) < \frac{1}{r}$ .

b) Hence, given that $n$ is a positive integer greater than one, show that

(i) $r = 1 \sum n \frac{1}{r} > ln (1 + n)$

(ii) $r = 1 \sum n \frac{1}{r} < 1 + ln n$

Let $U_{n} = r = 1 \sum n \frac{1}{r} - ln n$ .

c) Hence, given that $n$ is a positive integer greater than one, show that

(i) $U_{n} > 0$

(ii) $U_{n + 1} < U_{n}$

d) Explain why these two results prove that ${U_{n}}$ is a convergent sequence.

Show answer

Answer: proved.

Show working and answer

a) 1. For $x > 0$ , $y = \frac{1}{x}$ is decreasing. Over the interval $[r, r + 1]$ the area under the curve is $\int_{r}^{r + 1} \frac{1}{x} d x = ln (r + 1) - ln r = ln \frac{r + 1}{r}$ . 2. The rectangle of base $1$ and height $\frac{1}{r + 1}$ lies below the curve; the rectangle of base $1$ and height $\frac{1}{r}$ lies above it. 3. Therefore, $\frac{1}{r + 1} < ln \frac{r + 1}{r} < \frac{1}{r}$ .

Graph of y = 1/x with two rectangles of base 1 between x = r and x = r + 1: one of height 1/(r+1) below the curve and one of height 1/r above it

b) (i) 1. We use the right-hand inequality, $\frac{1}{r} > ln \frac{r + 1}{r} = ln (r + 1) - ln r$ , for $r = 1, \dots, n$ and add. 2. The sum on the right is telescoping: $r = 1 \sum n [ln (r + 1) - ln r] = ln (n + 1) - ln 1 = ln (n + 1)$ . 3. Hence $r = 1 \sum n \frac{1}{r} > ln (1 + n)$ .

(ii) 1. We use the left-hand inequality, $\frac{1}{r + 1} < ln (r + 1) - ln r$ , for $r = 1, \dots, n - 1$ and add: $r = 1 \sum n - 1 \frac{1}{r + 1} < ln n - ln 1 = ln n$ . 2. The left-hand side is $r = 2 \sum n \frac{1}{r} = r = 1 \sum n \frac{1}{r} - 1$ . 3. Therefore $r = 1 \sum n \frac{1}{r} - 1 < ln n$ , that is, $r = 1 \sum n \frac{1}{r} < 1 + ln n$ .

c) (i) 1. By b)(i), $r = 1 \sum n \frac{1}{r} > ln (n + 1) > ln n$ , because $ln$ is increasing. 2. Subtracting $ln n$ : $U_{n} > 0$ .

(ii) 1. $U_{n + 1} - U_{n} = \frac{1}{n + 1} - ln (n + 1) + ln n = \frac{1}{n + 1} - ln \frac{n + 1}{n}$ . 2. By the left-hand inequality of part a) with $r = n$ , $\frac{1}{n + 1} < ln \frac{n + 1}{n}$ , so $U_{n + 1} - U_{n} < 0$ and $U_{n + 1} < U_{n}$ .

d) 1. By c)(ii), the sequence ${U_{n}}$ is decreasing (monotonic), and by c)(i) it is bounded below by $0$ . 2. A monotonic bounded sequence is convergent.

Answer: proved.

Exercise 173 HardIB · P1 TZ1 May 2022

The function $f$ is defined by $f (x) = e^{x} sin x$ , where $x \in R$ .

a) Find the Maclaurin series for $f (x)$ up to and including the term in $x^{3}$ .

b) Using the above, find an approximate value for $\int_{0}^{1} e^{x^{2}} sin (x^{2}) d x$ .

The function $g$ is defined by $g (x) = e^{x} cos x$ , where $x \in R$ .

c) (i) Show that $g (x)$ satisfies the equation $g^{''} (x) = 2 (g^{'} (x) - g (x))$ .

(ii) Using the above, deduce that $g^{(4)} (x) = 2 (g^{'''} (x) - g^{''} (x))$ .

d) Using the result from part (c), find the Maclaurin series for $g (x)$ up to and including the term in $x^{4}$ .

e) Using the above, or otherwise, determine the value of $x \to 0 lim \frac{e ^{x} cos x - 1 - x}{x ^{3}}$ .

Show answer

Answer: a) $x + x^{2} + \frac{x ^{3}}{3}$ ; b) $\frac{61}{105}$ ; c) shown; d) $1 + x - \frac{x ^{3}}{3} - \frac{x ^{4}}{6}$ ; e) $- \frac{1}{3}$

Show working and answer

a) 1. We multiply the expansions: $e^{x} = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \dots$ and $sin x = x - \frac{x ^{3}}{6} + \dots$ . 2. Terms of $e^{x} sin x$ up to $x^{3}$ : $x$ ; $x \cdot x = x^{2}$ ; $x \cdot \frac{x ^{2}}{2} - \frac{x ^{3}}{6} = \frac{x ^{3}}{3}$ . 3. $f (x) \approx x + x^{2} + \frac{x ^{3}}{3}$ .

b) 1. We substitute $x^{2}$ for $x$ : $e^{x^{2}} sin x^{2} \approx x^{2} + x^{4} + \frac{x ^{6}}{3}$ . 2. $\int_{0}^{1} (x^{2} + x^{4} + \frac{x ^{6}}{3}) d x = \frac{1}{3} + \frac{1}{5} + \frac{1}{21} = \frac{35 + 21 + 5}{105} = \frac{61}{105}$ .

c) (i) 1. $g^{'} (x) = e^{x} cos x - e^{x} sin x = e^{x} (cos x - sin x)$ . 2. $g^{''} (x) = e^{x} (cos x - sin x) + e^{x} (- sin x - cos x) = - 2 e^{x} sin x$ . 3. $g^{'} (x) - g (x) = e^{x} (cos x - sin x - cos x) = - e^{x} sin x$ , so $2 (g^{'} - g) = - 2 e^{x} sin x = g^{''} (x)$ .

(ii) 1. Differentiating $g^{''} = 2 (g^{'} - g)$ : $g^{'''} = 2 (g^{''} - g^{'})$ . 2. Differentiating once more: $g^{(4)} = 2 (g^{'''} - g^{''})$ .

d) 1. $g (0) = 1$ , $g^{'} (0) = 1$ . 2. $g^{''} (0) = 2 (g^{'} (0) - g (0)) = 0$ . 3. $g^{'''} (0) = 2 (g^{''} (0) - g^{'} (0)) = - 2$ . 4. $g^{(4)} (0) = 2 (g^{'''} (0) - g^{''} (0)) = - 4$ . 5. $g (x) \approx 1 + x + \frac{0}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 4}{4 !} x^{4} = 1 + x - \frac{x ^{3}}{3} - \frac{x ^{4}}{6}$ .

e) 1. We substitute: $e^{x} cos x - 1 - x \approx - \frac{x ^{3}}{3} - \frac{x ^{4}}{6}$ . 2. Dividing by $x^{3}$ : $- \frac{1}{3} - \frac{x}{6} \to - \frac{1}{3}$ as $x \to 0$ .

Answer: a) $x + x^{2} + \frac{x ^{3}}{3}$ ; b) $\frac{61}{105}$ ; c) shown; d) $1 + x - \frac{x ^{3}}{3} - \frac{x ^{4}}{6}$ ; e) $- \frac{1}{3}$

Calculating areas

Exercise 174 BasicIB · P1 May 1997

Calculate, in terms of $k$ , the area enclosed by the curve $y = x$ , the $x$ -axis and the line $x = k$ , where $k > 0$ .

Show answer

Answer: $A = \frac{2}{3} k^{\frac{3}{2}}$

Show working and answer

The region lies under $y = x$ (which is positive) between $x = 0$ (where the curve touches the $x$ -axis) and $x = k$ .
$A = \int_{0}^{k} x^{\frac{1}{2}} d x = [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{k}$ .
$A = \frac{2}{3} k^{\frac{3}{2}}$ .

Answer: $A = \frac{2}{3} k^{\frac{3}{2}}$

Exercise 175 MediumIB · P1 May 2001

Let $f (x) = \frac{sin x}{x}$ , $π \leq x \leq 3 π$ . Calculate the area enclosed by the graph of $f$ and the $x$ -axis.

Show answer

Answer: area $\approx 0.690$ (square units)

Show working and answer

The sign of $f (x) = \frac{sin x}{x}$ on $[π, 3 π]$ is given by the sine: it is negative on $] π, 2 π [$ and positive on $] 2 π, 3 π [$ , and it crosses the axis at $x = 2 π$ .
The area is the sum of the absolute values of the two regions: $A = \int_{π}^{2 π} \frac{sin x}{x} d x + \int_{2 π}^{3 π} \frac{sin x}{x} d x$ , or equivalently $A = \int_{π}^{3 π} \frac{sin x}{x} d x$ .
This function has no elementary antiderivative, so using the GDC: $\int_{π}^{2 π} \frac{sin x}{x} d x \approx - 0.433$ and $\int_{2 π}^{3 π} \frac{sin x}{x} d x \approx 0.257$ .
Adding the absolute values: $0.433 + 0.257 \approx 0.690$ .

Answer: area $\approx 0.690$ (square units)

Exercise 176 MediumIB · P1 November 2001

Calculate the area enclosed between the curves $y = \frac{2}{1 + x ^{2}}$ and $y = e^{\frac{x}{3}}$ , given that $- 3 \leq x \leq 3$ .

Show answer

Answer: area $\approx 1.22$

Show working and answer

We look for the intersections of the curves: $\frac{2}{1 + x ^{2}} = e^{\frac{x}{3}}$ . It cannot be solved by hand; using the GDC, on $[- 3, 3]$ there are two solutions: $x_{1} \approx - 1.525$ and $x_{2} \approx 0.748$ .
Between them, the upper curve is $y = \frac{2}{1 + x ^{2}}$ (for example at $x = 0$ it equals $2$ , compared with $e^{0} = 1$ ). The region enclosed by the two curves is the one between $x_{1}$ and $x_{2}$ .
$A = \int_{x_{1}}^{x_{2}} (\frac{2}{1 + x ^{2}} - e^{\frac{x}{3}}) d x$ .
Using the GDC: $A \approx 1.22$ .

Answer: area $\approx 1.22$

Exercise 177 MediumIB · P1 November 2000

Calculate the area enclosed by the graph of $y = x sin (x^{2})$ and the $x$ -axis, between $x = 0$ and the smallest positive value of $x$ at which the graph meets the $x$ -axis.

Show answer

Answer: area $= 1$

Show working and answer

The intersections with the $x$ -axis satisfy $x sin (x^{2}) = 0$ . For $x > 0$ we need $sin (x^{2}) = 0$ , that is $x^{2} = π, 2 π, \dots$ ; the smallest is $x = π$ .
On $] 0, π [$ we have $0 < x^{2} < π$ , so $sin (x^{2}) > 0$ and the function is positive: $A = \int_{0}^{π} x sin (x^{2}) d x$ .
Substitution $u = x^{2}$ , $d u = 2 x d x$ , or $x d x = \frac{d u}{2}$ . Limits: $x = 0 \to u = 0$ ; $x = π \to u = π$ .
$A = \frac{1}{2} \int_{0}^{π} sin u d u = \frac{1}{2} [- cos u]_{0}^{π} = \frac{1}{2} (1 + 1) = 1$ .

Answer: area $= 1$

Exercise 178 MediumIB · P1 May 1998

Calculate the total area enclosed by the curve $y^{2} = x^{2} - x^{4}$ , $- 1 \leq x \leq 1$ .

Show answer

Answer: area $= \frac{4}{3}$

Show working and answer

Solving for $y$ : $y^{2} = x^{2} (1 - x^{2})$ , so $y = \pm ∣ x ∣ 1 - x^{2}$ . The curve is symmetric about both axes, so the total area is four times the area of the first quadrant.
$A = 4 \int_{0}^{1} x 1 - x^{2} d x$ .
Substitution $u = 1 - x^{2}$ , $d u = - 2 x d x$ . Limits: $x = 0 \to u = 1$ ; $x = 1 \to u = 0$ .
$A = 4 \int_{1}^{0} u (- \frac{1}{2}) d u = 2 \int_{0}^{1} u^{\frac{1}{2}} d u = 2 [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1} = \frac{4}{3}$ .

Answer: area $= \frac{4}{3}$

Exercise 179 MediumIB · P1 TZ0 May 2006

Let $f (x) = 2^{0.5 x}$ and $g (x) = 3^{- 0.5 x} + \frac{5}{3}$ . Let $R$ be the region completely enclosed between the graphs of $f$ and $g$ , and the $y$ -axis. Find the area of $R$ .

Show answer

Answer: $Area (R) \approx 1.66$

Show working and answer

We look for where they intersect: $2^{0.5 x} = 3^{- 0.5 x} + \frac{5}{3}$ . Using the GDC (or by trial), the solution is $x = 2$ : $f (2) = 2$ and $g (2) = \frac{1}{3} + \frac{5}{3} = 2$ . As $f$ is increasing and $g$ is decreasing, there are no other intersections.
The region lies between the $y$ -axis ( $x = 0$ ) and $x = 2$ . At $x = 0$ : $f (0) = 1$ and $g (0) = \frac{8}{3}$ , so $g$ lies above $f$ .
$A = \int_{0}^{2} (g (x) - f (x)) d x = \int_{0}^{2} (3^{- 0.5 x} + \frac{5}{3} - 2^{0.5 x}) d x$ .
Using the GDC: $A \approx 1.66$ . (Exactly: $\frac{4}{3 ln 3} + \frac{10}{3} - \frac{2}{ln 2}$ .)

Answer: $Area (R) \approx 1.66$

Exercise 180 MediumIB · P1 TZ0 November 2006

Let $R$ be the region bounded by the curves $y^{2} = k x$ and $x^{2} = k y$ , where $k > 0$ . Given that the area of $R$ is $12$ , find the value of $k$ .

Show answer

Answer: $k = 6$

Show working and answer

We rearrange both curves for $x \geq 0$ : $y = k x$ and $y = \frac{x ^{2}}{k}$ .
They intersect where $k x = \frac{x ^{2}}{k}$ . Squaring, $k x = \frac{x ^{4}}{k ^{2}}$ , that is $x (x^{3} - k^{3}) = 0$ , so $x = 0$ or $x = k$ .
Between $0$ and $k$ , $y = k x$ lies above (for example, at $x = \frac{k}{2}$ : $k^{2} /2 \approx 0.71 k$ compared with $\frac{k}{4}$ ).
$A = \int_{0}^{k} (k x^{\frac{1}{2}} - \frac{x ^{2}}{k}) d x = [\frac{2 k}{3} x^{\frac{3}{2}} - \frac{x ^{3}}{3 k}]_{0}^{k} = \frac{2 k ^{2}}{3} - \frac{k ^{2}}{3} = \frac{k ^{2}}{3}$ .
Setting $\frac{k ^{2}}{3} = 12$ , we get $k^{2} = 36$ and, as $k > 0$ , $k = 6$ .

Answer: $k = 6$

Exercise 181 MediumIB · P1 TZ1 May 2011

Find the area enclosed by the curve $y = arctan x$ , the $x$ -axis and the line $x = 3$ .

Show answer

Answer: $A = 3 \cdot \frac{π}{3} - \frac{1}{2} ln 4 \approx 1.12$

Show working and answer

On $[0, 3]$ we have $arctan x \geq 0$ , so the area is $A = \int_{0}^{3} arctan x d x$ .
By parts with $u = arctan x$ and $d v = d x$ : $d u = \frac{1}{1 + x ^{2}} d x$ and $v = x$ .
$\int arctan x d x = x arctan x - \int \frac{x}{1 + x ^{2}} d x = x arctan x - \frac{1}{2} ln (1 + x^{2})$ .
Evaluating between $0$ and $3$ : $3 arctan 3 - \frac{1}{2} ln 4 - 0 = 3 \cdot \frac{π}{3} - \frac{1}{2} ln 4$ .

Answer: $A = 3 \cdot \frac{π}{3} - \frac{1}{2} ln 4 \approx 1.12$

Exercise 182 MediumIB · P2 TZ0 November 2012

Consider the graph of $y = x + sin (x - 3)$ , $- π \leq x \leq π$ .

a) Sketch the graph approximately, clearly labelling the intercepts with the $x$ -axis and $y$ -axis and stating their values.

b) Find the area of the region bounded by the graph and the $x$ - and $y$ -axes.

View solution

a) 1. Intercept with the $y$ -axis ( $x = 0$ ): $y = 0 + sin (- 3) \approx - 0.141$ , that is, $(0; - 0.141)$ . 2. Intercept with the $x$ -axis: $x + sin (x - 3) = 0$ . Using the GDC, $x \approx 0.820$ , that is, $(0.820; 0)$ . 3. As $y^{'} = 1 + cos (x - 3) \geq 0$ , the function is increasing over the whole interval, so there are no further intercepts. The endpoints are $y (- π) \approx - 3.00$ and $y (π) \approx 3.28$ .

Graph of y = x + sin(x − 3) for x between −π and π: an increasing curve that cuts the y-axis below the origin and the x-axis at a point with a positive x-coordinate

b) 1. The region lies between the $y$ -axis ( $x = 0$ ), the $x$ -axis and the graph, for $0 \leq x \leq 0.820$ . On that interval the graph lies below the $x$ -axis (it goes from $- 0.141$ to $0$ ). 2. $A = - \int_{0}^{0.820} (x + sin (x - 3)) d x$ . 3. Using the GDC: $A \approx 0.0816$ .

Answer: a) intercepts $(0; - 0.141)$ and $(0.820; 0)$ ; b) $A \approx 0.0816$

Exercise 183 MediumIB · P1 TZ0 November 2004

Find the total area of the regions enclosed by the curve $y = x^{3} - 3 x^{2} - 9 x + 27$ and the line $y = x + 3$ .

Show answer

Answer: total area $= \frac{407}{4} = 101.75$

Show working and answer

The intersection points satisfy $x^{3} - 3 x^{2} - 9 x + 27 = x + 3$ , that is $x^{3} - 3 x^{2} - 10 x + 24 = 0$ .
With $x = 2$ : $8 - 12 - 20 + 24 = 0$ ✓. Factorising: $(x - 2) (x^{2} - x - 12) = (x - 2) (x - 4) (x + 3)$ . The intersections are at $x = - 3$ , $x = 2$ and $x = 4$ .
Let $h (x) = x^{3} - 3 x^{2} - 10 x + 24$ (curve minus line). At $x = 0$ , $h = 24 > 0$ : on $] - 3, 2 [$ the curve lies above the line; on $] 2, 4 [$ it lies below (at $x = 3$ , $h = 27 - 27 - 30 + 24 = - 6 < 0$ ).
An antiderivative of $h$ is $H (x) = \frac{x ^{4}}{4} - x^{3} - 5 x^{2} + 24 x$ . Values: $H (- 3) = - \frac{279}{4}$ , $H (2) = 24$ , $H (4) = 16$ .
First region: $\int_{- 3}^{2} h d x = H (2) - H (- 3) = 24 + \frac{279}{4} = \frac{375}{4}$ .
Second region: $- \int_{2}^{4} h d x = - (16 - 24) = 8$ .
Total area: $\frac{375}{4} + 8 = \frac{407}{4}$ .

Answer: total area $= \frac{407}{4} = 101.75$

Exercise 184 MediumIB · P1 TZ1 May 2008

Find the area enclosed between the curves $y = 2 + x - x^{2}$ and $y = 2 - 3 x + x^{2}$ .

Show answer

Answer: Area $= \frac{8}{3}$

Show working and answer

We find where the curves intersect: $2 + x - x^{2} = 2 - 3 x + x^{2}$ , i.e. $4 x - 2 x^{2} = 0$ , that is $2 x (2 - x) = 0$ . They intersect at $x = 0$ and $x = 2$ .
Between these limits, the first curve lies above the second (for example, at $x = 1$ : $2$ versus $0$ ). The difference is $(2 + x - x^{2}) - (2 - 3 x + x^{2}) = 4 x - 2 x^{2}$ .
The area is $\int_{0}^{2} (4 x - 2 x^{2}) d x = [2 x^{2} - \frac{2 x ^{3}}{3}]_{0}^{2} = 8 - \frac{16}{3} = \frac{8}{3} .$

Answer: Area $= \frac{8}{3}$

Exercise 185 MediumIB · P1 TZ1 May 2005

Find the area of the region enclosed by the graphs of $f (x) = 4 - x^{2}$ and $g (x) = (x + 1) cos x$ .

Show answer

Answer: Area $\approx 9.4$

Show working and answer

Using the GDC (Paper 1 with calculator) we find the points of intersection of $y = 4 - x^{2}$ and $y = (x + 1) cos x$ : $x_{1} \approx - 1.92$ and $x_{2} \approx 2.72$ .
Between these limits $4 - x^{2}$ lies above $(x + 1) cos x$ (at $x = 0$ : $4$ versus $1$ ).
The area is $\int_{- 1.92}^{2.72} [4 - x^{2} - (x + 1) cos x] d x \approx 9.41.$

Answer: Area $\approx 9.4$

Exercise 186 MediumIB · P1 May 2002

The diagram below shows part of the curve $y = x^{3} - 7 x^{2} + 14 x - 7$ .

Graph of a cubic polynomial function that meets the x-axis at the points A, B and C; the region between A and B, on the x-axis, is shaded

The curve meets the $x$ -axis at the points $A$ , $B$ and $C$ .

a) Find the $x$ -coordinate of $A$ .

b) Find the $x$ -coordinate of $B$ .

c) Find the area of the shaded region.

Show answer

Answer: a) $x \approx 0.753$ ; b) $x \approx 2.45$ ; c) Area $\approx 1.78$

Show working and answer

a) and b) 1. Using the GDC, we solve $x^{3} - 7 x^{2} + 14 x - 7 = 0$ : the three roots are $0.753$ , $2.45$ and $3.80$ , from left to right ( $A$ , $B$ , $C$ ). 2. Therefore, the $x$ -coordinate of $A$ is $0.753$ and that of $B$ is $2.45$ .

c) 1. The shaded region lies between $A$ and $B$ , where the curve is above the $x$ -axis. 2. Area $= \int_{0.753}^{2.45} (x^{3} - 7 x^{2} + 14 x - 7) d x \approx 1.78$ (using the GDC, with the exact values of the roots).

Answer: a) $x \approx 0.753$ ; b) $x \approx 2.45$ ; c) Area $\approx 1.78$

Exercise 187 MediumIB · P1 November 2003

Calculate the area enclosed by the curves $y = ln x$ and $y = e^{x} - e$ , $x > 0$ .

Show answer

Answer: Area $\approx 0.201$

Show working and answer

Using the GDC, we find the intersections of $y = ln x$ and $y = e^{x} - e$ : $x = 1$ (both equal $0$ ) and $x \approx 0.233$ .
In the interval between them, for example at $x = 0.5$ : $ln 0.5 \approx - 0.69$ compared with $e^{0.5} - e \approx - 1.07$ , so $y = ln x$ lies above.
Area $= \int_{0.233}^{1} (ln x - e^{x} + e) d x \approx 0.201$ .

Answer: Area $\approx 0.201$

Exercise 188 MediumIB · P2 May 1996

Calculate the area enclosed by the curve $y = x^{2} sin x$ and the $x$ -axis for $0 \leq x \leq 2 π$ , giving your answer exactly in terms of $π$ .

Show answer

Answer: Area $= 6 π^{2} - 8$

Show working and answer

Since $x^{2} \geq 0$ , the sign of the function is the sign of $sin x$ : positive on $] 0, π [$ and negative on $] π, 2 π [$ . The area is $\int_{0}^{π} x^{2} sin x d x - \int_{π}^{2 π} x^{2} sin x d x$ .
We integrate by parts twice. First, with $u = x^{2}$ , $d v = sin x d x$ : $\int x^{2} sin x d x = - x^{2} cos x + 2 \int x cos x d x$ . Then, with $u = x$ , $d v = cos x d x$ : $\int x cos x d x = x sin x + cos x$ .
$F (x) = - x^{2} cos x + 2 x sin x + 2 cos x$ (check by differentiating).
$F (0) = 2$ , $F (π) = π^{2} - 2$ , $F (2 π) = - 4 π^{2} + 2$ .
$\int_{0}^{π} = F (π) - F (0) = π^{2} - 4$ ; $\int_{π}^{2 π} = F (2 π) - F (π) = - 5 π^{2} + 4$ .
Area $= (π^{2} - 4) + (5 π^{2} - 4) = 6 π^{2} - 8$ .

Answer: Area $= 6 π^{2} - 8$

Exercise 189 MediumIB · P1 TZ2 May 2018

a) Show that $lo g_{r^{2}} x = \frac{1}{2} lo g_{r} x$ , where $r, x \in R^{+}$ .

It is given that $lo g_{2} y + lo g_{4} x + lo g_{4} 2 x = 0$ .

b) Express $y$ in terms of $x$ . Give your answer in the form $y = p x^{q}$ , where $p$ , $q$ are constants.

The region $R$ is bounded by the graph of the function found in part b), the $x$ -axis and the lines $x = 1$ and $x = α$ , where $α > 1$ . The area of $R$ is equal to $2$ .

c) Find the value of $α$ .

Show answer

Answer: a) shown; b) $y = \frac{1}{2} x^{- 1}$ ( $p = \frac{1}{2}$ , $q = - 1$ ); c) $α = e^{2}$

Show working and answer

a) By the change of base, $lo g_{r^{2}} x = \frac{lo g _{r} x}{lo g _{r} r ^{2}} = \frac{lo g _{r} x}{2} = \frac{1}{2} lo g_{r} x$ .

b) 1. By a) with $r = 2$ : $lo g_{4} x = \frac{1}{2} lo g_{2} x$ and $lo g_{4} (2 x) = \frac{1}{2} lo g_{2} (2 x) = \frac{1}{2} (1 + lo g_{2} x)$ . 2. The equation becomes $lo g_{2} y + \frac{1}{2} lo g_{2} x + \frac{1}{2} + \frac{1}{2} lo g_{2} x = 0$ , that is, $lo g_{2} y = - lo g_{2} x - \frac{1}{2}$ . 3. Hence $y = 2^{- \frac{1}{2}} \cdot x^{- 1} = \frac{1}{2} x^{- 1}$ , with $p = \frac{1}{2}$ and $q = - 1$ .

c) 1. Area $= \int_{1}^{α} \frac{1}{2} \cdot \frac{1}{x} d x = \frac{1}{2} ln α$ . 2. Setting this equal to $2$ : $ln α = 2 \cdot 2 = 2$ , so $α = e^{2}$ .

Answer: a) shown; b) $y = \frac{1}{2} x^{- 1}$ ( $p = \frac{1}{2}$ , $q = - 1$ ); c) $α = e^{2}$

Exercise 190 MediumIB · P2 TZ1 May 2017

Region $A$ is enclosed by the graph of $y = 2 arcsin (x - 1) - \frac{π}{4}$ , the $y$ -axis and the line $y = \frac{π}{4}$ .

a) Write down a definite integral that represents the area of $A$ .

b) Calculate the area of $A$ .

Show answer

Answer: a) $\int_{0}^{1.71} [\frac{π}{4} - (2 arcsin (x - 1) - \frac{π}{4})] d x$ ; b) area $= 3.3$

Show working and answer

a) 1. The curve $y = 2 arcsin (x - 1) - \frac{π}{4}$ is defined for $0 \leq x \leq 2$ . It meets the line $y = \frac{π}{4}$ when $2 arcsin (x - 1) = \frac{π}{2}$ , i.e. $x - 1 = sin \frac{π}{4}$ , so $x = 1 + \frac{2}{2} \approx 1.71$ . 2. At the $y$ -axis ( $x = 0$ ) the curve has value $2 arcsin (- 1) - \frac{π}{4} = - \frac{5 π}{4}$ , which is below $\frac{π}{4}$ . Between $x = 0$ and $x \approx 1.71$ the curve lies below the line, so the line $y = \frac{π}{4}$ is the upper boundary. 3. The area is the integral of (upper function minus lower function): $\int_{0}^{1.71} [\frac{π}{4} - (2 arcsin (x - 1) - \frac{π}{4})] d x$ .

b) 1. Simplify the integrand: $\frac{π}{2} - 2 arcsin (x - 1)$ . 2. Using the GDC: area $\approx 3.3$ . 3. (Exact value, with $u = x - 1$ and $\int arcsin u d u = u arcsin u + 1 - u^{2}$ : area $= \frac{3 π}{2} - 2 \approx 3.30$ .)

Answer: a) $\int_{0}^{1.71} [\frac{π}{4} - (2 arcsin (x - 1) - \frac{π}{4})] d x$ ; b) area $= 3.3$

Exercise 191 MediumIB · P2 May 1999

Consider the function $f : x \mapsto x - x^{2}$ for $- 1 \leq x \leq k$ , where $1 < k \leq 3$ .

a) Sketch the graph of the function $f$ .

b) Calculate the total finite area enclosed by the graph of $f$ , the $x$ -axis and the line $x = k$ .

View solution

a) 1. $f (x) = x - x^{2} = x (1 - x)$ is a parabola opening downwards, which cuts the $x$ -axis at $0$ and $1$ and has its maximum at $(\frac{1}{2}, \frac{1}{4})$ . 2. We draw it from $x = - 1$ (where $f (- 1) = - 2$ ) to the vertical line $x = k$ , with $1 < k \leq 3$ .

Graph of the function f(x) = x − x² for −1 ≤ x ≤ k, with the vertical line x = k

b) 1. The region enclosed by the graph, the $x$ -axis and the line $x = k$ has two parts: the part between $0$ and $1$ (above the $x$ -axis) and the part between $1$ and $k$ (below the $x$ -axis, where $f < 0$ ). 2. First part: $\int_{0}^{1} (x - x^{2}) d x = [\frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{1} = \frac{1}{6}$ . 3. Second part (we change the sign because $f < 0$ ): $- \int_{1}^{k} (x - x^{2}) d x = - [\frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{1}^{k} = - (\frac{k ^{2}}{2} - \frac{k ^{3}}{3}) + \frac{1}{6} = \frac{k ^{3}}{3} - \frac{k ^{2}}{2} + \frac{1}{6}$ . 4. Total area: $\frac{1}{6} + \frac{k ^{3}}{3} - \frac{k ^{2}}{2} + \frac{1}{6} = \frac{k ^{3}}{3} - \frac{k ^{2}}{2} + \frac{1}{3}$ .

Answer: b) Area $= \frac{k ^{3}}{3} - \frac{k ^{2}}{2} + \frac{1}{3}$

Exercise 192 MediumIB · P1 November 1999

In the diagram, $P T Q$ is an arc of the parabola $y = a^{2} - x^{2}$ , where $a$ is a positive constant, and $P QR S$ is a rectangle. The area of the rectangle $P QR S$ is equal to the area between the arc $P T Q$ of the parabola and the $x$ -axis.

Arc of the parabola y = a² − x² with its vertex T, the points P and Q on the x-axis and the rectangle PQRS whose vertices R and S lie at height h

Calculate, in terms of $a$ , the dimensions of the rectangle.

Show answer

Answer: base $= 2 a$ ; height $= \frac{2}{3} a^{2}$

Show working and answer

The points $P$ and $Q$ are the intercepts of the parabola with the $x$ -axis: $a^{2} - x^{2} = 0$ , that is, $P (- a, 0)$ and $Q (a, 0)$ . The base of the rectangle is $P Q = 2 a$ .
Area under the arc $P T Q$ : $\int_{- a}^{a} (a^{2} - x^{2}) d x = [a^{2} x - \frac{x ^{3}}{3}]_{- a}^{a} = 2 (a^{3} - \frac{a ^{3}}{3}) = \frac{4 a ^{3}}{3}$ .
Let $h$ be the height of the rectangle. Its area is $2 a \cdot h$ and it must be equal to $\frac{4 a ^{3}}{3}$ .
So $h = \frac{4 a ^{3}}{3 \cdot 2 a} = \frac{2 a ^{2}}{3}$ .

Answer: base $= 2 a$ ; height $= \frac{2}{3} a^{2}$

Exercise 193 MediumIB · P2 TZ2 May 2004

Let $f (x) = x cos x$ , for $0 \leq x \leq π$ .

The curve of $f (x)$ has a local maximum at $x = a$ and a point of inflexion at $x = b$ .

a) Sketch an approximate graph of $f (x)$ , indicating the approximate positions of $a$ and $b$ .

b) Find the value of

(i) $a$ .

(ii) $b$ .

c) Using integration by parts, find an expression for $\int x cos x d x$ .

d) Hence find the exact value of the area enclosed by the curve and the $x$ -axis, for $0 \leq x \leq \frac{π}{2}$ .

View solution

a) Key features: $f (0) = 0$ , it rises to a local maximum at $x = a$ (a little less than $1$ ), cuts the $x$ -axis at $\frac{π}{2}$ , is concave down until the point of inflexion at $x = b$ (around $2.3$ ), and ends at $(π, - π)$ .

Approximate graph of f(x) = x cos x on (0, π), with the local maximum at x = a and the point of inflexion at x = b

b) i) 1. $f^{'} (x) = cos x - x sin x$ . 2. Using the GDC we solve $f^{'} (x) = 0$ on $[0, π]$ : $a \approx 0.860$ .

(ii) 1. $f^{''} (x) = - 2 sin x - x cos x$ . 2. Using the GDC, $f^{''} (x) = 0$ gives $b \approx 2.29$ (it changes sign).

c) 1. By parts with $u = x$ and $d v = cos x d x$ ( $d u = d x$ , $v = sin x$ ): $\int x cos x d x = x sin x - \int sin x d x$ . 2. Therefore $\int x cos x d x = x sin x + cos x + C$ .

d) 1. On $[0, \frac{π}{2}]$ , $f \geq 0$ , and the area is $\int_{0}^{\frac{π}{2}} x cos x d x$ . 2. $= [x sin x + cos x]_{0}^{\frac{π}{2}} = (\frac{π}{2} + 0) - (0 + 1) = \frac{π}{2} - 1$ .

Answer: b) i) $a \approx 0.860$ ; ii) $b \approx 2.29$ ; c) $x sin x + cos x + C$ ; d) $\frac{π}{2} - 1$

Exercise 194 MediumIB · P1 TZ2 May 2007

Let $f (x) = x^{2} ln (x + 1)$ and $g (x) = 2 x - 1$ , for $x \geq \frac{1}{2}$ .

a) Sketch approximately the graphs of $f$ and $g$ on the following grid.

Empty grid with x and y axes, for sketching the graphs of f and g: x from 0 to 2 with marks every 0.5 and y from 0 to 4 with marks every unit

b) Let $A$ be the region completely enclosed by the graphs of $f$ and $g$ . Find the area of $A$ .

View solution

a) Without a calculator there are no exact values, but using the GDC: $g$ starts at $(0.5; 0)$ with an almost vertical tangent and increases slowly; $f$ starts at $(0.5; 0.10)$ and increases faster. They intersect at $x \approx 0.505$ and $x \approx 1.23$ ; after that $f$ lies above $g$ and reaches a value above $4$ at $x = 2$ .

Graphs of f(x) = x² ln(x + 1) and g(x) = √(2x − 1) for x ≥ 1/2 on a grid from 0 to 2 in x and from 0 to 4 in y; they intersect at two points, near (0.51; 0.1) and (1.23; 1.21)

b) 1. Using the GDC, find the $x$ -coordinates of the points where $f (x) = g (x)$ : $x_{1} \approx 0.505$ and $x_{2} \approx 1.23$ . 2. Between them, $g$ lies above $f$ (for example, at $x = 1$ : $g = 1$ and $f = ln 2 \approx 0.69$ ). 3. Area $A = \int_{0.505}^{1.23} (2 x - 1 - x^{2} ln (x + 1)) d x \approx 0.201$ .

Answer: a) sketch of $f$ and $g$ (intersections at $x \approx 0.505$ and $x \approx 1.23$ ); b) $A \approx 0.201$

Exercise 195 MediumIB · P1 TZ1 May 2012

The following graph shows the two curves $y = \frac{1}{x}$ and $y = \frac{k}{x}$ , where $k > 1$ .

Graph of the curves y = 1/x and y = k/x (k > 1) in the first quadrant; region A lies between the two curves and the lines x = 1/6 and x = 1, and region B (shaded darker) lies between the two curves and the lines x = 1 and x = √6

a) Find the area of region $A$ in terms of $k$ .

b) Find the area of region $B$ in terms of $k$ .

c) Find the ratio of the area of region $A$ to the area of region $B$ .

Show answer

Answer: a) $Area (A) = (k - 1) ln 6$ ; b) $Area (B) = (k - 1) ln 6$ ; c) $\frac{Area ( A )}{Area ( B )} = 2$

Show working and answer

The two curves are $y = \frac{1}{x}$ (the lower one) and $y = \frac{k}{x}$ (the upper one, since $k > 1$ ). The height of the region between them is $\frac{k}{x} - \frac{1}{x} = \frac{k - 1}{x}$ .

a) 1. Region $A$ lies between the two curves from $x = \frac{1}{6}$ to $x = 1$ . 2. $Area (A) = \int_{1/6}^{1} \frac{k - 1}{x} d x = (k - 1) [ln x]_{1/6}^{1}$ . 3. We evaluate: $(k - 1) (ln 1 - ln \frac{1}{6}) = (k - 1) ln 6$ .

b) 1. Region $B$ lies between the same curves from $x = 1$ to $x = 6$ . 2. $Area (B) = (k - 1) [ln x]_{1}^{6} = (k - 1) (ln 6 - ln 1) = (k - 1) ln 6$ .

c) 1. As $ln 6 = \frac{1}{2} ln 6$ , the ratio is $\frac{Area ( A )}{Area ( B )} = \frac{( k - 1 ) ln 6}{( k - 1 ) \cdot \frac{1}{2} ln 6}$ . 2. Cancelling $(k - 1) ln 6$ (which is not zero since $k > 1$ ) leaves $2$ .

Answer: a) $Area (A) = (k - 1) ln 6$ ; b) $Area (B) = (k - 1) ln 6$ ; c) $\frac{Area ( A )}{Area ( B )} = 2$

Exercise 196 MediumIB · P2 TZ0 November 2012

Find the area of the region enclosed by the curves $y = x^{3}$ and $x = y^{2} - 3$ .

Show answer

Answer: Area $= 7.34$

Show working and answer

Paper 2: we use the GDC for the intersection points and for the integral.

To integrate it is more convenient to take $y$ as the variable: $y = x^{3}$ is equivalent to $x = 3 y$ , and the other curve is the parabola $x = y^{2} - 3$ .
Intersections: $3 y = y^{2} - 3$ . Using the GDC, the solutions are $y \approx - 1.374$ and $y \approx 2.067$ (i.e. $x \approx - 1.112$ and $x \approx 1.274$ ).
Between those values of $y$ the curve $x = 3 y$ lies to the right of the parabola: for example, at $y = 0$ we have $x = 0 > - 3$ .
$Area = \int_{- 1.374}^{2.067} (3 y - (y^{2} - 3)) d y$ .
Using the GDC: $Area \approx 7.34$ .

Answer: Area $= 7.34$

Exercise 197 MediumIB · P1 TZ2 May 2021

Consider the function $f$ given by $f (x) = 6 + 6 cos x$ , for $0 \leq x \leq 4 π$ .

The following figure shows the graph of $y = f (x)$ .

Graph of f(x) = 6 + 6cos x for 0 ≤ x ≤ 4π, touching the x-axis at A and B; the region under the curve between A and B is shaded

The graph of $f$ touches the $x$ -axis at the points $A$ and $B$ , as shown in the figure.

The shaded region is bounded by the graph of $y = f (x)$ and the $x$ -axis, between the points $A$ and $B$ .

a) Find the $x$ -coordinates of $A$ and $B$ .

b) Show that the area of the shaded region is equal to $12 π$ .

The following figure shows a right circular cone. The total surface area of the cone is $12 π$ , the same as the shaded area in the previous figure. In this cone, the radius of the base is $2$ , the height is $h$ and the slant height is $l$ .

Right cone, with the height h shown dashed from the apex to the centre of the base, base radius equal to 2 and slant height l

c) Find the value of $l$ .

d) Hence, find the volume of the cone.

Show answer

Answer: a) $A (π, 0)$ , $B (3 π, 0)$ ; b) proved ( $= 12 π$ ); c) $l = 4$ ; d) $V = \frac{8 π 3}{3}$

Show working and answer

a) 1. The graph touches the $x$ -axis when $6 + 6 cos x = 0$ , that is, $cos x = - 1$ . 2. On $[0, 4 π]$ : $x = π$ and $x = 3 π$ . $A (π, 0)$ and $B (3 π, 0)$ .

b) 1. $f \geq 0$ , so the area is $\int_{π}^{3 π} (6 + 6 cos x) d x = [6 x + 6 sin x]_{π}^{3 π}$ . 2. $= (18 π + 0) - (6 π + 0) = 12 π$ .

c) 1. Total surface area of the cone: base $π r^{2}$ plus lateral surface $π r l$ , with $r = 2$ : $4 π + 2 π l$ . 2. $4 π + 2 π l = 12 π$ , so $l = 4$ .

d) 1. Pythagoras: $l^{2} = r^{2} + h^{2}$ , so $h = 16 - 4 = 12 = 23$ . 2. $V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \cdot 4 \cdot 23 = \frac{8 3}{3} π$ .

Answer: a) $A (π, 0)$ , $B (3 π, 0)$ ; b) proved ( $= 12 π$ ); c) $l = 4$ ; d) $V = \frac{8 π 3}{3}$

Exercise 198 MediumIB · P2 TZ0 November 2022

The function $f$ is given by $f (x) = ln (x e^{x} + 1) - x^{4}$ , for $0 \leq x \leq 2$ . The following figure shows the graph of $f$ .

Graph of f(x) = ln(x·e^x + 1) − x⁴ for 0 ≤ x ≤ 2, with a local maximum A and an intersection B with the x-axis

The graph of $f$ has a local maximum at the point $A$ , and crosses the $x$ -axis at the origin and at the point $B$ .

a) Find the coordinates of $A$ .

b) Find the $x$ -coordinate of $B$ .

c) Find the total area of the region bounded by the graph of $f$ , the $x$ -axis and the line $x = 2$ .

Show answer

Answer: a) $A (0.709, 0.640)$ ; b) $x = 1.10$ ; c) Area $= 4.61$

Show working and answer

Paper 2: we use the GDC.

a) 1. The local maximum is a point where $f^{'} (x) = 0$ . Using the GDC (the maximum of the graph between $0$ and $1$ ): $A (0.709, 0.640)$ .

b) 1. $B$ is the second intersection with the $x$ -axis (the first is the origin): $f (x) = 0$ gives $x \approx 1.10$ .

c) 1. The graph lies above the axis between $x = 0$ and $x \approx 1.10$ , and below the axis between $x \approx 1.10$ and $x = 2$ . For the total area we need the absolute value, or we add the two parts with the sign of the second one changed. 2. $Area = \int_{0}^{2} ∣ f (x) ∣ d x = \int_{0}^{1.10} f (x) d x - \int_{1.10}^{2} f (x) d x$ . 3. Using the GDC: $0.436 + 4.175 \approx 4.61$ .

Answer: a) $A (0.709, 0.640)$ ; b) $x = 1.10$ ; c) Area $= 4.61$

Exercise 199 HardIB · P1 TZ1 May 2007

The function $f$ is defined as $f (x) = sin x ln x$ for $x \in [0.5; 3.5]$ .

a) Write down the intercepts with the $x$ -axis.

b) The area above the $x$ -axis is $A$ and the total area below the $x$ -axis is $B$ . If $A = k B$ , calculate $k$ .

Show answer

Answer: a) $x = 1$ , $x = π$ ; b) $k \approx 5.17$

Show working and answer

a) 1. The intercepts satisfy $sin x ln x = 0$ , that is, $sin x = 0$ or $ln x = 0$ . 2. On $[0.5; 3.5]$ : $sin x = 0$ gives $x = π$ , and $ln x = 0$ gives $x = 1$ .

b) 1. We check the sign on each interval. On $[0.5; 1 [$ : $sin x > 0$ and $ln x < 0$ , so $f < 0$ . On $] 1, π [$ : both are positive, so $f > 0$ . On $] π; 3.5]$ : $sin x < 0$ and $ln x > 0$ , so $f < 0$ . 2. Area above the axis: $A = \int_{1}^{π} sin x ln x d x \approx 0.881$ (GDC). 3. Area below the axis (the sum of the two regions, as absolute values): $B = - \int_{0.5}^{1} f (x) d x - \int_{π}^{3.5} f (x) d x \approx 0.1705$ (GDC). 4. From $A = k B$ : $k = \frac{A}{B} = \frac{0.881}{0.1705} \approx 5.17$ .

Answer: a) $x = 1$ , $x = π$ ; b) $k \approx 5.17$

Exercise 200 HardIB · P1 TZ2 May 2009

A function is defined by $f (x) = k x$ with $k > 0$ and $x \geq 0$ .

a) Sketch the graph of $y = f (x)$ .

b) Verify that $f$ is an injective function.

c) Find the inverse function, $f^{- 1} (x)$ , and state its domain.

d) Given that the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ intersect at the point $(4, 4)$ , find the value of $k$ .

e) Consider the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ , using the value of $k$ found in part (d).

(i) Find the area enclosed by the two graphs.

(ii) The line $x = c$ meets the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ at the points $P$ and $Q$ , respectively. Given that the tangent to $y = f (x)$ at the point $P$ is parallel to the tangent to $y = f^{- 1} (x)$ at the point $Q$ , find the value of $c$ .

View solution

a) 1. $f (x) = k x$ starts at the origin, with a vertical tangent, and is increasing and concave down, passing through $(1, k)$ .

Graph of y = k√x for x ≥ 0 (drawn with k = 2): starts at the origin with a vertical tangent, increasing and concave down

b) 1. We differentiate: $f^{'} (x) = \frac{k}{2 x} > 0$ for all $x > 0$ (since $k > 0$ ). 2. $f$ is strictly increasing, so each value of $y$ is reached only once: $f$ is injective.

c) 1. We write $y = k x$ and solve for $x$ : $x = \frac{y}{k}$ , so $x = \frac{y ^{2}}{k ^{2}}$ . 2. Interchanging the variables, $f^{- 1} (x) = \frac{x ^{2}}{k ^{2}}$ . 3. The domain of $f^{- 1}$ is the range of $f$ : $[0, + \infty [$ .

d) 1. $(4, 4)$ lies on $y = f (x)$ : $4 = k 4 = 2 k$ , so $k = 2$ . 2. Check on the inverse: $f^{- 1} (4) = \frac{16}{4} = 4$ ✓.

e) (i) 1. With $k = 2$ : $f (x) = 2 x$ and $f^{- 1} (x) = \frac{x ^{2}}{4}$ . They meet where $2 x = \frac{x ^{2}}{4}$ : at $x = 0$ and at $x = 4$ . 2. Between $0$ and $4$ , $f$ lies above (at $x = 1$ : $2$ compared with $\frac{1}{4}$ ). 3. $A = \int_{0}^{4} (2 x - \frac{x ^{2}}{4}) d x = [\frac{4}{3} x^{\frac{3}{2}} - \frac{x ^{3}}{12}]_{0}^{4} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$ .

(ii) 1. Gradient of the tangent to $f$ at $P$ : $f^{'} (c) = \frac{1}{c}$ . Gradient of the tangent to $f^{- 1}$ at $Q$ : $(f^{- 1})^{'} (c) = \frac{c}{2}$ . 2. The tangents are parallel if the gradients are equal: $\frac{1}{c} = \frac{c}{2}$ , so $c^{\frac{3}{2}} = 2$ . 3. $c = 2^{\frac{2}{3}} = 34$ .

Answer: b) $f^{'} (x) > 0$ , strictly increasing; c) $f^{- 1} (x) = \frac{x ^{2}}{k ^{2}}$ , domain $[0, + \infty [$ ; d) $k = 2$ ; e) (i) $\frac{16}{3}$ ; (ii) $c = 34$

Exercise 201 HardIB · P2 TZ2 May 2010

The function $f$ is given by $f (x) = (x^{3} + 6 x^{2} + 3 x - 10)^{\frac{1}{2}}$ , for $x \in D$ , where $D \subseteq R$ is the largest possible domain of $f$ .

a) Find the roots of $f (x) = 0$ .

b) Hence, state the set $D$ .

c) Find the coordinates of the local maximum on the graph of $y = f (x)$ .

d) Solve the equation $f (x) = 3$ .

e) Sketch the graph of $∣ f (x) ∣$ , for $x \in D$ .

f) Find the area of the region that is completely enclosed by the graph of $∣ f (x) ∣$ and the $X$ -axis.

View solution

a) 1. Trying small integers, $x = 1$ satisfies $1 + 6 + 3 - 10 = 0$ , so $(x - 1)$ is a factor. 2. Dividing: $x^{3} + 6 x^{2} + 3 x - 10 = (x - 1) (x^{2} + 7 x + 10) = (x - 1) (x + 2) (x + 5)$ . 3. The roots are $x = 1$ , $x = - 2$ and $x = - 5$ .

b) 1. The square root requires $p (x) = (x - 1) (x + 2) (x + 5) \geq 0$ . 2. We study the sign: for $x < - 5$ it is negative; on $] - 5, - 2 [$ it is positive; on $] - 2, 1 [$ it is negative; for $x > 1$ it is positive. 3. $D = [- 5, - 2] \cup [1, + \infty [$ .

c) 1. The extrema of $f$ occur where $p^{'} (x) = 0$ (in the interior of the domain): $3 x^{2} + 12 x + 3 = 0 ⟺ x^{2} + 4 x + 1 = 0$ , so $x = - 2 \pm 3$ , that is $x \approx - 3.73$ and $x \approx - 0.268$ . 2. Only $x = - 2 - 3 \approx - 3.73$ lies in $D$ (on the interval $[- 5, - 2]$ , where $f$ rises from $0$ and returns to $0$ , so it is a maximum). 3. $f (- 3.73) \approx 3.22$ .

d) 1. $f (x) = 3 ⟺ p (x) = 9 ⟺ x^{3} + 6 x^{2} + 3 x - 19 = 0$ . 2. Using the GDC: $x \approx - 4.23$ ; $x \approx - 3.18$ ; $x \approx 1.41$ (all three lie in $D$ ).

e) 1. On $D$ , $f (x) \geq 0$ , so $∣ f (x) ∣ = f (x)$ . The graph has an arc between $x = - 5$ and $x = - 2$ with a maximum at $(- 3.73; 3.22)$ , and an increasing branch leaving $(1, 0)$ with a vertical tangent.

Graph of |f(x)| = √(x³ + 6x² + 3x − 10): an arc between x = −5 and x = −2 with a local maximum, and an increasing branch leaving the point (1, 0)

f) 1. The only region completely enclosed by the graph and the $X$ -axis is the arc between $x = - 5$ and $x = - 2$ . 2. $A = \int_{- 5}^{- 2} x^{3} + 6 x^{2} + 3 x - 10 d x \approx 7.47$ (GDC).

Answer: a) $x = 1$ , $x = - 2$ , $x = - 5$ ; b) $D = [- 5, - 2] \cup [1, + \infty [$ ; c) maximum $(- 3.73; 3.22)$ ; d) $x \approx - 4.23$ ; $x \approx - 3.18$ ; $x \approx 1.41$ ; e) graph as described; f) $A \approx 7.47$

Exercise 202 HardIB · P1 TZ2 May 2012

The function $f$ , defined on the domain $[0, \frac{3 π}{2}]$ , is given by $f (x) = e^{- x} cos x$ .

a) State the two zeros of $f$ . Sketch the graph of $f$ .

b) Let $A$ be the region enclosed by the graph, the $x$ -axis and the $y$ -axis, and let $B$ be the region enclosed by the graph and the $x$ -axis. Verify that the ratio of the area $A$ to the area $B$ is equal to $\frac{e ^{π} ( e ^{\frac{π}{2}} + 1 )}{e ^{π} + 1}$ .

View solution

a) 1. The zeros satisfy $e^{- x} cos x = 0$ . Since $e^{- x} > 0$ , we need $cos x = 0$ : $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ . 2. The function starts at $(0, 1)$ , is positive on $[0, \frac{π}{2} [$ and negative on $] \frac{π}{2}, \frac{3 π}{2} [$ (with a small minimum near $x = \frac{3 π}{4}$ ).

Graph of f(x) = e^(−x) cos x on the interval (0, 3π/2): starts at (0, 1), decreases to cross the x-axis at π/2, dips slightly below the axis to a very small minimum and returns to the axis at 3π/2

b) 1. We find an antiderivative of $e^{- x} cos x$ by integrating by parts twice. With $u = cos x$ , $d v = e^{- x} d x$ : $I = - e^{- x} cos x - \int e^{- x} sin x d x$ . 2. For the last integral, with $u = sin x$ , $d v = e^{- x} d x$ : $\int e^{- x} sin x d x = - e^{- x} sin x + I$ . 3. Substituting: $I = - e^{- x} cos x + e^{- x} sin x - I$ , so $I = F (x) = \frac{e ^{- x} ( sin x - cos x )}{2}$ . 4. Values: $F (0) = - \frac{1}{2}$ , $F (\frac{π}{2}) = \frac{e ^{- π /2}}{2}$ , $F (\frac{3 π}{2}) = \frac{e ^{- 3 π /2} ( - 1 )}{2} = - \frac{e ^{- 3 π /2}}{2}$ . 5. Area $A$ (from $0$ to $\frac{π}{2}$ , positive): $A = F (\frac{π}{2}) - F (0) = \frac{e ^{- π /2} + 1}{2}$ . 6. Area $B$ (from $\frac{π}{2}$ to $\frac{3 π}{2}$ , where the function is negative): $B = - [F (\frac{3 π}{2}) - F (\frac{π}{2})] = \frac{e ^{- 3 π /2} + e ^{- π /2}}{2}$ . 7. Ratio: $\frac{A}{B} = \frac{1 + e ^{- π /2}}{e ^{- π /2} + e ^{- 3 π /2}}$ . Multiplying numerator and denominator by $e^{3 π /2}$ : $\frac{e ^{3 π /2} + e ^{π}}{e ^{π} + 1} = \frac{e ^{π} ( e ^{π /2} + 1 )}{e ^{π} + 1}$ ✓.

Answer: a) zeros $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ ; b) verified: $\frac{A}{B} = \frac{e ^{π} ( e ^{\frac{π}{2}} + 1 )}{e ^{π} + 1}$ .

Exercise 203 HardIB · P1 TZ0 November 2010

Consider the curve $y = x e^{x}$ and the line $y = k x$ , $k \in R$ .

a) Let $k = 0$ .

(i) Verify that the curve and the line intersect once.

(ii) Find the angle between the tangent to the curve and the line at the point of intersection.

b) Let $k = 1$ . Verify that the line is a tangent to the curve.

c) (i) Find the values of $k$ for which the curve $y = x e^{x}$ and the line $y = k x$ intersect at two distinct points.

(ii) Write down the coordinates of the points of intersection.

(iii) Write down an integral that represents the area $A$ of the region bounded by the curve and the line.

(iv) Hence, given that $0 < k < 1$ , verify that $A < 1$ .

Show answer

Answer: a) (i) only one point, $(0, 0)$ ; (ii) angle $= \frac{π}{4}$ ; b) verified; c) (i) $k > 0$ and $k \neq = 1$ ; (ii) $(0, 0)$ and $(ln k, k ln k)$ ; (iii) $k > 1$ : $A = \int_{0}^{l n k} (k x - x e^{x}) d x$ ; $0 < k < 1$ : $A = \int_{l n k}^{0} (k x - x e^{x}) d x$ ; (iv) verified: $A < 1$ .

Show working and answer

a) (i) 1. With $k = 0$ : $x e^{x} = 0$ . As $e^{x} > 0$ always, only $x = 0$ . 2. They intersect only at $(0, 0)$ : once.

(ii) 1. The line is $y = 0$ (the $x$ -axis), with gradient $0$ . 2. Gradient of the tangent to the curve: $y^{'} = e^{x} + x e^{x} = e^{x} (1 + x)$ , which at $x = 0$ equals $1$ . 3. The angle $θ$ between them satisfies $tan θ = 1$ , so $θ = \frac{π}{4}$ .

b) 1. With $k = 1$ : $x e^{x} = x ⟺ x (e^{x} - 1) = 0 ⟺ x = 0$ . There is only one common point, $(0, 0)$ . 2. The gradient of the curve at $x = 0$ is $e^{0} \cdot 1 = 1$ , the same as the gradient of the line $y = x$ . 3. Same point and same gradient: the line is tangent to the curve at $(0, 0)$ .

c) (i) 1. $x e^{x} = k x ⟺ x (e^{x} - k) = 0 ⟺ x = 0$ or $e^{x} = k$ . 2. $e^{x} = k$ has a solution only if $k > 0$ , and it is $x = ln k$ . For this to differ from $x = 0$ we need $k \neq = 1$ . 3. Two distinct points for $k > 0$ and $k \neq = 1$ .

(ii) 1. The points are $(0, 0)$ and $(ln k, k ln k)$ (the $y$ -coordinate comes from the line $y = k x$ ).

(iii) 1. For $k > 1$ we have $ln k > 0$ . Between $0$ and $ln k$ the line lies above the curve (the line has a greater gradient at the origin). $A = \int_{0}^{l n k} (k x - x e^{x}) d x$ . 2. For $0 < k < 1$ , $ln k < 0$ and the region runs from $ln k$ to $0$ ; the line is still above the curve. $A = \int_{l n k}^{0} (k x - x e^{x}) d x$ .

(iv) 1. For $0 < k < 1$ we calculate the integral. Integrating by parts, $\int x e^{x} d x = (x - 1) e^{x}$ , and $\int k x d x = \frac{k x ^{2}}{2}$ . An antiderivative of $k x - x e^{x}$ is $G (x) = \frac{k x ^{2}}{2} - (x - 1) e^{x}$ . 2. $G (0) = 1$ and $G (ln k) = \frac{k ( ln k ) ^{2}}{2} - (ln k - 1) k$ . 3. $A = G (0) - G (ln k) = 1 - [\frac{k ln ^{2} k}{2} + k - k ln k]$ . 4. Therefore $1 - A = \frac{k ln ^{2} k}{2} + k - k ln k = \frac{k}{2} [(ln k - 1)^{2} + 1]$ , which is positive because $k > 0$ . 5. Hence $A < 1$ .

Exercise 204 HardIB · P1 TZ2 May 2011

a) (i) Sketch, on the same axes, the graphs of $y = sin x$ and $y = sin 2 x$ , for $0 \leq x \leq \frac{π}{2}$ .

(ii) Find the $x$ -coordinates of the points of intersection of both graphs, within the domain $0 \leq x \leq \frac{π}{2}$ .

(iii) Find the area bounded by the graphs.

b) Find the value of $\int_{0}^{1} \frac{x}{4 - x} d x$ . To do so, use the substitution $x = 4 sin^{2} θ$ .

c) The increasing function $f$ satisfies $f (0) = 0$ and $f (a) = b$ , where $a > 0$ and $b > 0$ .

(i) With reference to a rough sketch, show that $\int_{0}^{a} f (x) d x = ab - \int_{0}^{b} f^{- 1} (x) d x$ .

(ii) Hence, find the value of $\int_{0}^{2} arcsin (\frac{x}{4}) d x$ .

View solution

a) (i) The graph of $y = sin x$ increases from $0$ to $1$ on $[0, \frac{π}{2}]$ . The graph of $y = sin 2 x$ has period $π$ : it rises to $1$ at $x = \frac{π}{4}$ and falls to $0$ at $x = \frac{π}{2}$ . Both start at the origin and they intersect at $x = \frac{π}{3}$ .

Graphs of y = sin x and y = sin 2x for x between 0 and π/2, drawn on the same axes: they start at the origin and intersect at an intermediate point

(ii) 1. $sin x = sin 2 x = 2 sin x cos x$ , so $sin x (1 - 2 cos x) = 0$ . 2. Either $sin x = 0$ , i.e. $x = 0$ , or $cos x = \frac{1}{2}$ , i.e. $x = \frac{π}{3}$ .

(iii) 1. The region enclosed by the two graphs is the one between the two points of intersection, $0 \leq x \leq \frac{π}{3}$ . There $sin 2 x \geq sin x$ (for example, at $x = \frac{π}{4}$ : $1$ versus $\frac{2}{2}$ ). 2. $Area = \int_{0}^{π /3} (sin 2 x - sin x) d x = [- \frac{cos 2 x}{2} + cos x]_{0}^{π /3}$ . 3. At $\frac{π}{3}$ : $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$ ; at $0$ : $- \frac{1}{2} + 1 = \frac{1}{2}$ . The area is $\frac{3}{4} - \frac{1}{2} = \frac{1}{4}$ .

b) Integration by substitution, $x = 4 sin^{2} θ$ .

$d x = 8 sin θ cos θ d θ$ . Also $\frac{x}{4 - x} = \frac{4 sin ^{2} θ}{4 cos ^{2} θ} = tan^{2} θ$ , and with $θ \in [0, \frac{π}{2}]$ , $\frac{x}{4 - x} = tan θ$ .
Limits: $x = 0 \Rightarrow θ = 0$ ; $x = 1 \Rightarrow sin^{2} θ = \frac{1}{4} \Rightarrow θ = \frac{π}{6}$ .
The integral becomes $\int_{0}^{π /6} tan θ \cdot 8 sin θ cos θ d θ = \int_{0}^{π /6} 8 sin^{2} θ d θ = \int_{0}^{π /6} 4 (1 - cos 2 θ) d θ$ .
$= 4 [θ - \frac{sin 2 θ}{2}]_{0}^{π /6} = 4 (\frac{π}{6} - \frac{3}{4}) = \frac{2 π}{3} - 3$ .

c) (i) 1. As $f$ is increasing with $f (0) = 0$ and $f (a) = b$ , its graph runs from $(0, 0)$ to $(a, b)$ inside the rectangle with vertices $(0, 0)$ and $(a, b)$ , whose area is $ab$ . 2. The curve divides the rectangle into two regions. The region below the curve has area $\int_{0}^{a} f (x) d x$ . 3. The region to the left of the curve (between the curve and the $y$ -axis) is found by integrating with respect to $y$ the function $x = f^{- 1} (y)$ from $y = 0$ to $y = b$ : its area is $\int_{0}^{b} f^{- 1} (x) d x$ . 4. The sum of the two areas is $ab$ , so $\int_{0}^{a} f (x) d x = ab - \int_{0}^{b} f^{- 1} (x) d x$ .

(ii) 1. The function $arcsin (\frac{x}{4})$ is the inverse of $f (x) = 4 sin x$ on $[0, \frac{π}{2}]$ , which is increasing with $f (0) = 0$ . 2. We take $b = 2$ : $4 sin a = 2 \Rightarrow a = \frac{π}{6}$ . 3. $\int_{0}^{2} arcsin (\frac{x}{4}) d x = ab - \int_{0}^{π /6} 4 sin x d x = \frac{π}{6} \cdot 2 - [- 4 cos x]_{0}^{π /6} = \frac{π}{3} - (4 - 23)$ . 4. That is, $\frac{π}{3} + 23 - 4$ .

Answer: a) (i) graph; (ii) $x = 0$ and $x = \frac{π}{3}$ ; (iii) Area $= \frac{1}{4}$ ; b) $\frac{2 π}{3} - 3$ ; c) (i) verified; (ii) $\frac{π}{3} + 23 - 4$

Exercise 205 HardIB · P1 TZ0 November 2013

The function $f$ is given by $f (x) = x e^{- x}$ , $x \geq 0$ .

a) (i) Find an expression for $f^{'} (x)$ .

(ii) Hence, determine the coordinates of $A$ , the point where $f^{'} (x) = 0$ .

b) Find an expression for $f^{''} (x)$ and, hence, verify that the point $A$ is a maximum.

c) Find the coordinates of $B$ , the point of inflexion.

d) The graph of the function $g$ is obtained from the graph of $f$ by a stretch of scale factor $2$ in the direction of the $x$ -axis.

(i) Write down an expression for $g (x)$ .

(ii) State the coordinates of $C$ , the maximum point of $g$ .

(iii) Find the $x$ -coordinates of $D$ and $E$ , the two points where $f (x) = g (x)$ .

e) Sketch the graphs of $y = f (x)$ and $y = g (x)$ on the same axes, clearly showing the points $A$ , $B$ , $C$ , $D$ and $E$ .

f) Find an exact value for the area of the region bounded by the curve $y = g (x)$ , the $x$ -axis and the line $x = 1$ .

View solution

a) (i) Product rule: $f^{'} (x) = e^{- x} + x (- e^{- x}) = e^{- x} (1 - x)$ .

(ii) $f^{'} (x) = 0$ with $e^{- x} > 0$ gives $x = 1$ ; $f (1) = \frac{1}{e}$ . Therefore $A (1, \frac{1}{e})$ .

b) 1. $f^{''} (x) = - e^{- x} (1 - x) + e^{- x} (- 1) = e^{- x} (x - 2)$ . 2. $f^{''} (1) = - e^{- 1} < 0$ , so there is a maximum at $A$ .

c) 1. The point of inflexion satisfies $f^{''} (x) = 0$ : $x = 2$ , and $f^{''}$ changes sign (from $-$ to $+$ ). 2. $f (2) = 2 e^{- 2}$ . Therefore $B (2, \frac{2}{e ^{2}})$ .

d) (i) A horizontal stretch of scale factor $2$ replaces $x$ by $\frac{x}{2}$ : $g (x) = f (\frac{x}{2}) = \frac{x}{2} e^{- \frac{x}{2}}$ .

(ii) The maximum of $f$ at $(1, e^{- 1})$ moves to $(2 \cdot 1, e^{- 1})$ : $C (2, \frac{1}{e})$ .

(iii) 1. $x e^{- x} = \frac{x}{2} e^{- \frac{x}{2}}$ . One solution is $x = 0$ . 2. If $x \neq = 0$ , we divide by $x$ : $e^{- x} = \frac{1}{2} e^{- \frac{x}{2}}$ , so $e^{- \frac{x}{2}} = \frac{1}{2}$ and $x = 2 ln 2$ . 3. The points are $D (0, 0)$ and $E (2 ln 2, \frac{ln 2}{2})$ (with $g (2 ln 2) = ln 2 \cdot e^{- l n 2} = \frac{ln 2}{2}$ ). The $x$ -coordinates are $0$ and $2 ln 2$ .

e) The graph of $f$ starts at $(0, 0)$ , has the maximum $A (1, \frac{1}{e})$ , the point of inflexion $B (2, \frac{2}{e ^{2}})$ , and tends to $0$ as $x \to \infty$ . The graph of $g$ also starts at $(0, 0)$ , is wider, has the maximum $C (2, \frac{1}{e})$ and also tends to $0$ . They intersect at $D (0, 0)$ and at $E$ ( $x = 2 ln 2 \approx 1.39$ ).

Graphs of f(x) = x·e^(−x) and g(x) = (x/2)·e^(−x/2) for x ≥ 0 on the same axes, with their maxima, the point of inflexion of f and the two points where they intersect

f) Integration by substitution, $u = \frac{x}{2}$ .

Area $= \int_{0}^{1} \frac{x}{2} e^{- \frac{x}{2}} d x$ . With $u = \frac{x}{2}$ , $d x = 2 d u$ , and the limits become $0$ and $\frac{1}{2}$ : $= \int_{0}^{1/2} 2 u e^{- u} d u$ .
Integration by parts, with $u$ and $d v = e^{- u} d u$ (the derivative of $u$ is $1$ , $v = - e^{- u}$ ): $\int u e^{- u} d u = - u e^{- u} - e^{- u}$ .
Area $= 2 [- u e^{- u} - e^{- u}]_{0}^{1/2} = 2 [(- \frac{3}{2} e^{- \frac{1}{2}}) - (- 1)] = 2 - 3 e^{- \frac{1}{2}}$ .

Answer: a) (i) $f^{'} (x) = e^{- x} (1 - x)$ ; (ii) $A (1, \frac{1}{e})$ ; b) $f^{''} (x) = e^{- x} (x - 2)$ , $f^{''} (1) < 0$ ; c) $B (2, \frac{2}{e ^{2}})$ ; d) (i) $g (x) = \frac{x}{2} e^{- \frac{x}{2}}$ ; (ii) $C (2, \frac{1}{e})$ ; (iii) $x = 0$ and $x = 2 ln 2$ ; e) graph; f) Area $= 2 - 3 e^{- \frac{1}{2}}$

Exercise 206 HardIB · P1 TZ0 November 2014

The function $f$ is given by $f (x) = e^{3 x + 1}$ , $x \in R$ .

a) i) Find $f^{- 1} (x)$ .

(ii) State the domain of $f^{- 1}$ .

The function $g$ is given by $g (x) = ln x$ , $x \in R^{+}$ .

The graph of $y = g (x)$ and the graph of $y = f^{- 1} (x)$ intersect at the point $P$ .

b) Find the coordinates of $P$ .

The graph of $y = g (x)$ meets the $x$ -axis at the point $Q$ .

c) Show that the equation of the tangent $T$ to the graph of $y = g (x)$ at the point $Q$ is $y = x - 1$ .

A region $R$ is enclosed by the graph of $y = g (x)$ , the tangent $T$ and the line $x = e$ .

d) Find the area of the region $R$ .

e) i) Show that $g (x) \leq x - 1$ , $x \in R^{+}$ .

(ii) Substituting $x$ by $\frac{1}{x}$ in part e) (i), show that $\frac{x - 1}{x} \leq g (x)$ , $x \in R^{+}$ .

Show answer

Answer: a) (i) $f^{- 1} (x) = \frac{ln x - 1}{3}$ ; (ii) $] 0, + \infty [$ ; b) $P (\frac{1}{e}, - \frac{1}{2})$ ; c) $y = x - 1$ ; d) Area $= \frac{e ^{2}}{2} - e - \frac{1}{2}$ ; e) (i) and (ii) proved

Show working and answer

a) (i) 1. We write $y = e^{3 x + 1}$ and solve for $x$ : $ln y = 3 x + 1$ , so $x = \frac{ln y - 1}{3}$ . 2. Interchanging the variables: $f^{- 1} (x) = \frac{ln x - 1}{3}$ .

(ii) The domain of $f^{- 1}$ is the range of $f$ . As $e^{3 x + 1} > 0$ and takes all positive values, the domain is $] 0, + \infty [$ .

b) 1. Equating $ln x = \frac{ln x - 1}{3}$ : $3 ln x = ln x - 1$ , so $ln x = - \frac{1}{2}$ . 2. $x = e^{- \frac{1}{2}} = \frac{1}{e}$ and $y = - \frac{1}{2}$ . Thus $P (\frac{1}{e}, - \frac{1}{2})$ .

c) 1. $g (x) = 0 \Rightarrow ln x = 0 \Rightarrow x = 1$ , so $Q (1, 0)$ . 2. $g^{'} (x) = \frac{1}{x}$ , so the gradient at $Q$ is $g^{'} (1) = 1$ . 3. The tangent is $y - 0 = 1 (x - 1)$ , that is, $y = x - 1$ .

d) 1. Between $x = 1$ (where the tangent touches the curve) and $x = e$ , the line lies above the curve: at $x = e$ , $e - 1 \approx 1.72 > 1 = ln e$ . 2. Area $= \int_{1}^{e} (x - 1 - ln x) d x$ . An antiderivative of $ln x$ (by parts, $u = ln x$ , $d v = d x$ ) is $x ln x - x$ . 3. Area $= [\frac{x ^{2}}{2} - x - x ln x + x]_{1}^{e} = [\frac{x ^{2}}{2} - x ln x]_{1}^{e} = (\frac{e ^{2}}{2} - e) - \frac{1}{2}$ .

e) (i) 1. Let $h (x) = x - 1 - ln x$ , $x > 0$ . Proving $g (x) \leq x - 1$ is equivalent to proving $h (x) \geq 0$ . 2. $h^{'} (x) = 1 - \frac{1}{x} = \frac{x - 1}{x}$ : negative if $x < 1$ , positive if $x > 1$ , and zero at $x = 1$ . Hence $h$ has a minimum at $x = 1$ . 3. $h (1) = 0$ , so $h (x) \geq 0$ for all $x > 0$ , that is, $ln x \leq x - 1$ .

(ii) 1. Substituting $x$ by $\frac{1}{x}$ (valid because $\frac{1}{x} > 0$ ): $ln \frac{1}{x} \leq \frac{1}{x} - 1$ . 2. As $ln \frac{1}{x} = - ln x$ : $- ln x \leq \frac{1 - x}{x}$ . 3. Multiplying by $- 1$ (which reverses the inequality): $ln x \geq \frac{x - 1}{x}$ , that is, $\frac{x - 1}{x} \leq g (x)$ .

Exercise 207 HardIB · P1 TZ2 May 2014

The graph of the function $f (x) = \frac{x + 1}{x ^{2} + 1}$ is shown below.

Graph of a rational function that meets the x-axis at x = −1, passes through (0, 1), has a maximum near x = 0.4 and an inflexion point labelled A (1, 1); it tends to 0 at both ends; the region between the curve and the x-axis for x between −1 and 0 is shaded

a) Find $f^{'} (x)$ .

b) Hence, find the $x$ -coordinates of the points at which the gradient of the graph of $f$ is equal to zero.

c) Find $f^{''} (x)$ , expressing the answer in the form $\frac{p ( x )}{( x ^{2} + 1 ) ^{3}}$ , where $p (x)$ is a polynomial of degree $3$ .

The point $(1, 1)$ is a point of inflexion. There are two other points of inflexion.

d) Find the $x$ -coordinates of the other two points of inflexion.

e) Find the area of the shaded region. Give your answer in the form $\frac{π}{a} - ln b$ , where $a$ and $b$ are integers.

Show answer

Answer: a) $f^{'} (x) = \frac{- x ^{2} - 2 x + 1}{( x ^{2} + 1 ) ^{2}}$ ; b) $x = - 1 \pm 2$ ; c) $f^{''} (x) = \frac{2 x ^{3} + 6 x ^{2} - 6 x - 2}{( x ^{2} + 1 ) ^{3}}$ ; d) $x = - 2 \pm 3$ ; e) Area $= \frac{π}{4} - ln 2$

Show working and answer

a) 1. Using the quotient rule with $u = x + 1$ , $v = x^{2} + 1$ : $f^{'} (x) = \frac{1 \cdot ( x ^{2} + 1 ) - ( x + 1 ) \cdot 2 x}{( x ^{2} + 1 ) ^{2}}$ . 2. Numerator: $x^{2} + 1 - 2 x^{2} - 2 x = - x^{2} - 2 x + 1$ . Therefore $f^{'} (x) = \frac{- x ^{2} - 2 x + 1}{( x ^{2} + 1 ) ^{2}}$ .

b) 1. $f^{'} (x) = 0 \Rightarrow - x^{2} - 2 x + 1 = 0 \Rightarrow x^{2} + 2 x - 1 = 0$ . 2. $x = \frac{- 2 \pm 8}{2} = - 1 \pm 2$ .

c) 1. Differentiating the quotient with $N = - x^{2} - 2 x + 1$ and $D = (x^{2} + 1)^{2}$ : $f^{''} = \frac{N ^{'} D - N D ^{'}}{D ^{2}}$ , with $D^{'} = 2 (x^{2} + 1) \cdot 2 x$ . 2. Taking out the common factor $(x^{2} + 1)$ and simplifying: $f^{''} (x) = \frac{N ^{'} ( x ^{2} + 1 ) - 4 x N}{( x ^{2} + 1 ) ^{3}}$ . 3. Numerator: $(- 2 x - 2) (x^{2} + 1) - 4 x (- x^{2} - 2 x + 1) = - 2 x^{3} - 2 x^{2} - 2 x - 2 + 4 x^{3} + 8 x^{2} - 4 x = 2 x^{3} + 6 x^{2} - 6 x - 2$ . 4. Hence $f^{''} (x) = \frac{2 x ^{3} + 6 x ^{2} - 6 x - 2}{( x ^{2} + 1 ) ^{3}}$ .

d) 1. Points of inflexion satisfy $f^{''} (x) = 0$ : $2 x^{3} + 6 x^{2} - 6 x - 2 = 0$ , that is, $x^{3} + 3 x^{2} - 3 x - 1 = 0$ . 2. $x = 1$ is a root (given in the question): $1 + 3 - 3 - 1 = 0$ . Dividing: $x^{3} + 3 x^{2} - 3 x - 1 = (x - 1) (x^{2} + 4 x + 1)$ . 3. $x^{2} + 4 x + 1 = 0 \Rightarrow x = \frac{- 4 \pm 12}{2} = - 2 \pm 3$ .

e) 1. Between $x = - 1$ and $x = 0$ the curve lies above the $x$ -axis (from $f (- 1) = 0$ to $f (0) = 1$ ). Area $= \int_{- 1}^{0} \frac{x + 1}{x ^{2} + 1} d x$ . 2. Splitting: $\int \frac{x}{x ^{2} + 1} d x + \int \frac{1}{x ^{2} + 1} d x = \frac{1}{2} ln (x^{2} + 1) + arctan x$ (the first with $u = x^{2} + 1$ ). 3. $[\frac{1}{2} ln (x^{2} + 1) + arctan x]_{- 1}^{0} = 0 - (\frac{1}{2} ln 2 - \frac{π}{4}) = \frac{π}{4} - \frac{1}{2} ln 2 = \frac{π}{4} - ln 2$ .

Exercise 208 HardIB · P2 May 2001

Let $f (x) = x cos 3 x$ .

a) Use integration by parts to show that $\int f (x) d x = \frac{1}{3} x sin 3 x + \frac{1}{9} cos 3 x + C$ .

b) Use your answer to part a) to calculate the exact value of the area enclosed by $f (x)$ and the $x$ -axis in each of the following cases. Give your answers in terms of $π$ .

(i) $\frac{π}{6} \leq x \leq \frac{3 π}{6}$

(ii) $\frac{3 π}{6} \leq x \leq \frac{5 π}{6}$

(iii) $\frac{5 π}{6} \leq x \leq \frac{7 π}{6}$

c) Given that the areas calculated above are the first three terms of an arithmetic sequence, calculate an expression for the total area enclosed by $f (x)$ and the $x$ -axis for $\frac{π}{6} \leq x \leq \frac{( 2 n + 1 ) π}{6}$ , with $n \in Z^{+}$ . Give your answer in terms of $n$ and $π$ .

Show answer

Answer: a) verified; b) (i) $\frac{2 π}{9}$ ; (ii) $\frac{4 π}{9}$ ; (iii) $\frac{6 π}{9}$ ; c) Area $= \frac{n ( n + 1 ) π}{9}$

Show working and answer

a) Integration by parts with $u = x$ and $d v = cos 3 x d x$ .

Then $d u = d x$ and $v = \frac{1}{3} sin 3 x$ .
$\int x cos 3 x d x = \frac{1}{3} x sin 3 x - \frac{1}{3} \int sin 3 x d x = \frac{1}{3} x sin 3 x + \frac{1}{9} cos 3 x + C$ .

b) Let $F (x) = \frac{1}{3} x sin 3 x + \frac{1}{9} cos 3 x$ . The area is the absolute value of the integral over each interval, because $f$ does not change sign within each one ( $cos 3 x$ changes sign at $\frac{π}{6}$ , $\frac{3 π}{6}$ , $\frac{5 π}{6}$ , $\frac{7 π}{6}$ , ...).

(i) 1. $F (\frac{π}{2}) = \frac{1}{3} \cdot \frac{π}{2} \cdot (- 1) + 0 = - \frac{π}{6}$ and $F (\frac{π}{6}) = \frac{1}{3} \cdot \frac{π}{6} \cdot 1 + 0 = \frac{π}{18}$ . 2. $\int_{π /6}^{π /2} f = - \frac{π}{6} - \frac{π}{18} = - \frac{2 π}{9}$ , so the area is $\frac{2 π}{9}$ .

(ii) 1. $F (\frac{5 π}{6}) = \frac{1}{3} \cdot \frac{5 π}{6} \cdot sin \frac{5 π}{2} + 0 = \frac{5 π}{18}$ . 2. $\int_{π /2}^{5 π /6} f = \frac{5 π}{18} + \frac{π}{6} = \frac{8 π}{18} = \frac{4 π}{9}$ . The area is $\frac{4 π}{9}$ .

(iii) 1. $F (\frac{7 π}{6}) = \frac{1}{3} \cdot \frac{7 π}{6} \cdot sin \frac{7 π}{2} + 0 = - \frac{7 π}{18}$ . 2. $\int_{5 π /6}^{7 π /6} f = - \frac{7 π}{18} - \frac{5 π}{18} = - \frac{2 π}{3}$ . The area is $\frac{6 π}{9} = \frac{2 π}{3}$ .

c) 1. The areas $\frac{2 π}{9}, \frac{4 π}{9}, \frac{6 π}{9}, \dots$ form an arithmetic sequence with first term $u_{1} = \frac{2 π}{9}$ and common difference $d = \frac{2 π}{9}$ , with $u_{k} = \frac{2 k π}{9}$ . 2. The interval from $\frac{π}{6}$ to $\frac{( 2 n + 1 ) π}{6}$ contains exactly the first $n$ intervals. The total area is the sum of the first $n$ terms: $S_{n} = \frac{n}{2} (u_{1} + u_{n}) = \frac{n}{2} (\frac{2 π}{9} + \frac{2 nπ}{9}) = \frac{n ( n + 1 ) π}{9}$ .

Answer: a) verified; b) (i) $\frac{2 π}{9}$ ; (ii) $\frac{4 π}{9}$ ; (iii) $\frac{6 π}{9}$ ; c) Area $= \frac{n ( n + 1 ) π}{9}$

Exercise 209 HardIB · P2 November 2002

a) On the same axes, sketch the graphs of the functions $f (x)$ and $g (x)$ , where

$f (x) = 4 - (1 - x)^{2}$ , for $- 2 \leq x \leq 4$ $g (x) = ln (x + 3) - 2$ , for $- 3 \leq x \leq 5$ .

b) i) Write down the equation of every vertical asymptote.

(ii) Write down the intercepts of $g (x)$ with the $x$ -axis and with the $y$ -axis.

c) Find the values of $x$ for which $f (x) = g (x)$ .

d) Let $A$ be the region in which $f (x) \geq g (x)$ , $x \geq 0$ .

(i) Shade the region $A$ on the graph.

(ii) Write down an integral that represents the shaded area $A$ .

(iii) Evaluate this integral.

e) Find, in the region $A$ , the maximum vertical distance between $f (x)$ and $g (x)$ .

View solution

a) The parabola $f (x) = 4 - (1 - x)^{2}$ has its vertex at $(1, 4)$ , opens downwards and meets the $x$ -axis at $x = - 1$ and $x = 3$ . The curve $g (x) = ln (x + 3) - 2$ is a shifted logarithmic curve: it has the vertical asymptote $x = - 3$ and increases slowly.

Graphs of f(x) = 4 − (1 − x)², a parabola with vertex (1, 4), and of g(x) = ln(x + 3) − 2, with vertical asymptote x = −3, on the same axes

b) (i) The vertical asymptote occurs when $x + 3 = 0$ : $x = - 3$ .

(ii) 1. Intersection with the $y$ -axis: $g (0) = ln 3 - 2 \approx - 0.901$ , that is, $(0, - 0.901)$ . 2. Intersection with the $x$ -axis: $ln (x + 3) = 2 \Rightarrow x = e^{2} - 3 \approx 4.39$ , that is, $(4.39, 0)$ .

c) Using the GDC, we solve $f (x) = g (x)$ : $x \approx - 1.34$ and $x \approx 3.05$ .

d) (i) For $x \geq 0$ , $f (x) \geq g (x)$ between $x = 0$ and $x \approx 3.05$ . The region $A$ is the one between the parabola (above) and the logarithmic curve (below), bounded on the left by the $y$ -axis.

Graphs of f(x) = 4 − (1 − x)² and g(x) = ln(x + 3) − 2 with the region A shaded: the one between the parabola, above, and the logarithmic curve, below, for x between 0 and the second intersection point

(ii) $Area = \int_{0}^{3.05} [4 - (1 - x)^{2} - ln (x + 3) + 2] d x$ .

(iii) Using the GDC, the value of the integral is $\approx 10.6$ .

e) 1. The vertical distance is $D (x) = f (x) - g (x) = 6 - (1 - x)^{2} - ln (x + 3)$ for $0 \leq x \leq 3.05$ . 2. Using the GDC, the maximum of $D$ occurs at $x \approx 0.871$ (it can also be obtained from $D^{'} (x) = 2 (1 - x) - \frac{1}{x + 3} = 0$ ). 3. $D_{m a x} \approx 4.63$ .

Answer: a) graphs; b) (i) $x = - 3$ ; (ii) $(0, - 0.901)$ and $(4.39, 0)$ ; c) $x \approx - 1.34$ , $x \approx 3.05$ ; d) (ii) $\int_{0}^{3.05} [4 - (1 - x)^{2} - ln (x + 3) + 2] d x$ ; (iii) $\approx 10.6$ ; e) maximum vertical distance $\approx 4.63$

Exercise 210 HardIB · P1 TZ2 May 2019

Consider the functions $f$ and $g$ defined on the domain $0 < x < 2 π$ by $f (x) = 3 cos 2 x$ and $g (x) = 4 - 11 cos x$ . The following diagram shows the graph of $y = f (x)$ and the graph of $y = g (x)$ .

Graphs of y = 4 − 11cos x (a large arc that starts below the x-axis and reaches its maximum at the centre) and y = 3cos 2x (oscillates between −3 and 3), for 0 < x < 2π; the region between them is shaded between their two points of intersection; A on the first curve and B on the second, with the same x-coordinate

a) Find the $x$ -coordinate of each of the points of intersection of the two graphs.

b) Find the exact area of the shaded region; give your answer in the form $p π + q 3$ , where $p, q \in Q$ .

At the points $A$ and $B$ shown in the diagram, the two graphs have the same gradient.

c) Determine the $y$ -coordinate of $A$ , which lies on the graph of $g$ .

Show answer

Answer: a) $x = \frac{π}{3}$ , $x = \frac{5 π}{3}$ ; b) Area $= \frac{16}{3} π + \frac{25}{2} 3$ ; c) $y_{A} = \frac{169}{12}$

Show working and answer

a) 1. We equate the two functions: $3 cos 2 x = 4 - 11 cos x$ . Using $cos 2 x = 2 cos^{2} x - 1$ : $6 cos^{2} x - 3 = 4 - 11 cos x$ , that is, $6 cos^{2} x + 11 cos x - 7 = 0$ . 2. With $c = cos x$ : $c = \frac{- 11 \pm 121 + 168}{12} = \frac{- 11 \pm 17}{12}$ , so $c = \frac{1}{2}$ or $c = - \frac{7}{3}$ . 3. $cos x = - \frac{7}{3}$ is impossible. $cos x = \frac{1}{2}$ on $0 < x < 2 π$ gives $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$ .

b) 1. Between the points of intersection, $g$ lies above $f$ (at $x = π$ : $g = 15$ and $f = 3$ ). 2. Area $= \int_{π /3}^{5 π /3} (4 - 11 cos x - 3 cos 2 x) d x = [4 x - 11 sin x - \frac{3}{2} sin 2 x]_{π /3}^{5 π /3}$ . 3. At $\frac{5 π}{3}$ : $\frac{20 π}{3} + \frac{11 3}{2} + \frac{3 3}{4}$ . At $\frac{π}{3}$ : $\frac{4 π}{3} - \frac{11 3}{2} - \frac{3 3}{4}$ . 4. Subtracting: $\frac{16 π}{3} + 113 + \frac{3 3}{2} = \frac{16}{3} π + \frac{25}{2} 3$ .

c) 1. The gradients are equal at $A$ and $B$ (they have the same $x$ -coordinate), so $f^{'} (x) = g^{'} (x)$ . 2. $f^{'} (x) = - 6 sin 2 x = - 12 sin x cos x$ and $g^{'} (x) = 11 sin x$ . 3. $- 12 sin x cos x = 11 sin x$ . As $sin x \neq = 0$ on $] 0, π [$ (the points are to the left of the centre), $cos x = - \frac{11}{12}$ . 4. The $y$ -coordinate of $A$ (on $g$ ) is $4 - 11 cos x = 4 + \frac{121}{12} = \frac{169}{12}$ .

Answer: a) $x = \frac{π}{3}$ , $x = \frac{5 π}{3}$ ; b) Area $= \frac{16}{3} π + \frac{25}{2} 3$ ; c) $y_{A} = \frac{169}{12}$

Exercise 211 HardIB · P2 November 1996

The graph of $f (x) = \frac{1 + ln x}{x}$ is shown below.

Graph of f(x) = (1 + ln x)/x with a vertical asymptote on the y-axis: cuts the x-axis at A, has its maximum at B and a point of inflexion at C; the line OT passes through the origin and through the point T on the curve, and vertical lines are drawn from T, B and C to the x-axis (D is the foot of the vertical from T)

The line $(OT)$ , where $O$ is the origin and $T$ lies on the curve, is the tangent at $T$ .

The graph of $f (x)$ cuts the $x$ -axis at $A$ .

a) Calculate the coordinates of $A$ .

b) Calculate the coordinates of the point $B$ where $f (x)$ attains its maximum value.

c) Calculate the coordinates of $C$ , the point of inflexion of $f (x)$ .

d) If the $x$ -coordinate of $T$ is $x_{0}$ , calculate, in terms of $x_{0}$ , the equation of the tangent line $(OT)$ .

e) Calculate the value of $x_{0}$ .

f) Prove that the $x$ -coordinates of the points $A$ , $T$ , $B$ and $C$ , as defined above (and shown on the diagram), are four consecutive terms of a geometric sequence. Write down the common ratio of the sequence.

g) Calculate the area of the region $OTA$ .

Show answer

Answer: a) $A (e^{- 1}, 0)$ ; b) $B (1, 1)$ ; c) $C (e^{\frac{1}{2}}, \frac{3 e ^{- 1/2}}{2})$ ; d) $OT$ : $y = - \frac{ln x _{0}}{x _{0}^{2}} x$ ; e) $x_{0} = e^{- \frac{1}{2}}$ ; f) proved, ratio $r = e^{\frac{1}{2}}$ ; g) Area $OTA = \frac{1}{8}$

Show working and answer

a) $f (x) = 0 \Rightarrow 1 + ln x = 0 \Rightarrow x = e^{- 1}$ . So $A (e^{- 1}, 0)$ .

b) 1. Differentiating with the quotient rule: $f^{'} (x) = \frac{\frac{1}{x} \cdot x - ( 1 + ln x )}{x ^{2}} = \frac{- ln x}{x ^{2}}$ . 2. $f^{'} (x) = 0 \Rightarrow x = 1$ . It is a maximum: $f^{'} > 0$ for $x < 1$ and $f^{'} < 0$ for $x > 1$ . 3. $f (1) = 1$ , so $B (1, 1)$ .

c) 1. $f^{''} (x) = \frac{- \frac{1}{x} \cdot x ^{2} + ln x \cdot 2 x}{x ^{4}} = \frac{2 ln x - 1}{x ^{3}}$ . 2. $f^{''} (x) = 0 \Rightarrow ln x = \frac{1}{2} \Rightarrow x = e^{\frac{1}{2}}$ , and $f^{''}$ changes sign ( $-$ to $+$ ). 3. $f (e^{1/2}) = \frac{1 + \frac{1}{2}}{e ^{1/2}} = \frac{3}{2} e^{- \frac{1}{2}}$ . So $C (e^{\frac{1}{2}}, \frac{3 e ^{- 1/2}}{2})$ .

d) 1. The tangent at $x_{0}$ is $y - f (x_{0}) = f^{'} (x_{0}) (x - x_{0})$ with $f^{'} (x_{0}) = - \frac{ln x _{0}}{x _{0}^{2}}$ . 2. As it passes through the origin, $f (x_{0}) = x_{0} f^{'} (x_{0})$ , and the line is $y = f^{'} (x_{0}) x$ . 3. That is, $OT$ : $y = - \frac{ln x _{0}}{x _{0}^{2}} x$ .

e) 1. From the condition $f (x_{0}) = x_{0} f^{'} (x_{0})$ : $\frac{1 + ln x _{0}}{x _{0}} = - \frac{ln x _{0}}{x _{0}}$ . 2. $1 + ln x_{0} = - ln x_{0} \Rightarrow ln x_{0} = - \frac{1}{2} \Rightarrow x_{0} = e^{- \frac{1}{2}}$ .

f) 1. The $x$ -coordinates of $A$ , $T$ , $B$ and $C$ are $e^{- 1}$ , $e^{- \frac{1}{2}}$ , $1 = e^{0}$ and $e^{\frac{1}{2}}$ . 2. The ratio of each term to the previous one is $e^{\frac{1}{2}}$ ( $e^{- \frac{1}{2}} / e^{- 1} = e^{\frac{1}{2}}$ , $1/ e^{- \frac{1}{2}} = e^{\frac{1}{2}}$ , $e^{\frac{1}{2}} /1 = e^{\frac{1}{2}}$ ). This ratio is constant, so they are consecutive terms of a geometric sequence with common ratio $r = e^{\frac{1}{2}}$ .

g) 1. The foot of the vertical from $T$ is $D (e^{- 1/2}, 0)$ , and $y_{T} = f (e^{- 1/2}) = \frac{1 - \frac{1}{2}}{e ^{- 1/2}} = \frac{1}{2} e^{\frac{1}{2}}$ . 2. Triangle $ODT$ has area $\frac{1}{2} \cdot e^{- \frac{1}{2}} \cdot \frac{1}{2} e^{\frac{1}{2}} = \frac{1}{4}$ . 3. Region $OTA$ is that triangle minus the area under the curve between $A$ and $D$ (the curve lies below its tangent $OT$ because $f^{''} < 0$ there): $\int_{e^{- 1}}^{e^{- 1/2}} \frac{1 + ln x}{x} d x = [ln x + \frac{( ln x ) ^{2}}{2}]_{e^{- 1}}^{e^{- 1/2}} = (- \frac{1}{2} + \frac{1}{8}) - (- 1 + \frac{1}{2}) = \frac{1}{8}$ . 4. Area $OTA = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$ .

Exercise 212 HardIB · P1 TZ0 November 2018

a) Use integration by parts to show that $\int e^{x} cos 2 x d x = \frac{2 e ^{x}}{5} sin 2 x + \frac{e ^{x}}{5} cos 2 x + c$ , $c \in R$ .

b) Hence, show that $\int e^{x} cos^{2} x d x = \frac{e ^{x}}{5} sin 2 x + \frac{e ^{x}}{10} cos 2 x + \frac{e ^{x}}{2} + c$ , $c \in R$ .

The function $f$ is defined as $f (x) = e^{x} cos^{2} x$ , where $0 \leq x \leq 5$ .

The following graph shows the curve $y = f (x)$ .

This curve has local maxima at $A$ and $C$ , and touches the $x$ -axis at $B$ and at $D$ .

Graph of y = eˣcos²x for x between 0 and 5: small local maximum at A, touches the x-axis at B, large local maximum at C, touches the axis at D and rises again up to x = 5; the region between the curve and the x-axis between B and D is shaded

c) Find the $x$ -coordinate of $A$ and of $C$ , giving your answers in the form $a + arctan b$ , where $a, b \in R$ .

d) Find the area of the region bounded by this curve and the $x$ -axis between $B$ and $D$ , as shaded in the figure.

View solution

a) Integration by parts twice. Let $I = \int e^{x} cos 2 x d x$ .

First time, with $u = cos 2 x$ and $d v = e^{x} d x$ : $I = e^{x} cos 2 x + 2 \int e^{x} sin 2 x d x$ .
Second time, with $u = sin 2 x$ and $d v = e^{x} d x$ : $\int e^{x} sin 2 x d x = e^{x} sin 2 x - 2 I$ .
Substituting: $I = e^{x} cos 2 x + 2 e^{x} sin 2 x - 4 I$ , so $5 I = e^{x} (cos 2 x + 2 sin 2 x)$ .
$I = \frac{2 e ^{x}}{5} sin 2 x + \frac{e ^{x}}{5} cos 2 x + c$ .

b) 1. As $cos^{2} x = \frac{1 + cos 2 x}{2}$ : $\int e^{x} cos^{2} x d x = \frac{1}{2} \int e^{x} d x + \frac{1}{2} \int e^{x} cos 2 x d x$ . 2. By a): $= \frac{e ^{x}}{2} + \frac{1}{2} (\frac{2 e ^{x}}{5} sin 2 x + \frac{e ^{x}}{5} cos 2 x) + c = \frac{e ^{x}}{5} sin 2 x + \frac{e ^{x}}{10} cos 2 x + \frac{e ^{x}}{2} + c$ .

c) 1. $f^{'} (x) = e^{x} cos^{2} x - 2 e^{x} cos x sin x = e^{x} cos x (cos x - 2 sin x)$ . 2. $f^{'} (x) = 0$ with $cos x = 0$ gives $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ : there $f = 0$ , so these are the points $B$ and $D$ that touch the axis. 3. The local maxima satisfy $cos x = 2 sin x$ , that is, $tan x = \frac{1}{2}$ : $x = arctan \frac{1}{2} \approx 0.464$ and $x = π + arctan \frac{1}{2} \approx 3.61$ (both in $[0, 5]$ ). 4. Hence $A$ : $x = arctan \frac{1}{2}$ ( $a = 0$ , $b = \frac{1}{2}$ ) and $C$ : $x = π + arctan \frac{1}{2}$ ( $a = π$ , $b = \frac{1}{2}$ ).

d) 1. $B$ and $D$ have $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ , and between them $f (x) = e^{x} cos^{2} x \geq 0$ , so the area is the integral. 2. Using the antiderivative from b), $G (x) = \frac{e ^{x}}{5} sin 2 x + \frac{e ^{x}}{10} cos 2 x + \frac{e ^{x}}{2}$ : $G (\frac{3 π}{2}) = e^{\frac{3 π}{2}} (0 - \frac{1}{10} + \frac{1}{2}) = \frac{2}{5} e^{\frac{3 π}{2}}$ and $G (\frac{π}{2}) = e^{\frac{π}{2}} (0 - \frac{1}{10} + \frac{1}{2}) = \frac{2}{5} e^{\frac{π}{2}}$ . 3. Area $= \frac{2}{5} (e^{\frac{3 π}{2}} - e^{\frac{π}{2}})$ .

Answer: a) and b) proved; c) $A$ : $x = arctan \frac{1}{2}$ ; $C$ : $x = π + arctan \frac{1}{2}$ ; d) Area $= \frac{2}{5} (e^{\frac{3 π}{2}} - e^{\frac{π}{2}})$

Exercise 213 HardIB · P1 TZ1 May 2017

a) i) Express $x^{2} + 3 x + 2$ in the form $(x + h)^{2} + k$ .

(ii) Factorise $x^{2} + 3 x + 2$ .

Consider the function $f (x) = \frac{1}{x ^{2} + 3 x + 2}$ , $x \in R$ , $x \neq = - 2$ , $x \neq = - 1$ .

b) Sketch the graph of $f (x)$ , indicating on it the equations of the asymptotes, the coordinates of the $y$ -intercept and the local maximum.

c) Show that $\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{1}{x ^{2} + 3 x + 2}$ .

d) Using the above, calculate the value of $p$ if $\int_{0}^{1} f (x) d x = ln (p)$ .

e) Sketch the graph of $y = f (∣ x ∣)$ .

f) Find the area of the region enclosed by the graph of $y = f (∣ x ∣)$ , the $x$ -axis and the lines $x = - 1$ and $x = 1$ .

View solution

a) (i) 1. Complete the square: $x^{2} + 3 x + 2 = (x + \frac{3}{2})^{2} - \frac{9}{4} + 2 = (x + \frac{3}{2})^{2} - \frac{1}{4}$ .

(ii) 1. Look for two numbers with product $2$ and sum $3$ : $x^{2} + 3 x + 2 = (x + 1) (x + 2)$ .

b) 1. The vertical asymptotes are where the denominator is zero: $x = - 1$ and $x = - 2$ . As $x \to \pm \infty$ , $f \to 0$ : horizontal asymptote $y = 0$ . 2. $y$ -intercept: $f (0) = \frac{1}{2}$ , i.e. $(0, \frac{1}{2})$ . It does not cut the $x$ -axis because $f$ is never $0$ . 3. Since $f = \frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{1}{4}}$ , the denominator has a minimum at $x = - \frac{3}{2}$ with value $- \frac{1}{4}$ . There $f = - 4$ , and it is a local maximum, $(- \frac{3}{2}, - 4)$ . 4. To the right of $x = - 1$ the curve falls from $+ \infty$ towards $0$ , passing through $(0, \frac{1}{2})$ . Between the two asymptotes the branch is an inverted arch (below the $x$ -axis) with that maximum. To the left of $x = - 2$ the curve comes from $0$ and rises towards $+ \infty$ .

Graph of f(x) = 1/(x² + 3x + 2) with vertical asymptotes x = −2 and x = −1 and horizontal asymptote y = 0: to the right of x = −1 the curve falls from above, passing through (0, 1/2); between the asymptotes there is a branch with local maximum at (−3/2, −4); to the left of x = −2 the curve rises upwards next to the asymptote

c) 1. Reduce to a common denominator: $\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{( x + 2 ) - ( x + 1 )}{( x + 1 ) ( x + 2 )} = \frac{1}{x ^{2} + 3 x + 2}$ .

d) 1. Use the decomposition from part c): $\int_{0}^{1} f (x) d x = \int_{0}^{1} (\frac{1}{x + 1} - \frac{1}{x + 2}) d x = [ln (x + 1) - ln (x + 2)]_{0}^{1}$ . 2. $= (ln 2 - ln 3) - (ln 1 - ln 2) = 2 ln 2 - ln 3 = ln \frac{4}{3}$ . 3. Hence $p = \frac{4}{3}$ .

e) 1. $y = f (∣ x ∣)$ is obtained by keeping the part of the graph with $x \geq 0$ and reflecting it in the $y$ -axis. The curve is symmetric, with maximum $(0, \frac{1}{2})$ , and tends to $0$ as $x \to \pm \infty$ (it has no vertical asymptotes).

Graph of y = f(|x|) = 1/(x² + 3|x| + 2): curve symmetric about the y-axis, with maximum (0, 1/2) and tending to 0 as x tends to ±∞; the lines x = −1 and x = 1 are marked with dashed lines

f) 1. Between $x = - 1$ and $x = 1$ the function $f (∣ x ∣)$ is positive and symmetric about the $y$ -axis. 2. Area $= 2 \int_{0}^{1} f (x) d x = 2 ln \frac{4}{3}$ .

Answer: a) (i) $(x + \frac{3}{2})^{2} - \frac{1}{4}$ ; (ii) $(x + 1) (x + 2)$ ; b) vertical asymptotes $x = - 1$ , $x = - 2$ , horizontal $y = 0$ ; intercept $(0, \frac{1}{2})$ ; local maximum $(- \frac{3}{2}, - 4)$ ; d) $p = \frac{4}{3}$ ; f) area $= 2 ln (\frac{4}{3})$

Exercise 214 HardIB · P1 TZ0 November 2016

Let $y = e^{x} sin x$ .

a) Find an expression for $\frac{d y}{d x}$ .

b) Show that $\frac{d ^{2} y}{d x ^{2}} = 2 e^{x} cos x$ .

Consider the function $f$ , defined by $f (x) = e^{x} sin x$ , $0 \leq x \leq π$ .

c) Show that the function $f$ has a local maximum value at $x = \frac{3 π}{4}$ .

d) Find the $x$ -coordinate of the point of inflexion of the graph of $f$ .

e) Sketch the graph of $f$ , clearly indicating the position of the local maximum point, the point of inflexion and the intercepts with the axes.

f) Find the area of the region bounded by the graph of $f$ and the $x$ -axis.

The curvature of a graph at any point $(x, y)$ is defined as $k = \frac{\frac{d ^{2} y}{d x ^{2}}}{( 1 + ( \frac{d y}{d x} ) ^{2} ) ^{\frac{3}{2}}}$ .

g) Find the value of the curvature of the graph of $f$ at the local maximum point.

h) Find the value of $k$ for $x = \frac{π}{2}$ and comment on its meaning with regard to the shape of the graph.

View solution

a) 1. Product rule: $\frac{d y}{d x} = e^{x} sin x + e^{x} cos x = e^{x} (sin x + cos x)$ .

b) 1. Differentiate again: $\frac{d ^{2} y}{d x ^{2}} = e^{x} (sin x + cos x) + e^{x} (cos x - sin x) = 2 e^{x} cos x$ .

c) 1. A local extremum satisfies $f^{'} (x) = 0$ : $e^{x} (sin x + cos x) = 0$ , so $tan x = - 1$ , and in $[0, π]$ the solution is $x = \frac{3 π}{4}$ . 2. Second derivative: $f^{''} (\frac{3 π}{4}) = 2 e^{3 π /4} cos \frac{3 π}{4} < 0$ , because $cos \frac{3 π}{4} = - \frac{2}{2} < 0$ . 3. Therefore it is a local maximum.

d) 1. A point of inflexion satisfies $f^{''} (x) = 0$ : $2 e^{x} cos x = 0$ , so $cos x = 0$ and $x = \frac{π}{2}$ in $[0, π]$ . 2. $f^{''}$ changes sign at $\frac{π}{2}$ (positive before, negative after), so it is a point of inflexion.

e) 1. Intercepts with the axes: $f (x) = 0$ when $sin x = 0$ , i.e. $(0, 0)$ and $(π, 0)$ . 2. The curve starts at $(0, 0)$ and increases, concave up until the point of inflexion at $x = \frac{π}{2}$ , reaches the local maximum at $x = \frac{3 π}{4}$ , and falls to cut the $x$ -axis at $(π, 0)$ .

Graph of f(x) = e^x sin x on (0, π), showing the intercepts with the axes, the local maximum and the point of inflexion

f) 1. On $[0, π]$ , $f \geq 0$ , so the area is $\int_{0}^{π} e^{x} sin x d x$ . 2. Integrating by parts twice (or by differentiating the candidate) gives the antiderivative $\frac{e ^{x} ( sin x - cos x )}{2}$ . 3. $[\frac{e ^{x} ( sin x - cos x )}{2}]_{0}^{π} = \frac{e ^{π} \cdot 1}{2} - \frac{1 \cdot ( - 1 )}{2} = \frac{e ^{π} + 1}{2} \approx 12.07$ .

g) 1. At the local maximum $f^{'} (\frac{3 π}{4}) = 0$ , so the denominator of $k$ equals $(1 + 0)^{3/2} = 1$ . 2. $f^{''} (\frac{3 π}{4}) = 2 e^{3 π /4} \cdot \frac{2}{2} = 2 e^{3 π /4}$ . 3. $k = 2 e^{3 π /4}$ .

h) 1. At $x = \frac{π}{2}$ we have $f^{''} = 0$ , so $k = 0$ . 2. Zero curvature means that at the point of inflexion the graph does not bend: locally it is approximated by a straight line.

Answer: a) $\frac{d y}{d x} = e^{x} (sin x + cos x)$ ; d) $x = \frac{π}{2}$ ; e) intercepts $(0, 0)$ and $(π, 0)$ ; f) area $= \frac{e ^{π} + 1}{2}$ ; g) $k = 2 e^{\frac{3 π}{4}}$ ; h) $k = 0$ : at $x = \frac{π}{2}$ the graph of $f$ is approximated by a straight line.

Exercise 215 HardIB · P2 November 1998

Consider the polynomial $f (x) = x^{3} - c x^{2} - 4 x + 4 c$ , where $- 2 < c < 2$ .

a) Show that $f (c) = 0$ for all values of $c$ .

b) Express $f (x)$ as a product of three linear factors.

c) Sketch the graph of $y = f (x)$ , clearly indicating the $x$ -intercepts.

d) Show that the total area, $A$ , enclosed by the curve $y = f (x)$ and the $x$ -axis is given by $A = 8 + 4 c^{2} - \frac{1}{6} c^{4}$ .

e) Calculate the value of $c$ , $- 2 < c < 2$ , for which $A$ is a minimum.

f) Calculate the volume generated when the region enclosed by the graph of $y = f (x)$ and the $x$ -axis between $x = c$ and $x = 2$ , for the value of $c$ found in part (e), is rotated $36 0^{\circ}$ about the $x$ -axis.

View solution

a) 1. $f (c) = c^{3} - c \cdot c^{2} - 4 c + 4 c = 0$ for all $c$ .

b) 1. Factorise by grouping: $f (x) = x^{2} (x - c) - 4 (x - c) = (x - c) (x^{2} - 4)$ . 2. $f (x) = (x - c) (x - 2) (x + 2)$ .

c) 1. The $x$ -intercepts are $x = - 2$ , $x = c$ and $x = 2$ , with $- 2 < c < 2$ . It is a cubic with positive leading coefficient: it comes from $- \infty$ , cuts the axis at $- 2$ , has a local maximum between $- 2$ and $c$ , cuts at $c$ , has a local minimum between $c$ and $2$ , and rises, cutting at $2$ . 2. The general shape is the same for any $c$ ; only the position of the point $c$ relative to $0$ changes.

If $c < 0$ :

Graph of y = f(x) with intercepts at x = -2, x = c and x = 2, for c < 0

If $c > 0$ :

Graph of y = f(x) with intercepts at x = -2, x = c and x = 2, for c > 0

d) 1. The sign of $f$ is positive on $] - 2, c [$ and negative on $] c, 2 [$ (just look at the sign of the factors). So the total area is $A = \int_{- 2}^{c} f (x) d x - \int_{c}^{2} f (x) d x$ . 2. An antiderivative is $F (x) = \frac{x ^{4}}{4} - \frac{c x ^{3}}{3} - 2 x^{2} + 4 c x$ . 3. Values: $F (2) = - 4 + \frac{16 c}{3}$ ; $F (- 2) = - 4 - \frac{16 c}{3}$ ; $F (c) = \frac{c ^{4}}{4} - \frac{c ^{4}}{3} - 2 c^{2} + 4 c^{2} = 2 c^{2} - \frac{c ^{4}}{12}$ . 4. $A = [F (c) - F (- 2)] - [F (2) - F (c)] = 2 F (c) - F (- 2) - F (2) = 4 c^{2} - \frac{c ^{4}}{6} + 8$ . 5. That is, $A = 8 + 4 c^{2} - \frac{1}{6} c^{4}$ .

e) 1. Differentiate with respect to $c$ : $\frac{d A}{d c} = 8 c - \frac{2}{3} c^{3} = c (8 - \frac{2}{3} c^{2})$ . 2. This is zero at $c = 0$ (the other option, $c^{2} = 12$ , lies outside $- 2 < c < 2$ ). 3. $\frac{d ^{2} A}{d c ^{2}} = 8 - 2 c^{2} > 0$ at $c = 0$ , so it is a minimum. 4. $c = 0$ .

f) 1. With $c = 0$ , $f (x) = x (x^{2} - 4) = x^{3} - 4 x$ . The region between $x = 0$ and $x = 2$ is rotated about the $x$ -axis. 2. $V = π \int_{0}^{2} (x^{3} - 4 x)^{2} d x = π \int_{0}^{2} (x^{6} - 8 x^{4} + 16 x^{2}) d x$ . 3. $= π [\frac{x ^{7}}{7} - \frac{8 x ^{5}}{5} + \frac{16 x ^{3}}{3}]_{0}^{2} = π (\frac{128}{7} - \frac{256}{5} + \frac{128}{3}) = \frac{1024 π}{105} \approx 30.6$ .

Answer: b) $f (x) = (x - c) (x + 2) (x - 2)$ ; e) $c = 0$ ; f) $V = \frac{1024}{105} π = 30.6$

Exercise 216 HardIB · P2 November 1999

Consider the functions $f_{k} (x) = {x ln x - k x, 0, x > 0 x = 0$ , where $k \in N$ .

a) Calculate the derivative of $f_{k} (x)$ , $x > 0$ .

b) Find the open interval where $f_{0} (x)$ is increasing.

The graph of the function $f_{k} (x)$ is shown below.

Graph of the function f_k(x): starts at the origin, goes down to a minimum and cuts the x-axis again at point A; the tangent line at A is also drawn, cutting the y-axis below the origin

c) i) Show that a stationary point of $f_{k} (x)$ occurs at $x = e^{k - 1}$ .

(ii) One $x$ -intercept is $(0, 0)$ . Find the coordinates of the other $x$ -intercept.

d) Calculate the area enclosed by the curve and the $x$ -axis.

e) Calculate the equation of the tangent to the curve at $A$ .

f) Show that the area of the triangular region formed by the tangent and the coordinate axes is twice the area enclosed by the curve and the $x$ -axis.

g) Show that the $x$ -intercepts of $f_{k} (x)$ for consecutive values of $k$ form a geometric sequence.

Show answer

Answer: a) $f_{k}^{'} (x) = ln x + 1 - k$ ; b) $] \frac{1}{e}, + \infty [$ ; c) i) proved; ii) $A (e^{k}, 0)$ ; d) $\frac{e ^{2 k}}{4}$ ; e) $y = x - e^{k}$ ; f) proved; g) geometric sequence with common ratio $r = e$ .

Show working and answer

a) 1. Differentiating $f_{k} (x) = x ln x - k x$ with the product rule: $f_{k}^{'} (x) = ln x + x \cdot \frac{1}{x} - k$ . 2. Therefore $f_{k}^{'} (x) = ln x + 1 - k$ .

b) 1. With $k = 0$ , $f_{0}^{'} (x) = ln x + 1$ . 2. $f_{0}$ is increasing if $f_{0}^{'} (x) > 0$ : $ln x > - 1$ , that is, $x > \frac{1}{e}$ . 3. The open interval is $] \frac{1}{e}, + \infty [$ .

c) i) 1. A stationary point satisfies $f_{k}^{'} (x) = 0$ : $ln x + 1 - k = 0$ , so $ln x = k - 1$ . 2. Therefore $x = e^{k - 1}$ .

(ii) 1. We solve $f_{k} (x) = 0$ with $x > 0$ : $x (ln x - k) = 0$ , and as $x \neq = 0$ , $ln x = k$ . 2. The other intercept is $x = e^{k}$ : $A (e^{k}, 0)$ .

d) 1. Between $x = 0$ and $x = e^{k}$ the curve lies below the $x$ -axis (it starts at $O$ , goes down to the minimum at $x = e^{k - 1}$ and rises again to $A$ ), so the area is $- \int_{0}^{e^{k}} (x ln x - k x) d x$ . 2. Antiderivative of $x ln x$ by parts, with $u = ln x$ and $d v = x d x$ (so $d u = \frac{1}{x} d x$ , $v = \frac{x ^{2}}{2}$ ): $\int x ln x d x = \frac{x ^{2}}{2} ln x - \int \frac{x}{2} d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4}$ . 3. Then $\int (x ln x - k x) d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} - \frac{k x ^{2}}{2} = \frac{x ^{2}}{2} (ln x - k) - \frac{x ^{2}}{4}$ . 4. At $x = e^{k}$ this equals $0 - \frac{e ^{2 k}}{4}$ , and at $x = 0$ the limit is $0$ . The integral equals $- \frac{e ^{2 k}}{4}$ . 5. The area is $\frac{e ^{2 k}}{4}$ .

e) 1. The gradient at $A$ is $f_{k}^{'} (e^{k}) = k + 1 - k = 1$ . 2. The tangent: $y - 0 = 1 \cdot (x - e^{k})$ , that is, $y = x - e^{k}$ .

f) 1. The tangent cuts the $x$ -axis at $x = e^{k}$ and the $y$ -axis at $y = - e^{k}$ . 2. The triangle is right-angled with legs $e^{k}$ and $e^{k}$ : area $= \frac{1}{2} e^{2 k}$ . 3. And $\frac{1}{2} e^{2 k} = 2 \cdot \frac{e ^{2 k}}{4}$ , twice the area in (d).

g) 1. For $k$ and $k + 1$ the $x$ -intercepts are $e^{k}$ and $e^{k + 1}$ . 2. The ratio of two consecutive ones is $\frac{e ^{k + 1}}{e ^{k}} = e$ , a constant. 3. Hence $e, e^{2}, e^{3}, \dots$ form a geometric sequence with common ratio $e$ .

Exercise 217 HardIB · P2 May 2000

Let $f (x) = ln x^{5} - 3 x^{2}$ , $- 0.5 < x < 2$ , $x \neq = a$ , $x \neq = b$ ; $a$ , $b$ are values of $x$ for which $f (x)$ is not defined.

a) i) Sketch the graph of $f (x)$ , indicating the zeros of $f (x)$ on your sketch.

Also show the position of any asymptotes.

(ii) Find all zeros of $f (x)$ (solutions of $f (x) = 0$ ).

b) Calculate the exact values of $a$ and $b$ .

c) Calculate $f^{'} (x)$ , and state clearly where $f^{'} (x)$ is not defined.

d) Calculate the exact value of the $x$ -coordinate of the local maximum of $f (x)$ , for $0 < x < 1.5$ .

(You may assume that there are no points of inflexion.)

e) Write down the definite integral that represents the area of the region enclosed by $f (x)$ and the $x$ -axis. (Do not evaluate the integral.)

View solution

a) i) 1. Domain: $f$ is not defined where $x^{5} - 3 x^{2} = 0$ , that is, $x = 0$ and $x = 33 \approx 1.44$ ; there are vertical asymptotes there ( $f \to - \infty$ ). 2. The zeros come from $x^{5} - 3 x^{2} = 1$ (part ii). 3. Shape of the graph: for $- 0.5 < x < 0$ and for $0 < x < 0.6$ it is negative; between $0.6$ and $1.35$ it is positive with a local maximum; it falls to $- \infty$ at $x = 33$ ; to the right it comes back from $- \infty$ , cuts the axis at $1.51$ , and increases afterwards.

Graph of f(x) = ln|x^5 − 3x²| for −0.5 < x < 2, with vertical asymptotes at x = 0 and x = ∛3 and the three points where the graph cuts the x-axis

(ii) 1. $f (x) = 0$ is equivalent to $x^{5} - 3 x^{2} = 1$ , that is, $x^{5} - 3 x^{2} = 1$ or $x^{5} - 3 x^{2} = - 1$ . 2. Using the GDC, in the given interval, we obtain three solutions: $x \approx 0.599$ , $x \approx 1.35$ and $x \approx 1.51$ (for $- 0.5 < x < 0$ the value of $x^{5} - 3 x^{2}$ is less than $1$ and there are no zeros).

b) 1. $f$ is not defined if $x^{5} - 3 x^{2} = x^{2} (x^{3} - 3) = 0$ . 2. So $x = 0$ or $x^{3} = 3$ : $a = 0$ and $b = 33$ .

c) 1. Derivative of the logarithm of an absolute value: $\frac{d}{d x} ln ∣ u ∣ = \frac{u ^{'}}{u}$ , with $u = x^{5} - 3 x^{2}$ . 2. $f^{'} (x) = \frac{5 x ^{4} - 6 x}{x ^{5} - 3 x ^{2}} = \frac{6 x - 5 x ^{4}}{3 x ^{2} - x ^{5}}$ . 3. It is not defined at $x = 0$ and $x = 33$ , where the denominator vanishes.

d) 1. $f^{'} (x) = 0$ when the numerator vanishes: $6 x - 5 x^{4} = x (6 - 5 x^{3}) = 0$ . 2. For $0 < x < 1.5$ the solution is $5 x^{3} = 6$ , that is, $x = 3 \frac{6}{5} \approx 1.06$ , which is the local maximum (it lies between $0.6$ and $1.35$ ).

e) 1. The area enclosed by the curve and the $x$ -axis is the region where $f > 0$ , between the two first zeros. 2. Area $= \int_{0.599}^{1.35} ln x^{5} - 3 x^{2} d x$ .

Answer: a) ii) $x \approx 0.599$ , $x \approx 1.35$ , $x \approx 1.51$ ; b) $a = 0$ , $b = 33$ ; c) $f^{'} (x) = \frac{6 x - 5 x ^{4}}{3 x ^{2} - x ^{5}}$ , not defined at $x = 0$ and $x = 33$ ; d) $x = 3 \frac{6}{5}$ ; e) $\int_{0.599}^{1.35} ln x^{5} - 3 x^{2} d x$

Exercise 218 HardIB · P2 November 2000

a) Let $y = \frac{a + b sin x}{b + a sin x}$ , where $0 < a < b$ .

(i) Show that $\frac{d y}{d x} = \frac{( b ^{2} - a ^{2} ) cos x}{( b + a sin x ) ^{2}}$ .

(ii) Calculate the maximum and minimum values of $y$ .

(iii) Show that the graph of $y = \frac{a + b sin x}{b + a sin x}$ , $0 < a < b$ cannot have a vertical asymptote.

b) For the graph of $y = \frac{4 + 5 sin x}{5 + 4 sin x}$ for $0 \leq x \leq 2 π$ ,

(i) Write down the $y$ -intercept.

(ii) Find the $x$ -intercepts $m$ and $n$ ( $m < n$ ), correct to four significant figures.

(iii) Sketch the graph.

c) The area enclosed by the graph of $y = \frac{4 + 5 sin x}{5 + 4 sin x}$ and the $x$ -axis from $x = 0$ to $x = n$ is denoted by $A$ .

Write down, but do not evaluate, an expression for the area $A$ .

View solution

a) i) 1. Quotient rule with $u = a + b sin x$ and $v = b + a sin x$ : $\frac{d y}{d x} = \frac{b cos x ( b + a sin x ) - ( a + b sin x ) a cos x}{( b + a sin x ) ^{2}}$ . 2. Numerator: $b^{2} cos x + ab sin x cos x - a^{2} cos x - ab sin x cos x = (b^{2} - a^{2}) cos x$ . 3. Therefore $\frac{d y}{d x} = \frac{( b ^{2} - a ^{2} ) cos x}{( b + a sin x ) ^{2}}$ .

(ii) 1. As $0 < a < b$ , $b^{2} - a^{2} > 0$ and the denominator is positive (see iii), so the sign of $\frac{d y}{d x}$ is that of $cos x$ , and it vanishes when $cos x = 0$ . 2. At $x = \frac{π}{2}$ (mod $2 π$ ), $cos x$ changes from $+$ to $-$ : maximum, with $y = \frac{a + b}{b + a} = 1$ . 3. At $x = \frac{3 π}{2}$ , $cos x$ changes from $-$ to $+$ : minimum, with $y = \frac{a - b}{b - a} = - 1$ . 4. Maximum $= 1$ , minimum $= - 1$ .

(iii) 1. A vertical asymptote would require $b + a sin x = 0$ (with the numerator non-zero). 2. But $b + a sin x \geq b - a > 0$ because $a < b$ , so the denominator never vanishes and there is no vertical asymptote.

b) i) $x = 0$ : $y = \frac{4}{5}$ , the point $(0, \frac{4}{5})$ .

(ii) 1. $y = 0$ if $4 + 5 sin x = 0$ , that is, $sin x = - \frac{4}{5}$ . 2. In $[0, 2 π]$ the solutions are $x = π + arcsin 0.8 \approx 4.069$ and $x = 2 π - arcsin 0.8 \approx 5.356$ . 3. So $m \approx 4.069$ and $n \approx 5.356$ .

(iii) Key features: it starts at $(0, \frac{4}{5})$ , rises to the maximum $y = 1$ at $x = \frac{π}{2}$ , falls crossing the $x$ -axis at $m$ , reaches the minimum $y = - 1$ at $x = \frac{3 π}{2}$ and rises crossing the axis at $n$ up to $y = \frac{4}{5}$ at $x = 2 π$ .

Graph of y = (4 + 5 sin x)/(5 + 4 sin x) for 0 ≤ x ≤ 2π, with the y-intercept and the two points m and n where the graph cuts the x-axis

c) 1. On $[0, m]$ the graph lies above the $x$ -axis and on $[m, n]$ below it. 2. The area is the integral of the absolute value: $A = \int_{0}^{4.069} \frac{4 + 5 sin x}{5 + 4 sin x} d x - \int_{4.069}^{5.356} \frac{4 + 5 sin x}{5 + 4 sin x} d x$ .

Answer: a) ii) maximum $1$ , minimum $- 1$ ; b) i) $(0, \frac{4}{5})$ ; ii) $m \approx 4.069$ , $n \approx 5.356$ ; c) $A = \int_{0}^{4.069} \frac{4 + 5 sin x}{5 + 4 sin x} d x - \int_{4.069}^{5.356} \frac{4 + 5 sin x}{5 + 4 sin x} d x$

Exercise 219 HardIB · P2 TZ2 May 2005

The function $f$ is defined by $f (x) = e^{p x} (x + 1)$ , with $p \in R$ .

a) i) Verify that $f^{'} (x) = e^{p x} (p (x + 1) + 1)$ .

(ii) Let $f^{(n)} (x)$ be the $n$ th derivative of $f (x)$ with respect to $x$ . Prove by mathematical induction that $f^{(n)} (x) = p^{n - 1} e^{p x} (p (x + 1) + n)$ , $n \in Z^{+}$ .

b) When $p = 3$ the curve of $f$ has a minimum and a point of inflexion.

Find the exact value of the $x$ -coordinate of

(i) the minimum.

(ii) the point of inflexion.

c) Let $p = \frac{1}{2}$ . Let $R$ be the region enclosed by the curve, the $x$ -axis and the lines $x = - 2$ and $x = 2$ . Find the area of $R$ .

Show answer

Answer: a) proved; b) i) $x = \frac{- 1 - 3}{3}$ ; ii) $x = \frac{- 2 - 3}{3}$ ; c) $8 e^{- \frac{1}{2}} - 6 e^{- 1} + 2 e$

Show working and answer

a) i) 1. Product rule: $f^{'} (x) = p e^{p x} (x + 1) + e^{p x} \cdot 1$ . 2. Taking out the common factor, $f^{'} (x) = e^{p x} (p (x + 1) + 1)$ .

(ii) Induction on $n \in Z^{+}$ .

Base case $n = 1$ : the formula gives $p^{0} e^{p x} (p (x + 1) + 1) = e^{p x} (p (x + 1) + 1) = f^{'} (x)$ by (i). It holds.
Hypothesis: for $n = k$ , $f^{(k)} (x) = p^{k - 1} e^{p x} (p (x + 1) + k)$ .
Step: we differentiate, $f^{(k + 1)} (x) = p^{k - 1} [p e^{p x} (p (x + 1) + k) + e^{p x} \cdot p]$ .
Taking out the common factor $p e^{p x}$ : $f^{(k + 1)} (x) = p^{k} e^{p x} (p (x + 1) + k + 1)$ , which is the formula with $n = k + 1$ .
By induction, the formula is true for all $n \in Z^{+}$ .

b) i) 1. With $p = 3$ : $f^{'} (x) = e^{3 x} (3 (x + 1) + 1) = 0$ when $3 (x + 1) = - 1$ . 2. $x = - 1 - \frac{1}{3} = \frac{- 1 - 3}{3}$ . 3. It is a minimum: $f^{''} (x) = 3 e^{3 x} (3 (x + 1) + 2)$ , and at that point $3 (x + 1) + 2 = 1 > 0$ .

(ii) 1. $f^{''} (x) = 0$ when $3 (x + 1) + 2 = 0$ , that is, $x = - 1 - \frac{2}{3} = \frac{- 2 - 3}{3}$ . 2. It is a point of inflexion: $f^{'''} (x) = 3 e^{3 x} (3 (x + 1) + 3)$ , and at that point this equals $3 e^{3 x} \neq = 0$ .

c) 1. With $p = \frac{1}{2}$ , $f (x) = e^{x /2} (x + 1)$ is zero at $x = - 1$ : it is negative on $[- 2, - 1 [$ and positive on $] - 1, 2]$ . 2. Antiderivative by parts with $u = x + 1$ and $d v = e^{x /2} d x$ : $\int e^{x /2} (x + 1) d x = 2 e^{x /2} (x + 1) - 2 \int e^{x /2} d x = 2 e^{x /2} (x + 1) - 4 e^{x /2} = 2 e^{x /2} (x - 1) =: F (x)$ . 3. $F (- 2) = - \frac{6}{e}$ , $F (- 1) = - 4 e^{- 1/2}$ , $F (2) = 2 e$ . 4. Area $= - (F (- 1) - F (- 2)) + (F (2) - F (- 1)) = F (- 2) + F (2) - 2 F (- 1) = - 6 e^{- 1} + 2 e + 8 e^{- 1/2}$ .

Answer: a) proved; b) i) $x = \frac{- 1 - 3}{3}$ ; ii) $x = \frac{- 2 - 3}{3}$ ; c) $8 e^{- \frac{1}{2}} - 6 e^{- 1} + 2 e$

Exercise 220 HardIB · P1 TZ0 November 2011

The graphs of $f (x) = - x^{2} + 2$ and $g (x) = x^{3} - x^{2} - b x + 2$ , $b > 0$ intersect, forming two closed regions. Verify that these two regions have the same area.

Graphs of the parabola f(x) = −x² + 2 and the cubic g(x) = x³ − x² − bx + 2 (b > 0): they intersect at three points, one of them on the y-axis at (0, 2), and bound two closed regions, one on each side of the y-axis

Show answer

Answer: shown (each region has area $\frac{b ^{2}}{4}$ ).

Show working and answer

The graphs intersect where $f (x) = g (x)$ : $- x^{2} + 2 = x^{3} - x^{2} - b x + 2$ , that is, $x^{3} - b x = 0$ , or $x (x^{2} - b) = 0$ .
The intersections are at $x = - b$ , $x = 0$ and $x = b$ (with $b > 0$ ).
Left region, $- b \leq x \leq 0$ : there $g (x) - f (x) = x^{3} - b x = x (x^{2} - b) \geq 0$ ( $x \leq 0$ and $x^{2} - b \leq 0$ ), so $g$ lies above.
Right region, $0 \leq x \leq b$ : $x \geq 0$ and $x^{2} - b \leq 0$ , so $g - f \leq 0$ and $f$ lies above.
Left area: $\int_{- b}^{0} (x^{3} - b x) d x = [\frac{x ^{4}}{4} - \frac{b x ^{2}}{2}]_{- b}^{0} = 0 - (\frac{b ^{2}}{4} - \frac{b ^{2}}{2}) = \frac{b ^{2}}{4}$ .
Right area: $\int_{0}^{b} (b x - x^{3}) d x = [\frac{b x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{b} = \frac{b ^{2}}{2} - \frac{b ^{2}}{4} = \frac{b ^{2}}{4}$ .
Both areas equal $\frac{b ^{2}}{4}$ , so they are the same.

Answer: shown (each region has area $\frac{b ^{2}}{4}$ ).

Exercise 221 HardIB · P2 TZ1 May 2010

The following diagram shows the graphs of $y = \frac{3}{2} x - 3$ , $y = 3$ and a quadratic function such that all three intersect at the same two points.

Graphs of y = |(3/2)x − 3| (V shape with vertex on the x-axis), of the line y = 3 and of a parabola with minimum −3 that meets the other two at the same points (0, 3) and (4, 3); the region between the V and the parabola is shaded

Given that the minimum value of the quadratic function is $- 3$ , find an expression for the area of the shaded region in the form $\int_{0}^{t} (a x^{2} + b x + c) d x$ , where the constants $a$ , $b$ , $c$ and $t$ are to be determined. (Note: The integral does not need to be evaluated.)

Show answer

Answer: $a = - 3$ , $b = 9$ , $c = 0$ , $t = 2$

Show working and answer

The quadratic passes through the intersection points $(0, 3)$ and $(4, 3)$ (where the V meets $y = 3$ : $\frac{3}{2} x - 3 = 3$ gives $x = 0$ and $x = 4$ ).
By symmetry its vertex is at $x = 2$ and equals $- 3$ : $y = m (x - 2)^{2} - 3$ .
With $(0, 3)$ : $4 m - 3 = 3$ , so $m = \frac{3}{2}$ . Thus $y = \frac{3}{2} (x - 2)^{2} - 3 = \frac{3}{2} x^{2} - 6 x + 3$ .
The shaded region lies between the V (above) and the parabola (below). For $0 \leq x \leq 2$ the V is $y = 3 - \frac{3}{2} x$ .
Difference: $(3 - \frac{3}{2} x) - (\frac{3}{2} x^{2} - 6 x + 3) = - \frac{3}{2} x^{2} + \frac{9}{2} x$ .
The region is symmetric about $x = 2$ , so its area is twice that over $[0, 2]$ : $2 \int_{0}^{2} (- \frac{3}{2} x^{2} + \frac{9}{2} x) d x = \int_{0}^{2} (- 3 x^{2} + 9 x) d x$ .

Answer: $a = - 3$ , $b = 9$ , $c = 0$ , $t = 2$

Exercise 222 HardIB · P1 TZ1 May 2014

Consider the function $f (x) = \frac{ln x}{x}$ , $x > 0$ .

The following diagram shows the graph of $y = f (x)$ and its tangent at the point $A$ .

Graph of y = ln x / x for x > 0 with its tangent at the point A, where it meets the x-axis; the maximum B and the point of inflexion C are marked

a) Prove that $f^{'} (x) = \frac{1 - ln x}{x ^{2}}$ .

b) Find the coordinates of $B$ , the point where the curve reaches its maximum value.

c) Find the coordinates of $C$ , the point of inflexion on the curve.

The graph of $y = f (x)$ cuts the $x$ -axis at the point $A$ .

d) Find the equation of the tangent to the graph of $f$ at the point $A$ .

e) Calculate the area enclosed by the curve $y = f (x)$ , the tangent at $A$ and the line $x = e$ .

Show answer

Answer: b) $B (e, \frac{1}{e})$ ; c) $C (e^{3/2}, \frac{3 e ^{- 3/2}}{2})$ ; d) $y = x - 1$ ; e) $\frac{e ^{2}}{2} - e$

Show working and answer

a) 1. Quotient rule with $u = ln x$ (so $u^{'} = \frac{1}{x}$ ) and $v = x$ (so $v^{'} = 1$ ). 2. $f^{'} (x) = \frac{\frac{1}{x} \cdot x - ln x \cdot 1}{x ^{2}} = \frac{1 - ln x}{x ^{2}}$ .

b) 1. Maximum: $f^{'} (x) = 0$ , i.e. $1 - ln x = 0$ , so $x = e$ . 2. $f^{'}$ changes from positive to negative at that point, so it is a maximum, and $f (e) = \frac{1}{e}$ . 3. $B (e, \frac{1}{e})$ .

c) 1. We differentiate again using the quotient rule: $f^{''} (x) = \frac{- \frac{1}{x} \cdot x ^{2} - ( 1 - ln x ) \cdot 2 x}{x ^{4}} = \frac{- 3 x + 2 x ln x}{x ^{4}} = \frac{2 ln x - 3}{x ^{3}}$ . 2. $f^{''} (x) = 0$ when $ln x = \frac{3}{2}$ , i.e. $x = e^{3/2}$ . There $f^{''}$ changes sign (negative before, positive after): it is a point of inflexion. 3. $f (e^{3/2}) = \frac{3/2}{e ^{3/2}} = \frac{3}{2} e^{- 3/2}$ . 4. $C (e^{3/2}, \frac{3 e ^{- 3/2}}{2})$ .

d) 1. $A$ is the point with $ln x = 0$ , i.e. $A (1, 0)$ . The gradient of the tangent is $f^{'} (1) = \frac{1 - 0}{1} = 1$ . 2. Tangent: $y - 0 = 1 \cdot (x - 1)$ , i.e. $y = x - 1$ .

e) 1. Between $x = 1$ and $x = e$ the line lies above the curve (for example, at $x = e$ : $e - 1 \approx 1.72 > \frac{1}{e} \approx 0.37$ ; the curve is concave and lies below its tangent). 2. $Area = \int_{1}^{e} (x - 1 - \frac{ln x}{x}) d x$ . 3. The antiderivative of $\frac{ln x}{x}$ is $\frac{( ln x ) ^{2}}{2}$ (substitution $u = ln x$ ). Hence $Area = [\frac{x ^{2}}{2} - x - \frac{( ln x ) ^{2}}{2}]_{1}^{e}$ . 4. $= (\frac{e ^{2}}{2} - e - \frac{1}{2}) - (\frac{1}{2} - 1 - 0) = \frac{e ^{2}}{2} - e$ .

Answer: b) $B (e, \frac{1}{e})$ ; c) $C (e^{3/2}, \frac{3 e ^{- 3/2}}{2})$ ; d) $y = x - 1$ ; e) $\frac{e ^{2}}{2} - e$

Exercise 223 HardIB · P1 TZ2 May 2015

Consider the functions $f (x) = tan x$ , $0 \leq x < \frac{π}{2}$ and $g (x) = \frac{x + 1}{x - 1}$ , $x \in R$ , $x \neq = 1$ .

a) Find an expression for $(g \circ f) (x)$ ; state its domain.

b) Hence, show that $(g \circ f) (x) = \frac{sin x + cos x}{sin x - cos x}$ .

c) Let $y = (g \circ f) (x)$ . Find the exact value of $\frac{d y}{d x}$ at the point on the graph of $y = (g \circ f) (x)$ where $x = \frac{π}{6}$ ; give your answer in the form $a + b 3$ , $a, b \in Z$ .

d) Show that the area enclosed by the graph of $y = (g \circ f) (x)$ , the $x$ -axis and the lines $x = 0$ and $x = \frac{π}{6}$ is $ln (1 + 3)$ .

Show answer

Answer: a) $\frac{tan x + 1}{tan x - 1}$ , $D = [0, \frac{π}{2} [- {\frac{π}{4}}$ ; c) $\frac{d y}{d x} = - 8 - 43$ ; d) proved ( $Area = ln (1 + 3)$ ).

Show working and answer

a) 1. $(g \circ f) (x) = g (f (x)) = g (tan x) = \frac{tan x + 1}{tan x - 1}$ . 2. Domain: we need $x \in [0, \frac{π}{2} [$ (domain of $f$ ) and $tan x \neq = 1$ (so that $g$ is defined), that is, $x \neq = \frac{π}{4}$ . 3. $D_{g \circ f} = [0, \frac{π}{2} [- {\frac{π}{4}}$ .

b) 1. On the interval $[0, \frac{π}{2} [$ we have $cos x \neq = 0$ . We write $tan x = \frac{sin x}{cos x}$ and multiply numerator and denominator by $cos x$ . 2. $\frac{tan x + 1}{tan x - 1} = \frac{\frac{s i n x}{c o s x} + 1}{\frac{s i n x}{c o s x} - 1} = \frac{sin x + cos x}{sin x - cos x}$ .

c) 1. Quotient rule with $u = sin x + cos x$ ( $u^{'} = cos x - sin x$ ) and $v = sin x - cos x$ ( $v^{'} = cos x + sin x$ ). 2. $\frac{d y}{d x} = \frac{( cos x - sin x ) ( sin x - cos x ) - ( sin x + cos x ) ( cos x + sin x )}{( sin x - cos x ) ^{2}} = \frac{- ( sin x - cos x ) ^{2} - ( sin x + cos x ) ^{2}}{( sin x - cos x ) ^{2}}$ . 3. As $(sin x - cos x)^{2} + (sin x + cos x)^{2} = 2$ , we get $\frac{d y}{d x} = \frac{- 2}{( sin x - cos x ) ^{2}}$ . 4. At $x = \frac{π}{6}$ : $sin x - cos x = \frac{1 - 3}{2}$ , and its square is $\frac{4 - 2 3}{4} = \frac{2 - 3}{2}$ . 5. $\frac{d y}{d x} = \frac{- 2}{\frac{2 - 3}{2}} = \frac{- 4}{2 - 3} = \frac{- 4 ( 2 + 3 )}{( 2 - 3 ) ( 2 + 3 )} = - 4 (2 + 3) = - 8 - 43$ .

d) 1. On $[0, \frac{π}{6}]$ we have $sin x < cos x$ , so the denominator is negative and the numerator positive: $y < 0$ and the curve lies below the $x$ -axis. The area is $- \int_{0}^{π /6} y d x$ . 2. Technique: we recognise that the numerator is the derivative of the denominator, $\frac{d}{d x} (sin x - cos x) = cos x + sin x$ . Hence $\int \frac{sin x + cos x}{sin x - cos x} d x = ln ∣ sin x - cos x ∣$ . 3. $Area = - [ln ∣ sin x - cos x ∣]_{0}^{π /6} = - (ln \frac{3 - 1}{2} - ln 1) = ln \frac{2}{3 - 1}$ . 4. Rationalising: $\frac{2}{3 - 1} = \frac{2 ( 3 + 1 )}{2} = 3 + 1$ . The area is $ln (1 + 3)$ .

Answer: a) $\frac{tan x + 1}{tan x - 1}$ , $D = [0, \frac{π}{2} [- {\frac{π}{4}}$ ; c) $\frac{d y}{d x} = - 8 - 43$ ; d) proved ( $Area = ln (1 + 3)$ ).

Exercise 224 HardIB · P1 TZ0 November 2015

Consider the function $f (x) = x 1 - x^{2}$ defined on the domain $- 1 \leq x \leq 1$ .

a) Show that $f$ is an odd function.

b) Find $f^{'} (x)$ .

c) Hence, find the $x$ -coordinates of all the local maximum or minimum points.

d) Find the range of $f$ .

e) Sketch the graph of $y = f (x)$ , clearly indicating the coordinates of the $x$ -intercepts and of all the local maximum and minimum points.

f) Find the area of the region enclosed by the graph of $y = f (x)$ and the $x$ -axis for $x \geq 0$ .

g) Show that $\int_{- 1}^{1} x 1 - x^{2} d x > \int_{- 1}^{1} x 1 - x^{2} d x$ .

View solution

a) 1. A function is odd if $f (- x) = - f (x)$ . 2. $f (- x) = (- x) 1 - (- x)^{2} = - x 1 - x^{2} = - f (x)$ . The domain $[- 1, 1]$ is symmetric, so $f$ is odd.

b) 1. Product rule with $u = x$ and $v = 1 - x^{2}$ , whose derivative is $v^{'} = \frac{- 2 x}{2 1 - x ^{2}} = \frac{- x}{1 - x ^{2}}$ . 2. $f^{'} (x) = 1 - x^{2} - \frac{x ^{2}}{1 - x ^{2}} = \frac{1 - x ^{2} - x ^{2}}{1 - x ^{2}} = \frac{1 - 2 x ^{2}}{1 - x ^{2}}$ .

c) 1. $f^{'} (x) = 0$ when $1 - 2 x^{2} = 0$ , that is, $x = \pm \frac{1}{2}$ (both inside $] - 1, 1 [$ ). 2. $f^{'} > 0$ if $∣ x ∣ < \frac{1}{2}$ and $f^{'} < 0$ if $\frac{1}{2} < ∣ x ∣ < 1$ : at $x = \frac{1}{2}$ there is a local maximum and, since $f$ is odd, at $x = - \frac{1}{2}$ a local minimum.

d) 1. $f (\frac{1}{2}) = \frac{1}{2} 1 - \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$ , and $f (- \frac{1}{2}) = - \frac{1}{2}$ . 2. At the endpoints $f (\pm 1) = 0$ . The absolute maximum is $\frac{1}{2}$ and the absolute minimum is $- \frac{1}{2}$ , so $R_{f} = [- \frac{1}{2}, \frac{1}{2}]$ .

e) 1. $x$ -intercepts: $x 1 - x^{2} = 0$ for $x = - 1$ , $x = 0$ and $x = 1$ . The $y$ -intercept is $(0, 0)$ . 2. Maximum $(\frac{1}{2}, \frac{1}{2})$ and minimum $(- \frac{1}{2}, - \frac{1}{2})$ . The graph is symmetric about the origin (odd) and has a vertical tangent at $x = \pm 1$ .

Graph of f(x) = x·√(1 − x²) on (−1, 1): an odd curve with intercepts at −1, 0 and 1, a maximum and a minimum

f) 1. For $x \geq 0$ the graph lies above the $x$ -axis on $[0, 1]$ , so $f \geq 0$ there: $Area = \int_{0}^{1} x 1 - x^{2} d x$ . 2. Technique: substitution $u = 1 - x^{2}$ , $d u = - 2 x d x$ . With $x = 0 \Rightarrow u = 1$ and $x = 1 \Rightarrow u = 0$ : $Area = - \frac{1}{2} \int_{1}^{0} u d u = \frac{1}{2} \int_{0}^{1} u^{1/2} d u$ . 3. $= \frac{1}{2} \cdot \frac{2}{3} [u^{3/2}]_{0}^{1} = \frac{1}{3}$ .

g) 1. Since $f$ is odd and the interval is symmetric, $\int_{- 1}^{1} f (x) d x = 0$ , so the right-hand side is $0$ . 2. $∣ f ∣$ is even and non-negative, so $\int_{- 1}^{1} ∣ f (x) ∣ d x = 2 \int_{0}^{1} f (x) d x = 2 \cdot \frac{1}{3} = \frac{2}{3}$ . 3. As $\frac{2}{3} > 0$ , the inequality is proved.

Answer: b) $f^{'} (x) = \frac{1 - 2 x ^{2}}{1 - x ^{2}}$ ; c) $x = \pm \frac{1}{2}$ ; d) $R_{f} = [- \frac{1}{2}, \frac{1}{2}]$ ; e) max. $(\frac{1}{2}, \frac{1}{2})$ , min. $(- \frac{1}{2}, - \frac{1}{2})$ ; f) $\frac{1}{3}$ ; g) $\frac{2}{3} > 0$

Exercise 225 HardIB · P1 TZ1 May 2021

Consider the function $f$ defined by $f (x) = 90 e^{- 0.5 x}$ for $x \in R^{+}$ . The graph of $f$ and the line $y = x$ intersect at the point $P$ .

a) Calculate the $x$ -coordinate of $P$ .

The line $L$ has gradient $- 1$ and is tangent to the graph of $f$ at the point $Q$ .

b) Calculate the exact coordinates of $Q$ .

c) Show that the equation of $L$ is $y = - x + 2 ln 45 + 2$ .

Graph of f(x) = 90e^(−0.5x), the line y = x which cuts it at P, and the line L of gradient −1 tangent at Q; the region A between the curve, y = x and L is shaded

The shaded region $A$ is enclosed by the graph of $f$ and the lines $y = x$ and $L$ .

d) (i) Calculate the $x$ -coordinate of the point where $L$ meets the line $y = x$ .

(ii) Using the above, calculate the area of $A$ .

The line $L$ is tangent to the graph of $f$ and to the graph of the inverse function $f^{- 1}$ .

Graphs of f and its inverse, symmetric about y = x, with the line L of gradient −1 tangent to both; the region enclosed between the two curves and L is shaded

e) Calculate the shaded area enclosed by the graphs of $f$ and $f^{- 1}$ and the line $L$ .

Show answer

Answer: a) $x = 5.57$ ; b) $Q (2 ln 45, 2)$ ; c) shown; d) (i) $x = 1 + ln 45$ ; (ii) $A = 1.52$ ; e) $3.04$

Show working and answer

a) 1. $90 e^{- 0.5 x} = x$ . Using the GDC, the positive solution is $x \approx 5.57$ .

b) 1. $f^{'} (x) = - 45 e^{- 0.5 x}$ . We impose gradient $- 1$ : $- 45 e^{- 0.5 x} = - 1$ , so $e^{- 0.5 x} = \frac{1}{45}$ . 2. $- 0.5 x = - ln 45$ , so $x = 2 ln 45$ . 3. $y = 90 \cdot \frac{1}{45} = 2$ . Therefore $Q (2 ln 45, 2)$ .

c) 1. Line of gradient $- 1$ through $Q$ : $y - 2 = - (x - 2 ln 45)$ . 2. $y = - x + 2 ln 45 + 2$ .

d) (i) 1. $L$ meets $y = x$ when $x = - x + 2 ln 45 + 2$ , that is, $2 x = 2 ln 45 + 2$ . 2. $x = ln 45 + 1 \approx 4.81$ .

(ii) 1. The region $A$ is bounded below by $L$ (from $x = 1 + ln 45$ to $x = 2 ln 45 \approx 7.61$ ) and above by $y = x$ (up to $x \approx 5.57$ ), and then by the curve $f$ (up to $Q$ ). We have checked that $f$ lies above $L$ on that stretch, for example at $x = 6$ : $f (6) \approx 4.48 > 3.61$ . 2. $Area (A) = \int_{1 + l n 45}^{5.57} (x - L (x)) d x + \int_{5.57}^{2 l n 45} (f (x) - L (x)) d x$ . 3. Using the GDC: $Area (A) \approx 1.52$ .

e) 1. $f$ and $f^{- 1}$ are symmetric about $y = x$ , and $L$ (gradient $- 1$ ) is symmetric to itself about that line. So the enclosed region is symmetric about $y = x$ . 2. Half of that region, on one side of $y = x$ , is exactly the region $A$ . 3. Area $= 2 \cdot 1.52 \approx 3.04$ .

Answer: a) $x = 5.57$ ; b) $Q (2 ln 45, 2)$ ; c) shown; d) (i) $x = 1 + ln 45$ ; (ii) $A = 1.52$ ; e) $3.04$

Exercise 226 HardIB · P2 TZ0 November 2021

Consider the function $f (x) = \frac{x ^{2} - x - 12}{2 x - 15}$ , where $x \in R$ , $x \neq = \frac{15}{2}$ .

a) Calculate the coordinates where the graph of $f$ crosses

(i) the $x$ -axis;

(ii) the $y$ -axis.

b) Calculate the equation of the vertical asymptote of the graph of $f$ .

c) The oblique asymptote of the graph of $f$ can be written $y = a x + b$ , $a, b \in Q$ .

Calculate the value of $a$ and the value of $b$ .

d) Sketch the graph of $f$ for $- 30 \leq x \leq 30$ , showing the intercepts with the axes and the asymptotes.

e) (i) Express $\frac{1}{f ( x )}$ in partial fractions.

(ii) Using the above, calculate the exact value of $\int_{0}^{3} \frac{1}{f ( x )} d x$ , giving your answer as a single logarithm.

View solution

a) (i) 1. Intercepts with the $x$ -axis: $f (x) = 0 ⟺ x^{2} - x - 12 = 0 ⟺ (x - 4) (x + 3) = 0$ (the denominator is not zero there). 2. $(- 3, 0)$ and $(4, 0)$ .

(ii) 1. $f (0) = \frac{- 12}{- 15} = \frac{4}{5}$ : $(0, \frac{4}{5})$ .

b) 1. The denominator is zero when $2 x - 15 = 0$ , and the numerator is not (it equals $\frac{225}{4} - \frac{15}{2} - 12 \neq = 0$ ). Vertical asymptote $x = \frac{15}{2}$ .

c) 1. Polynomial division: $x^{2} - x - 12 = (2 x - 15) (\frac{x}{2} + \frac{13}{4}) + \frac{147}{4}$ . 2. Check: $(2 x - 15) (\frac{x}{2} + \frac{13}{4}) = x^{2} + \frac{13}{2} x - \frac{15}{2} x - \frac{195}{4} = x^{2} - x - \frac{195}{4}$ , and $- 12 + \frac{195}{4} = \frac{147}{4}$ . 3. Therefore $f (x) = \frac{x}{2} + \frac{13}{4} + \frac{147/4}{2 x - 15}$ , and the oblique asymptote is $y = \frac{1}{2} x + \frac{13}{4}$ : $a = \frac{1}{2}$ , $b = \frac{13}{4}$ .

d) 1. With $f^{'} (x) = \frac{2 x ^{2} - 30 x + 39}{( 2 x - 15 ) ^{2}}$ , the local extrema are at $x \approx 1.44$ (local maximum, $y \approx 0.94$ ) and $x \approx 13.6$ (local minimum, $y \approx 13.1$ ). 2. To the left of the vertical asymptote the graph goes from $- \infty$ (as $x \to - \infty$ it follows the oblique asymptote) to $- \infty$ (as $x \to \frac{15}{2}^{-}$ ), passing through the intercepts $(- 3, 0)$ , $(0, \frac{4}{5})$ and $(4, 0)$ and the local maximum. On the right, it comes down from $+ \infty$ next to the asymptote, has the local minimum and rises close to the oblique asymptote.

Graph of f(x) = (x² − x − 12)/(2x − 15) for −30 ≤ x ≤ 30 with the vertical asymptote x = 15/2 and the oblique asymptote y = x/2 + 13/4 dashed

e) (i) 1. $\frac{1}{f ( x )} = \frac{2 x - 15}{( x - 4 ) ( x + 3 )} = \frac{A}{x + 3} + \frac{B}{x - 4}$ , that is, $2 x - 15 = A (x - 4) + B (x + 3)$ . 2. $x = - 3$ : $- 21 = - 7 A$ , so $A = 3$ . $x = 4$ : $- 7 = 7 B$ , so $B = - 1$ . 3. $\frac{1}{f ( x )} = \frac{3}{x + 3} - \frac{1}{x - 4}$ .

(ii) 1. On $[0, 3]$ there are no poles ( $x = - 3$ , $x = 4$ ), so the integral exists. 2. $\int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x = [3 ln ∣ x + 3∣ - ln ∣ x - 4∣]_{0}^{3} = (3 ln 6 - ln 1) - (3 ln 3 - ln 4)$ . 3. $= 3 ln 6 - 3 ln 3 + ln 4 = 3 ln 2 + 2 ln 2 = 5 ln 2 = ln 32$ .

Answer: a) (i) $(- 3, 0)$ , $(4, 0)$ ; (ii) $(0, \frac{4}{5})$ ; b) $x = \frac{15}{2}$ ; c) $y = \frac{1}{2} x + \frac{13}{4}$ ; e) (i) $\frac{3}{x + 3} - \frac{1}{x - 4}$ ; (ii) $ln 32$

Volumes of revolution

Exercise 227 BasicIB · P1 November 1999

The area between the graph of $y = e^{x}$ and the $x$ -axis from $x = 0$ to $x = k$ ( $k > 0$ ) is rotated through $36 0^{\circ}$ about the $x$ -axis. Calculate, in terms of $k$ and $e$ , the volume of the solid generated.

Show answer

Answer: $V = \frac{π}{2} (e^{2 k} - 1)$

Show working and answer

When rotated about the $x$ -axis, the volume is $V = π \int_{0}^{k} y^{2} d x$ , with $y = e^{x}$ .
$y^{2} = e^{2 x}$ , so $V = π \int_{0}^{k} e^{2 x} d x$ .
An antiderivative of $e^{2 x}$ is $\frac{1}{2} e^{2 x}$ : $V = π [\frac{1}{2} e^{2 x}]_{0}^{k} = \frac{π}{2} (e^{2 k} - 1)$ .

Answer: $V = \frac{π}{2} (e^{2 k} - 1)$

Exercise 228 MediumIB · P1 TZ1 May 2008

The region bounded by the curve $y = \frac{ln ( x )}{x}$ and the lines $x = 1$ , $x = e$ , $y = 0$ is rotated through $2 π$ radians about the $x$ -axis. Find the volume of the solid generated.

Show answer

Answer: $V = π (2 - \frac{5}{e})$

Show working and answer

Volume of revolution about the $x$ -axis: $V = π \int_{1}^{e} (\frac{ln x}{x})^{2} d x = π \int_{1}^{e} \frac{( ln x ) ^{2}}{x ^{2}} d x$ .
We integrate by parts with $u = (ln x)^{2}$ and $d v = \frac{1}{x ^{2}} d x$ : then $d u = \frac{2 ln x}{x} d x$ and $v = - \frac{1}{x}$ .
$\int \frac{( ln x ) ^{2}}{x ^{2}} d x = - \frac{( ln x ) ^{2}}{x} + 2 \int \frac{ln x}{x ^{2}} d x$ .
For the last integral, by parts again with $u = ln x$ and $d v = \frac{1}{x ^{2}} d x$ : $\int \frac{ln x}{x ^{2}} d x = - \frac{ln x}{x} + \int \frac{1}{x ^{2}} d x = - \frac{ln x}{x} - \frac{1}{x}$ .
Putting together: antiderivative $= - \frac{( ln x ) ^{2} + 2 ln x + 2}{x}$ .
Evaluating: at $x = e$ it equals $- \frac{1 + 2 + 2}{e} = - \frac{5}{e}$ ; at $x = 1$ it equals $- 2$ .
$V = π (- \frac{5}{e} + 2)$ .

Answer: $V = π (2 - \frac{5}{e})$

Exercise 229 MediumIB · P2 TZ1 May 2015

Region $R$ is enclosed by the graph of $y = e^{- x^{2}}$ , the $x$ -axis and the lines $x = - 1$ and $x = 1$ . Calculate the volume of the solid of revolution formed when $R$ is rotated through $2 π$ about the $x$ -axis.

Show answer

Answer: $V \approx 3.76$

Show working and answer

When rotating about the $x$ -axis: $V = π \int_{- 1}^{1} y^{2} d x = π \int_{- 1}^{1} e^{- 2 x^{2}} d x$ .
This function has no elementary antiderivative; using the GDC: $\int_{- 1}^{1} e^{- 2 x^{2}} d x \approx 1.196$ .
$V \approx π \cdot 1.196 \approx 3.76$ .

Answer: $V \approx 3.76$

Exercise 230 MediumIB · P1 TZ2 May 2009

Consider the part of the curve $4 x^{2} + y^{2} = 4$ shown below.

Arc of the curve 4x² + y² = 4 in the first quadrant, decreasing, from the point where it meets the y-axis to the point where it meets the x-axis

a) Find an expression for $\frac{d y}{d x}$ in terms of $x$ and $y$ .

b) Find the gradient of the tangent at the point $(\frac{2}{5}, \frac{2}{5})$ .

c) The curve is rotated through $2 π$ radians about the $x$ -axis, forming a bowl. Calculate the volume of this bowl.

Show answer

Answer: a) $\frac{d y}{d x} = - \frac{4 x}{y}$ ; b) gradient $= - 4$ ; c) $V = \frac{8 π}{3}$

Show working and answer

a) 1. Differentiating implicitly the equation $4 x^{2} + y^{2} = 4$ with respect to $x$ : $8 x + 2 y \frac{d y}{d x} = 0$ . 2. Rearranging: $\frac{d y}{d x} = - \frac{4 x}{y}$ .

b) 1. We check that the point lies on the curve: $4 \cdot \frac{4}{5} + \frac{4}{5} = 4$ ✓. 2. Gradient: $- \frac{4 \cdot \frac{2}{5}}{\frac{2}{5}} = - 4$ .

c) 1. Volume about the $x$ -axis: $V = π \int_{0}^{1} y^{2} d x$ , with $y^{2} = 4 - 4 x^{2}$ (the curve meets the $x$ -axis at $x = 1$ ). 2. $V = π \int_{0}^{1} (4 - 4 x^{2}) d x = π [4 x - \frac{4 x ^{3}}{3}]_{0}^{1} = π (4 - \frac{4}{3}) = \frac{8 π}{3}$ .

Answer: a) $\frac{d y}{d x} = - \frac{4 x}{y}$ ; b) gradient $= - 4$ ; c) $V = \frac{8 π}{3}$

Exercise 231 MediumIB · P1 TZ2 May 2007

The graph of $y = sin (3 x)$ for $0 \leq x \leq \frac{π}{4}$ is rotated through $2 π$ radians about the $x$ -axis. Find the exact volume of the solid of revolution thus generated.

Show answer

Answer: Volume $= \frac{π}{2} (\frac{π}{4} + \frac{1}{6})$

Show working and answer

The volume of revolution about the $x$ -axis is $V = π \int_{0}^{π /4} sin^{2} (3 x) d x$ .
We use $sin^{2} (3 x) = \frac{1 - cos 6 x}{2}$ .
$V = π [\frac{x}{2} - \frac{sin 6 x}{12}]_{0}^{π /4} = π (\frac{π}{8} - \frac{sin \frac{3 π}{2}}{12})$ .
As $sin \frac{3 π}{2} = - 1$ : $V = π (\frac{π}{8} + \frac{1}{12}) = \frac{π}{2} (\frac{π}{4} + \frac{1}{6})$ .

Answer: Volume $= \frac{π}{2} (\frac{π}{4} + \frac{1}{6})$

Exercise 232 MediumIB · P1 TZ1 May 2010

The region enclosed between the curves $y = x e^{x}$ and $y = e x$ is rotated through $2 π$ radians about the $x$ -axis. Calculate the volume of the solid obtained.

Show answer

Answer: Volume $= \frac{π}{4} (e^{2} - 1)$

Show working and answer

We equate the curves: $x e^{x} = e x \Rightarrow x (e^{x} - e) = 0 \Rightarrow x = 0$ or $x = 1$ .
For $0 < x < 1$ , $e^{x} < e$ , so $y = e x$ lies above (it is the outer curve) and $y = x e^{x}$ lies below.
Volume about the $x$ -axis (washers): $V = π \int_{0}^{1} [(e x)^{2} - (x e^{x})^{2}] d x = π \int_{0}^{1} (e^{2} x - x e^{2 x}) d x$ .
$\int_{0}^{1} e^{2} x d x = \frac{e ^{2}}{2}$ .
$\int x e^{2 x} d x$ by parts ( $u = x$ , $d v = e^{2 x} d x$ ): $\frac{x e ^{2 x}}{2} - \frac{e ^{2 x}}{4}$ . Between $0$ and $1$ : $(\frac{e ^{2}}{2} - \frac{e ^{2}}{4}) - (- \frac{1}{4}) = \frac{e ^{2}}{4} + \frac{1}{4}$ .
$V = π (\frac{e ^{2}}{2} - \frac{e ^{2}}{4} - \frac{1}{4}) = \frac{π}{4} (e^{2} - 1)$ .

Answer: Volume $= \frac{π}{4} (e^{2} - 1)$

Exercise 233 MediumIB · P2 TZ0 November 2015

A function is given by $f (x) = x^{2} + 2$ , $x \geq 0$ .

The region $R$ is bounded by $y = f (x)$ , the $y$ -axis and the line $y = 4$ .

a) i) Express the area of the region $R$ as an integral with respect to $y$ .

(ii) Determine the area of $R$ , giving your answer correct to four significant figures.

b) Find the exact volume generated when the region $R$ is rotated through $2 π$ radians about the $y$ -axis.

Show answer

Answer: a) (i) Area $= \int_{2}^{4} y - 2 d y$ ; (ii) Area $\approx 1.886$ ; b) Volume $= 2 π$

Show working and answer

a) (i) 1. To integrate with respect to $y$ , we make $x$ the subject: $y = x^{2} + 2 \Rightarrow x = y - 2$ (with $x \geq 0$ ). 2. The region lies between the $y$ -axis ( $x = 0$ ) and the curve, with $y$ between $f (0) = 2$ and $y = 4$ . 3. Area $= \int_{2}^{4} y - 2 d y$ .

(ii) $\int_{2}^{4} y - 2 d y = [\frac{2}{3} (y - 2)^{3/2}]_{2}^{4} = \frac{2}{3} \cdot 2^{3/2} = \frac{4 2}{3} \approx 1.886$ .

b) 1. When rotated about the $y$ -axis, the discs have radius $x$ : $V = π \int_{2}^{4} x^{2} d y = π \int_{2}^{4} (y - 2) d y$ . 2. $= π [\frac{( y - 2 ) ^{2}}{2}]_{2}^{4} = π \cdot \frac{4}{2} = 2 π$ .

Answer: a) (i) Area $= \int_{2}^{4} y - 2 d y$ ; (ii) Area $\approx 1.886$ ; b) Volume $= 2 π$

Exercise 234 MediumIB · P2 TZ2 May 2017

Consider the curve defined by the equation $4 x^{2} + y^{2} = 7$ .

a) Find the equation of the normal to the curve at the point $(1, 3)$ .

b) Find the volume of the solid formed when the region bounded by the curve, the $x$ -axis for $x \geq 0$ and the $y$ -axis for $y \geq 0$ is rotated through $2 π$ radians about the $x$ -axis.

Show answer

Answer: a) $y = \frac{3}{4} x + \frac{3 3}{4}$ ; b) $V = \frac{7 7 π}{3} \approx 19.4$

Show working and answer

a) 1. Differentiate implicitly: $8 x + 2 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{4 x}{y}$ . 2. At $(1, 3)$ (which lies on the curve: $4 + 3 = 7$ ) the gradient of the tangent is $- \frac{4}{3}$ , and that of the normal is $\frac{3}{4}$ . 3. Equation of the normal: $y - 3 = \frac{3}{4} (x - 1)$ , i.e. $y = \frac{3}{4} x + \frac{3 3}{4}$ .

b) 1. The region in the first quadrant bounded by the curve and the axes has $y^{2} = 7 - 4 x^{2}$ , with $x$ from $0$ to $\frac{7}{2}$ (where $y = 0$ ). 2. Rotating about the $x$ -axis: $V = π \int_{0}^{7 /2} y^{2} d x = π \int_{0}^{7 /2} (7 - 4 x^{2}) d x$ . 3. $V = π [7 x - \frac{4 x ^{3}}{3}]_{0}^{7 /2} = π (\frac{7 7}{2} - \frac{7 7}{6}) = \frac{7 7 π}{3} \approx 19.4$ .

Answer: a) $y = \frac{3}{4} x + \frac{3 3}{4}$ ; b) $V = \frac{7 7 π}{3} \approx 19.4$

Exercise 235 MediumIB · P2 May 1997

a) Draw a diagram showing the region $R$ enclosed by the curve $y = sin x$ , the $x$ -axis and the lines $x = 0$ and $x = k$ , $0 < k < π$ .

b) Calculate the area of the region $R$ .

c) Calculate the volume of the solid obtained by rotating the region $R$ through $36 0^{\circ}$ about the $x$ -axis.

View solution

a) 1. The region lies between the curve $y = sin x$ (which is positive on $] 0, π [$ ) and the $x$ -axis, from $x = 0$ to the vertical line $x = k$ with $0 < k < π$ .

Region bounded by the curve y = sin x, the x-axis and the line x = k, with 0 < k < π

b) 1. $A = \int_{0}^{k} sin x d x = [- cos x]_{0}^{k} = - cos k + 1$ .

c) 1. Rotating about the $x$ -axis: $V = π \int_{0}^{k} sin^{2} x d x$ . 2. Using $sin^{2} x = \frac{1 - cos 2 x}{2}$ : $V = \frac{π}{2} [x - \frac{sin 2 x}{2}]_{0}^{k}$ . 3. $V = \frac{π}{2} (k - \frac{sin 2 k}{2})$ .

Answer: b) $A = 1 - cos k$ ; c) $V = \frac{π}{2} (k - \frac{sin 2 k}{2})$

Exercise 236 MediumIB · P1 May 1999

The area of the closed region shown in the diagram is defined by $y \geq x^{2} + 2$ , $y \leq a x + 2$ , with $a > 0$ .

Closed region between the parabola y = x² + 2 and the line y = ax + 2 (a > 0), shaded, from x = 0 to x = a

This region is rotated $36 0^{\circ}$ about the $x$ -axis to form a solid of revolution. Calculate, in terms of $a$ , the volume of this solid of revolution.

Show answer

Answer: $V = \frac{2 π a ^{3}}{15} (a^{2} + 5)$

Show working and answer

The parabola $y = x^{2} + 2$ and the line $y = a x + 2$ intersect where $x^{2} = a x$ , i.e. at $x = 0$ and $x = a$ .
Between $0$ and $a$ the line lies above the parabola (for example, at $x = \frac{a}{2}$ : $\frac{a ^{2}}{2} + 2 > \frac{a ^{2}}{4} + 2$ ), as shown in the diagram.
When the region is rotated about the $x$ -axis, the solid is the difference between the solid generated by the line and the one generated by the parabola: $V = π \int_{0}^{a} [(a x + 2)^{2} - (x^{2} + 2)^{2}] d x$ .
Expand: $(a x + 2)^{2} - (x^{2} + 2)^{2} = a^{2} x^{2} + 4 a x + 4 - x^{4} - 4 x^{2} - 4 = (a^{2} - 4) x^{2} + 4 a x - x^{4}$ .
Integrate: $V = π [\frac{( a ^{2} - 4 ) x ^{3}}{3} + 2 a x^{2} - \frac{x ^{5}}{5}]_{0}^{a} = π [\frac{a ^{5} - 4 a ^{3}}{3} + 2 a^{3} - \frac{a ^{5}}{5}]$ .
Group terms: $V = π [\frac{2 a ^{5}}{15} + \frac{2 a ^{3}}{3}] = \frac{2 π a ^{3}}{15} (a^{2} + 5)$ .

Answer: $V = \frac{2 π a ^{3}}{15} (a^{2} + 5)$

Exercise 237 MediumIB · P1 TZ0 November 2007

The following diagram shows the shaded region $A$ , bounded by the axes and by part of the curve $y^{2} = 8 a (2 a - x)$ , $a > 0$ .

Shaded region A bounded by the axes and by the arc of the curve y² = 8a(2a − x), which goes from the point (0, 4a) to the point (2a, 0)

Find, in terms of $a$ , the volume of the solid generated when $A$ is rotated through $36 0^{\circ}$ about the $x$ -axis.

Show answer

Answer: Volume $= 16 π a^{3}$

Show working and answer

When $A$ is rotated about the $x$ -axis, the volume is $V = π \int_{0}^{2 a} y^{2} d x$ .
Substituting $y^{2} = 8 a (2 a - x)$ : $V = 8 π a \int_{0}^{2 a} (2 a - x) d x$ .
$\int_{0}^{2 a} (2 a - x) d x = [2 a x - \frac{x ^{2}}{2}]_{0}^{2 a} = 4 a^{2} - 2 a^{2} = 2 a^{2}$ .
$V = 8 π a \cdot 2 a^{2} = 16 π a^{3}$ .

Answer: Volume $= 16 π a^{3}$

Exercise 238 MediumIB · P1 TZ0 November 2011

Given that $f (x) = 1 + sin x$ , $0 \leq x \leq \frac{3 π}{2}$ ,

a) Sketch the graph of $f$ approximately.

b) Verify that $(f (x))^{2} = \frac{3}{2} + 2 sin x - \frac{1}{2} cos 2 x$ .

c) Find the volume of the solid generated when the graph of $f$ is rotated $2 π$ radians about the $x$ -axis.

View solution

a) Key features: it starts at $(0, 1)$ , rises to the maximum $2$ at $x = \frac{π}{2}$ , falls through $1$ at $x = π$ and reaches $0$ at $x = \frac{3 π}{2}$ , where it touches the $x$ -axis.

Graph of f(x) = 1 + sin x on (0, 3π/2): it starts at (0, 1), rises to the maximum 2 at x = π/2 and falls until it touches the x-axis at x = 3π/2

b) 1. $(1 + sin x)^{2} = 1 + 2 sin x + sin^{2} x$ . 2. As $cos 2 x = 1 - 2 sin^{2} x$ , $sin^{2} x = \frac{1 - cos 2 x}{2}$ . 3. Therefore $(f (x))^{2} = 1 + 2 sin x + \frac{1}{2} - \frac{1}{2} cos 2 x = \frac{3}{2} + 2 sin x - \frac{1}{2} cos 2 x$ .

c) 1. $V = π \int_{0}^{3 π /2} (f (x))^{2} d x = π \int_{0}^{3 π /2} (\frac{3}{2} + 2 sin x - \frac{1}{2} cos 2 x) d x$ . 2. Antiderivative: $\frac{3 x}{2} - 2 cos x - \frac{1}{4} sin 2 x$ . 3. At $x = \frac{3 π}{2}$ : $\frac{9 π}{4} - 0 - 0$ . At $x = 0$ : $0 - 2 - 0 = - 2$ . 4. $V = π (\frac{9 π}{4} + 2)$ .

Answer: a) sketch of $f$ ; b) shown; c) $V = π (\frac{9 π}{4} + 2)$

Exercise 239 MediumIB · P2 TZ2 May 2015

The graph of $y = ln (5 x + 10)$ is obtained from the graph of $y = ln x$ by a translation of $a$ units in the direction of the $x$ -axis followed by a translation of $b$ units in the direction of the $y$ -axis.

a) Find the value of $a$ and the value of $b$ .

b) The region bounded by the graph of $y = ln (5 x + 10)$ , the $x$ -axis and the lines $x = e$ and $x = 2 e$ is rotated $2 π$ radians about the $x$ -axis. Find the volume generated.

Show answer

Answer: a) $a = - 2$ , $b = ln 5$ ; b) $V = 99.2$

Show working and answer

a) 1. We write $ln (5 x + 10) = ln (5 (x + 2)) = ln 5 + ln (x + 2)$ . 2. $ln (x + 2)$ is obtained from $ln x$ by translating $2$ units to the left: $a = - 2$ . 3. Adding $ln 5$ translates the graph by $ln 5$ units upwards: $b = ln 5$ .

b) Method: volume of revolution about the $x$ -axis, $V = π \int y^{2} d x$ , using the GDC (Paper 2).

$V = π \int_{e}^{2 e} [ln (5 x + 10)]^{2} d x$ .
The function is positive on that interval, so there is no need to worry about signs.
Using the GDC: $V \approx 99.2$ .

Answer: a) $a = - 2$ , $b = ln 5$ ; b) $V = 99.2$

Exercise 240 MediumIB · P2 TZ1 May 2014

The shaded region $S$ is enclosed by the curve $y = x + 2 cos x$ , for $0 \leq x \leq 2 π$ , and the line $y = x$ , as shown in the following diagram.

Curve y = x + 2cos x for 0 ≤ x ≤ 2π and line y = x; the region S between them, where the curve lies below the line, is shaded

a) Find the coordinates of the points where the line meets the curve.

The region $S$ is rotated through $2 π$ about the $x$ -axis to generate a solid.

b) (i) Write down an integral that represents the volume $V$ of the solid.

(ii) Calculate the volume $V$ .

Show answer

Answer: a) $(\frac{π}{2}, \frac{π}{2})$ and $(\frac{3 π}{2}, \frac{3 π}{2})$ ; b) (i) $V = π \int_{π /2}^{3 π /2} [x^{2} - (x + 2 cos x)^{2}] d x$ ; (ii) $V = 6 π^{2} = 59.2$

Show working and answer

a) 1. The line meets the curve when $x + 2 cos x = x$ , that is, $cos x = 0$ . 2. On $[0, 2 π]$ : $x = \frac{π}{2}$ and $x = \frac{3 π}{2}$ , with $y = x$ in both cases. 3. Points: $(\frac{π}{2}, \frac{π}{2})$ and $(\frac{3 π}{2}, \frac{3 π}{2})$ .

b) (i) 1. Rotating $S$ about the $x$ -axis forms a solid with a hole: the line $y = x$ lies above the curve, and both are positive on the interval (the minimum of $x + 2 cos x$ there is $\frac{5 π}{6} - 3 \approx 0.89 > 0$ ). 2. Volume = (outer volume, with $y = x$ ) $-$ (inner volume, with $y = x + 2 cos x$ ): $V = π \int_{π /2}^{3 π /2} [x^{2} - (x + 2 cos x)^{2}] d x$ .

(ii) 1. We expand: $x^{2} - (x + 2 cos x)^{2} = - 4 x cos x - 4 cos^{2} x$ . 2. For $- 4 x cos x$ we use integration by parts ( $u = x$ , $d v = cos x d x$ ): $\int x cos x d x = x sin x + cos x$ . Between $\frac{π}{2}$ and $\frac{3 π}{2}$ : $(- \frac{3 π}{2}) - (\frac{π}{2}) = - 2 π$ , so this part equals $- 4 (- 2 π) = 8 π$ . 3. For $- 4 cos^{2} x = - 2 (1 + cos 2 x)$ : $\int_{π /2}^{3 π /2} - 2 (1 + cos 2 x) d x = - 2 [x + \frac{sin 2 x}{2}]_{π /2}^{3 π /2} = - 2 π$ . 4. The complete integral is $8 π - 2 π = 6 π$ , so $V = π \cdot 6 π = 6 π^{2} \approx 59.2$ .

Answer: a) $(\frac{π}{2}, \frac{π}{2})$ and $(\frac{3 π}{2}, \frac{3 π}{2})$ ; b) (i) $V = π \int_{π /2}^{3 π /2} [x^{2} - (x + 2 cos x)^{2}] d x$ ; (ii) $V = 6 π^{2} = 59.2$

Exercise 241 MediumIB · P2 TZ0 November 2022

The function $f$ is defined by $f (x) = e^{2 x} (3 x - 4)$ , where $x \in R$ .

a) Find $f^{'} (x)$ .

b) Hence, or otherwise, find the coordinates of the point on the graph of $y = f (x)$ at which the tangent is parallel to the line $y = x$ .

The region bounded by the curve $y = f (x)$ , the $x$ -axis and the $y$ -axis is rotated through $2 π$ radians about the $x$ -axis, generating a solid of revolution.

c) Find the volume of this solid.

Now consider a function $g$ such that $g (0) = 1$ and $g^{'} (0) = 2$ .

d) Find the value of:

(i) $(f \circ g) (0)$

(ii) $(f \circ g)^{'} (0)$

Show answer

Answer: a) $f^{'} (x) = e^{2 x} (6 x - 5)$ ; b) $(0.863; - 7.93)$ ; c) $V = 164.8 \approx 165$ ; d) (i) $- e^{2}$ ; (ii) $2 e^{2}$

Show working and answer

a) 1. Product rule with $u = e^{2 x}$ ( $u^{'} = 2 e^{2 x}$ ) and $v = 3 x - 4$ ( $v^{'} = 3$ ). 2. $f^{'} (x) = 2 e^{2 x} (3 x - 4) + 3 e^{2 x} = e^{2 x} (6 x - 8 + 3) = e^{2 x} (6 x - 5)$ .

b) 1. The tangent is parallel to $y = x$ when its gradient is $1$ : $e^{2 x} (6 x - 5) = 1$ . 2. Using the GDC: $x \approx 0.863$ . 3. $y = f (0.863) = e^{1.726} (3 \cdot 0.863 - 4) \approx - 7.93$ . The point is $(0.863; - 7.93)$ .

c) 1. $f (x) = 0$ when $x = \frac{4}{3}$ ; the curve cuts the $y$ -axis at $(0, - 4)$ . The region bounded by the curve and the two axes lies between $x = 0$ and $x = \frac{4}{3}$ . 2. $V = π \int_{0}^{4/3} [e^{2 x} (3 x - 4)]^{2} d x$ . 3. Using the GDC: $V \approx 164.8$ .

d) (i) 1. $(f \circ g) (0) = f (g (0)) = f (1) = e^{2} (3 - 4) = - e^{2}$ .

(ii) 1. Chain rule: $(f \circ g)^{'} (0) = f^{'} (g (0)) \cdot g^{'} (0) = f^{'} (1) \cdot 2$ . 2. $f^{'} (1) = e^{2} (6 - 5) = e^{2}$ , so $(f \circ g)^{'} (0) = 2 e^{2}$ .

Answer: a) $f^{'} (x) = e^{2 x} (6 x - 5)$ ; b) $(0.863; - 7.93)$ ; c) $V = 164.8 \approx 165$ ; d) (i) $- e^{2}$ ; (ii) $2 e^{2}$

Exercise 242 HardIB · P2 TZ0 May 2012

A cone has height $h$ and base radius $r$ . Deduce the formula for the volume of this cone by rotating the triangular region enclosed by the line $y = h - \frac{h}{r} x$ and the coordinate axes through $2 π$ about the $y$ -axis.

Show answer

Answer: $V = \frac{1}{3} π r^{2} h$

Show working and answer

We place the cone with its vertex at $(0, h)$ and its base on the $x$ -axis, with radius $r$ . The line $y = h - \frac{h}{r} x$ meets the axes at $(0, h)$ and $(r, 0)$ .
When rotating about the $y$ -axis we use $V = π \int_{0}^{h} x^{2} d y$ . Solving for $x$ : $x = \frac{r}{h} (h - y) = r (1 - \frac{y}{h})$ .
$V = π r^{2} \int_{0}^{h} (1 - \frac{y}{h})^{2} d y = π r^{2} [- \frac{h}{3} (1 - \frac{y}{h})^{3}]_{0}^{h}$ .
At $y = h$ the bracket equals $0$ ; at $y = 0$ it equals $- \frac{h}{3}$ . Therefore $V = π r^{2} \cdot \frac{h}{3}$ .

Answer: $V = \frac{1}{3} π r^{2} h$

Exercise 243 HardIB · P2 TZ1 May 2013

Let $f (x) = sin (x - 1)$ , $0 \leq x \leq \frac{π}{2} + 1$ . Find the volume of the solid formed when the region bounded by $y = f (x)$ and the lines $x = 0$ , $y = 0$ and $y = 1$ is rotated through $2 π$ about the $y$ -axis.

Show answer

Answer: $V \approx 8.20$

Show working and answer

The curve $y = sin (x - 1)$ equals $0$ at $x = 1$ and $1$ at $x = 1 + \frac{π}{2}$ , and it is increasing over that stretch. The region is bounded by $y = 0$ , $y = 1$ , the $y$ -axis ( $x = 0$ ) and the curve.
We rotate about the $y$ -axis, so we integrate with respect to $y$ using $V = π \int_{0}^{1} x^{2} d y$ .
Solving $y = sin (x - 1)$ for $x$ : $x = 1 + arcsin y$ . The outer radius is this curve and the inner radius is the $y$ -axis ( $x = 0$ ), so there is no hole.
$V = π \int_{0}^{1} (1 + arcsin y)^{2} d y$ .
Using the GDC: $V \approx 8.20$ .

Answer: $V \approx 8.20$

Exercise 244 HardIB · P1 TZ0 November 2009

The shape of a glass is defined by rotating the graph of $y = e^{x}$ about the $y$ -axis, for $1 \leq y \leq 5$ . Find the volume of the glass.

Show answer

Answer: $V = π [5 (ln 5)^{2} - 10 ln 5 + 8] \approx 15.3$

Show working and answer

When rotating about the $y$ -axis we use $V = π \int_{1}^{5} x^{2} d y$ . From $y = e^{x}$ we get $x = ln y$ .
$V = π \int_{1}^{5} (ln y)^{2} d y$ .
By parts with $u = (ln y)^{2}$ , $d v = d y$ : $d u = \frac{2 ln y}{y} d y$ , $v = y$ . Then $\int (ln y)^{2} d y = y (ln y)^{2} - 2 \int ln y d y$ .
And $\int ln y d y = y ln y - y$ (again by parts). This gives $\int (ln y)^{2} d y = y [(ln y)^{2} - 2 ln y + 2]$ .
We evaluate: at $y = 5$ it equals $5 [(ln 5)^{2} - 2 ln 5 + 2]$ ; at $y = 1$ it equals $2$ .
$V = π [5 (ln 5)^{2} - 10 ln 5 + 2 \cdot 5 - 2] = π [5 (ln 5)^{2} - 10 ln 5 + 8]$ .

Answer: $V = π [5 (ln 5)^{2} - 10 ln 5 + 8] \approx 15.3$

Exercise 245 HardIB · P2 TZ0 November 2013

A function $f$ is given by $f (x) = \frac{1}{2} (e^{x} + e^{- x})$ , $x \in R$ .

a) (i) Explain why the inverse function $f^{- 1}$ does not exist.

(ii) Verify that the equation of the normal to the curve at the point $P$ , where $x = ln 3$ , is given by $9 x + 12 y - 9 ln 3 - 20 = 0$ .

(iii) Find the $x$ -coordinates of the points $Q$ and $R$ on the curve such that the tangents at $Q$ and $R$ pass through $(0, 0)$ .

b) Now the domain of $f$ is restricted to $x \geq 0$ .

(i) Find an expression for $f^{- 1} (x)$ .

(ii) Find the volume generated when the region bounded by the curve $y = f (x)$ and the lines $x = 0$ and $y = 5$ is rotated through an angle of $2 π$ radians about the $y$ -axis.

Show answer

Answer: a) (i) $f$ is even, hence not one-to-one; (ii) verified; (iii) $x \approx \pm 1.20$ ; b) (i) $f^{- 1} (x) = ln (x + x^{2} - 1)$ ; (ii) $V \approx 37.1$

Show working and answer

a) (i) 1. $f (- x) = \frac{1}{2} (e^{- x} + e^{x}) = f (x)$ , so $f$ is even and its graph is symmetric about the $y$ -axis. 2. Therefore a horizontal line such as $y = 3$ cuts the curve at two distinct points: $f$ is not one-to-one and so has no inverse function.

(ii) 1. At $x = ln 3$ : $e^{l n 3} = 3$ and $e^{- l n 3} = \frac{1}{3}$ , so $f (ln 3) = \frac{1}{2} (3 + \frac{1}{3}) = \frac{5}{3}$ . Thus $P (ln 3, \frac{5}{3})$ . 2. $f^{'} (x) = \frac{1}{2} (e^{x} - e^{- x})$ , and $f^{'} (ln 3) = \frac{1}{2} (3 - \frac{1}{3}) = \frac{4}{3}$ . 3. The gradient of the normal is $- \frac{3}{4}$ , so $y - \frac{5}{3} = - \frac{3}{4} (x - ln 3)$ . 4. Multiplying by $12$ : $12 y - 20 = - 9 x + 9 ln 3$ , i.e. $9 x + 12 y - 9 ln 3 - 20 = 0$ .

(iii) 1. Let $x_{0}$ be the $x$ -coordinate of the point of tangency. The tangent is $y - f (x_{0}) = f^{'} (x_{0}) (x - x_{0})$ and passes through $(0, 0)$ if $- f (x_{0}) = - x_{0} f^{'} (x_{0})$ , i.e. $f (x_{0}) = x_{0} f^{'} (x_{0})$ . 2. This gives $\frac{1}{2} (e^{x} + e^{- x}) = \frac{x}{2} (e^{x} - e^{- x})$ . 3. Using the GDC (intersection of the two graphs) we obtain $x \approx 1.20$ and, by symmetry, $x \approx - 1.20$ .

b) (i) 1. With $y = \frac{1}{2} (e^{x} + e^{- x})$ , multiply by $2 e^{x}$ : $e^{2 x} - 2 y e^{x} + 1 = 0$ . 2. This is a quadratic equation in $e^{x}$ : $e^{x} = y \pm y^{2} - 1$ . 3. With $x \geq 0$ we have $e^{x} \geq 1$ , so we choose the $+$ sign (with the $-$ sign the value would be $\leq 1$ ). 4. Therefore $x = ln (y + y^{2} - 1)$ , and $f^{- 1} (x) = ln (x + x^{2} - 1)$ , with $x \geq 1$ .

(ii) 1. The region lies between $x = 0$ and the curve, with $y$ between $f (0) = 1$ and $y = 5$ . When rotating about the $y$ -axis we use discs of radius $x = f^{- 1} (y)$ . 2. $V = π \int_{1}^{5} [f^{- 1} (y)]^{2} d y = π \int_{1}^{5} [ln (y + y^{2} - 1)]^{2} d y$ . 3. Using the GDC, $V \approx 37.1$ .

Answer: a) (i) $f$ is even, hence not one-to-one; (ii) verified; (iii) $x \approx \pm 1.20$ ; b) (i) $f^{- 1} (x) = ln (x + x^{2} - 1)$ ; (ii) $V \approx 37.1$

Exercise 246 HardIB · P2 TZ1 May 2009

Let $f$ be the function defined by $f (x) = x + 2 cos x$ , $x \in [0, 2 π]$ . The following diagram shows the region $S$ enclosed by the graph of $f$ and the line $y = x$ .

Graph of the curve y = x + 2cos x for x between 0 and 2π and of the line y = x, which intersect at the points A and C; the minimum of the curve is the point B and the region S between both is shaded

$A$ and $C$ are the points of intersection of the line $y = x$ and the graph of $f$ , and $B$ is the minimum point of $f$ .

a) If $A$ , $B$ and $C$ have $x$ -coordinates $\frac{aπ}{2}$ , $\frac{bπ}{6}$ and $\frac{c π}{2}$ , where $a, b, c \in N$ , find the values of $a$ , $b$ and $c$ .

b) Find the range of $f$ .

c) Find the equation of the normal to the graph of $f$ at the point $C$ , giving your answer in the form $y = p x + q$ .

d) The region $S$ is rotated through $2 π$ radians about the $x$ -axis, generating a solid.

(i) Write down the integral that represents the volume $V$ of the solid.

(ii) Show that $V = 6 π^{2}$ .

Show answer

Answer: a) $a = 1$ , $b = 5$ , $c = 3$ ; b) $R_{f} = [\frac{5 π}{6} - 3, 2 π + 2]$ ; c) $y = - \frac{1}{3} x + 2 π$ ; d) (i) $V = π \int_{π /2}^{3 π /2} [x^{2} - (x + 2 cos x)^{2}] d x$ ; (ii) $V = 6 π^{2}$

Show working and answer

a) 1. Intersection of the curve with the line: $x + 2 cos x = x \Rightarrow cos x = 0 \Rightarrow x = \frac{π}{2}$ or $x = \frac{3 π}{2}$ (in $[0, 2 π]$ ). Thus $A$ has $x = \frac{π}{2}$ , that is, $a = 1$ , and $C$ has $x = \frac{3 π}{2}$ , that is, $c = 3$ . 2. Minimum: $f^{'} (x) = 1 - 2 sin x = 0 \Rightarrow sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}$ or $x = \frac{5 π}{6}$ . With $f^{''} (x) = - 2 cos x$ : $f^{''} (\frac{5 π}{6}) = 3 > 0$ (minimum) and $f^{''} (\frac{π}{6}) < 0$ (local maximum). 3. Hence $B$ has $x = \frac{5 π}{6}$ , that is, $b = 5$ .

b) 1. Absolute minimum: $f (\frac{5 π}{6}) = \frac{5 π}{6} + 2 cos \frac{5 π}{6} = \frac{5 π}{6} - 3 \approx 0.886$ . This is less than the value at the endpoint $f (0) = 2$ . 2. Absolute maximum: we compare $f (\frac{π}{6}) = \frac{π}{6} + 3 \approx 2.26$ with the endpoint $f (2 π) = 2 π + 2 \approx 8.28$ . The greater is $2 π + 2$ . 3. Range: $[\frac{5 π}{6} - 3, 2 π + 2]$ .

c) 1. At $C (\frac{3 π}{2}, \frac{3 π}{2})$ : $f^{'} (\frac{3 π}{2}) = 1 - 2 sin \frac{3 π}{2} = 3$ , so the gradient of the normal is $- \frac{1}{3}$ . 2. $y - \frac{3 π}{2} = - \frac{1}{3} (x - \frac{3 π}{2})$ , that is, $y = - \frac{1}{3} x + \frac{π}{2} + \frac{3 π}{2}$ . 3. $y = - \frac{1}{3} x + 2 π$ , with $p = - \frac{1}{3}$ and $q = 2 π$ .

d) (i) 1. For $\frac{π}{2} \leq x \leq \frac{3 π}{2}$ the line lies above the curve (there $cos x \leq 0$ ) and the curve lies above the $x$ -axis (its minimum is $0.886 > 0$ ). 2. Rotating about the $x$ -axis gives washers with outer radius $x$ and inner radius $f (x)$ : $V = π \int_{π /2}^{3 π /2} [x^{2} - (x + 2 cos x)^{2}] d x$ .

(ii) 1. $x^{2} - (x + 2 cos x)^{2} = - 4 x cos x - 4 cos^{2} x$ . 2. By parts ( $u = x$ , $d v = cos x d x$ ): $\int x cos x d x = x sin x + cos x$ . Between $\frac{π}{2}$ and $\frac{3 π}{2}$ : $(- \frac{3 π}{2} + 0) - (\frac{π}{2} + 0) = - 2 π$ . 3. $\int cos^{2} x d x = \frac{x}{2} + \frac{sin 2 x}{4}$ . Between the same limits: $\frac{3 π}{4} - \frac{π}{4} = \frac{π}{2}$ . 4. $V = π [- 4 (- 2 π) - 4 \cdot \frac{π}{2}] = π (8 π - 2 π) = 6 π^{2}$ .

Exercise 247 HardIB · P2 TZ1 May 2019

The function $f$ is defined by $f (x) = (x - 1)^{2}$ , $x \geq 1$ , and the function $g$ is defined by $g (x) = x^{2} + 1$ , $x \geq 0$ . The region $R$ is bounded by the curves $y = f (x)$ , $y = g (x)$ and the lines $y = 0$ , $x = 0$ and $y = 9$ , as shown in the following diagram.

Region R in the first quadrant between the curves y = x² + 1 (starting at (0, 1)) and y = (x − 1)² (starting at (1, 0)), bounded by the y-axis, the x-axis and the line y = 9; axes with grid

The shape of a clay vase can be modelled by rotating the region $R$ through $36 0^{\circ}$ about the $y$ -axis. Find the volume of clay used to make the vase.

Show answer

Answer: Volume $= \frac{107 π}{2} \approx 168$

Show working and answer

We rotate the region about the $y$ -axis, so it is convenient to express each curve as $x$ in terms of $y$ :
- $f (x) = (x - 1)^{2}$ , $x \geq 1$ : $y = (x - 1)^{2} \Rightarrow x = 1 + y$ (outer curve).
- $g (x) = x^{2} + 1$ , $x \geq 0$ : $y = x^{2} + 1 \Rightarrow x = y - 1$ , for $y \geq 1$ (inner curve).
For $0 \leq y \leq 1$ the region reaches the $y$ -axis (the curve $g$ starts at $y = 1$ ), so the solid has no hole: radius $1 + y$ . For $1 \leq y \leq 9$ it is hollow: outer radius $1 + y$ and inner radius $y - 1$ .
$V = π \int_{0}^{9} (1 + y)^{2} d y - π \int_{1}^{9} (y - 1) d y$ .
First integral: $(1 + y)^{2} = 1 + 2 y + y$ , and $\int_{0}^{9} (1 + 2 y + y) d y = [y + \frac{4}{3} y^{3/2} + \frac{y ^{2}}{2}]_{0}^{9} = 9 + 36 + \frac{81}{2} = \frac{171}{2}$ .
Second integral: $\int_{1}^{9} (y - 1) d y = [\frac{( y - 1 ) ^{2}}{2}]_{1}^{9} = 32$ .
$V = π (\frac{171}{2} - 32) = \frac{107 π}{2} \approx 168$ .

Answer: Volume $= \frac{107 π}{2} \approx 168$

Exercise 248 HardIB · P2 TZ1 May 2018

The following diagram shows the two parts of the curve defined by the equation $x^{2} y = 5 - y^{4}$ , and the normal to the curve at the point $P (2, 1)$ .

The two branches of the curve x²y = 5 − y⁴ (a bell shape above and an inverted one below) and the normal at the point P(2, 1), which cuts the curve again at Q; the region between the y-axis, the upper branch and the normal is shaded

a) Prove that there are exactly two points on the curve where the gradient is zero.

b) Find the equation of the normal to the curve at the point $P$ .

c) The normal at $P$ cuts the curve again at the point $Q$ . Find the $x$ -coordinate of $Q$ .

d) The shaded region is rotated through $2 π$ about the $y$ -axis. Find the volume of the solid formed.

Show answer

Answer: a) proved: $(0, 45)$ and $(0, - 45)$ ; b) $y = 2 x - 3$ ; c) $x \approx 0.724$ ; d) Volume $\approx 19.9$

Show working and answer

a) 1. Differentiating $x^{2} y = 5 - y^{4}$ implicitly: $2 x y + x^{2} \frac{d y}{d x} = - 4 y^{3} \frac{d y}{d x}$ . 2. Rearranging: $\frac{d y}{d x} = \frac{- 2 x y}{x ^{2} + 4 y ^{3}}$ . 3. The gradient is $0$ when $x y = 0$ (with a non-zero denominator). If $y = 0$ , the equation would give $0 = 5$ : impossible. So $x = 0$ and $y^{4} = 5$ , that is, $y = \pm 45$ . 4. At these points the denominator is $4 y^{3} \neq = 0$ . There are exactly two points with zero gradient: $(0, 45)$ and $(0, - 45)$ .

b) 1. At $P (2, 1)$ : $\frac{d y}{d x} = \frac{- 2 \cdot 2 \cdot 1}{4 + 4} = - \frac{1}{2}$ . 2. The gradient of the normal is $2$ : $y - 1 = 2 (x - 2)$ , that is, $y = 2 x - 3$ .

c) 1. Substituting $y = 2 x - 3$ into the curve: $x^{2} (2 x - 3) = 5 - (2 x - 3)^{4}$ . 2. Using the GDC, the solutions are $x = 2$ (the point $P$ ) and $x \approx 0.724$ . 3. The $x$ -coordinate of $Q$ is $0.724$ .

d) 1. The solid is generated by rotating about the $y$ -axis the region bounded by the $y$ -axis, the normal (from $(0, - 3)$ to $P$ ) and the upper branch (from $P$ to $(0, 45)$ ). 2. Part of the normal, $- 3 \leq y \leq 1$ : $x = \frac{y + 3}{2}$ , and $π \int_{- 3}^{1} (\frac{y + 3}{2})^{2} d y = \frac{π}{4} [\frac{( y + 3 ) ^{3}}{3}]_{- 3}^{1} = \frac{16 π}{3}$ . 3. Part of the curve, $1 \leq y \leq 45$ : from $x^{2} y = 5 - y^{4}$ we get $x^{2} = \frac{5 - y ^{4}}{y}$ , and $π \int_{1}^{45} (\frac{5}{y} - y^{3}) d y = π [5 ln y - \frac{y ^{4}}{4}]_{1}^{45} = π (\frac{5}{4} ln 5 - 1)$ . 4. $V = π (\frac{16}{3} + \frac{5}{4} ln 5 - 1) \approx 19.9$ .

Answer: a) proved: $(0, 45)$ and $(0, - 45)$ ; b) $y = 2 x - 3$ ; c) $x \approx 0.724$ ; d) Volume $\approx 19.9$

Exercise 249 HardIB · P2 TZ0 November 2017

Consider the function $f (x) = \frac{x}{sin x}$ , $0 < x < π$ .

a) i) Show that the $x$ -coordinate of the minimum point of the curve $y = f (x)$ satisfies the equation $tan x = 2 x$ .

(ii) Determine the values of $x$ for which the function $f (x)$ is decreasing.

b) Sketch the graph of $y = f (x)$ , clearly showing the minimum point and any asymptotic behaviour.

c) Calculate the coordinates of the point on the graph of $f$ where the normal to the graph is parallel to the line $y = - x$ .

Consider the region enclosed by the curve $y = f (x)$ , the $x$ -axis and the lines $x = \frac{π}{6}$ , $x = \frac{π}{3}$ .

d) The region is rotated $2 π$ radians about the $x$ -axis. Calculate the volume of revolution.

View solution

a) (i) 1. Differentiating with the quotient rule: $f^{'} (x) = \frac{\frac{1}{2 x} sin x - x cos x}{sin ^{2} x} = \frac{sin x - 2 x cos x}{2 x sin ^{2} x}$ . 2. At a minimum, $f^{'} (x) = 0$ : $sin x = 2 x cos x$ . 3. In $] 0, π [$ , $cos x \neq = 0$ in this equality (if it were $0$ , then $sin x$ would also be $0$ , which is impossible), so we can divide: $tan x = 2 x$ .

(ii) 1. The sign of $f^{'}$ is that of $sin x - 2 x cos x$ . Using the GDC, the equation $tan x = 2 x$ has exactly one solution in $] 0, π [$ , $x \approx 1.17$ (in $] \frac{π}{2}, π [$ the tangent is negative and $2 x$ is positive). 2. Near $0$ , $sin x - 2 x cos x \approx x - 2 x = - x < 0$ , so $f^{'} < 0$ to the left of that solution and $f^{'} > 0$ to the right. 3. $f$ is decreasing for $0 < x < 1.17$ .

b) 1. As $x \to 0^{+}$ , $f = \frac{x}{sin x} \approx \frac{1}{x} \to + \infty$ : vertical asymptote $x = 0$ . 2. As $x \to π^{-}$ , $sin x \to 0^{+}$ and the numerator tends to $π$ : $f \to + \infty$ , so vertical asymptote $x = π$ . 3. The minimum is at $x \approx 1.17$ , with $y = f (1.17) \approx 1.17$ . 4. The curve falls from $+ \infty$ next to the $y$ -axis, reaches the minimum and rises towards $+ \infty$ next to $x = π$ .

Graph of y = √x/sin x for 0 < x < π: vertical asymptotes x = 0 (the y-axis) and x = π; the curve falls from above next to the y-axis, has its minimum labelled at x ≈ 1.17 and rises next to the asymptote x = π

c) 1. The normal is parallel to $y = - x$ (gradient $- 1$ ), so the tangent has gradient $1$ (perpendicular to $- 1$ ). 2. Solving $f^{'} (x) = 1$ using the GDC: $x \approx 1.96$ . 3. $f (1.96) \approx 1.51$ . The point is $(1.96; 1.51)$ .

d) 1. Volume of revolution: $V = π \int_{π /6}^{π /3} f (x)^{2} d x = π \int_{π /6}^{π /3} \frac{x}{sin ^{2} x} d x$ . 2. Using the GDC: $V \approx 2.68$ . 3. (Exact: integrating by parts with $u = x$ , $d v = \frac{d x}{sin ^{2} x}$ , $\int \frac{x}{sin ^{2} x} d x = - x cot x + ln sin x$ , and we obtain $V = π (\frac{3 π}{18} + \frac{1}{2} ln 3)$ .)

Answer: a) (ii) $f$ is decreasing for $0 < x < 1.17$ ; b) vertical asymptotes $x = 0$ and $x = π$ , minimum at $(1.17; 1.17)$ ; c) $(1.96; 1.51)$ ; d) $V = 2.68$

Exercise 250 HardIB · P1 TZ1 May 2016

The following diagram shows the graph of $y = \frac{( ln x ) ^{2}}{x}$ , $x > 0$ .

Graph of y = (ln x)²/x for x > 0: the curve falls from above next to the y-axis, touches the x-axis at the point with x-coordinate a and rises again, more and more slowly towards the right

a) Given that the curve passes through the point $(a, 0)$ , state the value of $a$ .

Region $R$ is enclosed by the curve, the $x$ -axis and the line $x = e$ .

b) Use the substitution $u = ln x$ to find the area of region $R$ .

Let $I_{n} = \int_{1}^{e} \frac{( ln x ) ^{n}}{x ^{2}} d x$ , $n \in N$

c) i) Calculate the value of $I_{0}$ .

(ii) Prove that $I_{n} = - \frac{1}{e} + n I_{n - 1}$ , $n \in Z^{+}$ .

(iii) Using the above, calculate the value of $I_{1}$ .

d) Calculate the volume of the solid formed when region $R$ is rotated $2 π$ about the $x$ -axis.

Show answer

Answer: a) $a = 1$ ; b) area $= \frac{1}{3}$ ; c) (i) $I_{0} = 1 - \frac{1}{e}$ ; (ii) proven; (iii) $I_{1} = 1 - \frac{2}{e}$ ; d) $V = π (24 - \frac{65}{e})$

Show working and answer

a) 1. The curve touches the $x$ -axis when $(ln x)^{2} = 0$ , i.e. $ln x = 0$ : $x = 1$ . Therefore $a = 1$ .

b) 1. The curve is above the $x$ -axis, so the area is $\int_{1}^{e} \frac{( ln x ) ^{2}}{x} d x$ . 2. Substitution $u = ln x$ , $d u = \frac{d x}{x}$ ; the limits change from $x = 1$ to $u = 0$ and from $x = e$ to $u = 1$ . 3. $\int_{0}^{1} u^{2} d u = [\frac{u ^{3}}{3}]_{0}^{1} = \frac{1}{3}$ .

c) (i) 1. $I_{0} = \int_{1}^{e} \frac{d x}{x ^{2}} = [- \frac{1}{x}]_{1}^{e} = - \frac{1}{e} + 1$ .

(ii) 1. Integrate by parts with $u = (ln x)^{n}$ and $d v = \frac{d x}{x ^{2}}$ (so $d u = \frac{n ( ln x ) ^{n - 1}}{x} d x$ and $v = - \frac{1}{x}$ ). 2. $I_{n} = [- \frac{( ln x ) ^{n}}{x}]_{1}^{e} + n \int_{1}^{e} \frac{( ln x ) ^{n - 1}}{x ^{2}} d x$ . 3. At $x = e$ the bracket equals $- \frac{1}{e}$ , and at $x = 1$ it equals $0$ (because $ln 1 = 0$ and $n \geq 1$ ). 4. Therefore $I_{n} = - \frac{1}{e} + n I_{n - 1}$ .

(iii) 1. With $n = 1$ : $I_{1} = - \frac{1}{e} + I_{0} = - \frac{1}{e} - \frac{1}{e} + 1 = 1 - \frac{2}{e}$ .

d) 1. When $R$ is rotated about the $x$ -axis: $V = π \int_{1}^{e} (\frac{( ln x ) ^{2}}{x})^{2} d x = π \int_{1}^{e} \frac{( ln x ) ^{4}}{x ^{2}} d x = π I_{4}$ . 2. Apply the recurrence formula: $I_{2} = - \frac{1}{e} + 2 I_{1} = 2 - \frac{5}{e}$ . 3. $I_{3} = - \frac{1}{e} + 3 I_{2} = 6 - \frac{16}{e}$ . 4. $I_{4} = - \frac{1}{e} + 4 I_{3} = 24 - \frac{65}{e}$ . 5. $V = π (24 - \frac{65}{e})$ .

Answer: a) $a = 1$ ; b) area $= \frac{1}{3}$ ; c) (i) $I_{0} = 1 - \frac{1}{e}$ ; (ii) proven; (iii) $I_{1} = 1 - \frac{2}{e}$ ; d) $V = π (24 - \frac{65}{e})$

Exercise 251 HardIB · P2 TZ0 November 2016

Let $f$ be the function defined by $f (x) = \frac{2 - e ^{x}}{2 e ^{x} - 1}$ , $x \in D$ .

a) Determine $D$ , the largest possible domain of $f$ .

b) Show that the graph of $f$ has three asymptotes and state their equations.

c) Show that $f^{'} (x) = - \frac{3 e ^{x}}{( 2 e ^{x} - 1 ) ^{2}}$ .

d) Use your answers to parts (b) and (c) to justify that $f$ has an inverse and state its domain.

e) Find an expression for $f^{- 1} (x)$ .

f) Consider the region $R$ bounded by the graph of $y = f (x)$ and the axes.

Find the volume of the solid of revolution obtained when $R$ is rotated $2 π$ about the $y$ -axis.

Show answer

Answer: a) $D = R - {- ln 2}$ ; b) vertical asymptote $x = - ln 2$ , horizontal asymptote $y = - \frac{1}{2}$ (from the right) and $y = - 2$ (from the left); d) $f$ is strictly decreasing and the ranges of its two branches do not overlap; $D_{f^{- 1}} =] - \infty, - 2 [\cup] - \frac{1}{2}, + \infty [$ ; e) $f^{- 1} (x) = ln (\frac{2 + x}{2 x + 1})$ ; f) $V = 0.331$

Show working and answer

a) 1. The function is not defined when the denominator is zero: $2 e^{x} - 1 = 0$ , i.e. $e^{x} = \frac{1}{2}$ and $x = - ln 2$ . 2. $D = R - {- ln 2}$ .

b) 1. At $x = - ln 2$ the denominator is zero and the numerator equals $2 - \frac{1}{2} = \frac{3}{2} \neq = 0$ : vertical asymptote $x = - ln 2$ . 2. As $x \to + \infty$ , dividing by $e^{x}$ : $f (x) = \frac{2 e ^{- x} - 1}{2 - e ^{- x}} \to \frac{- 1}{2}$ : horizontal asymptote $y = - \frac{1}{2}$ (from the right). 3. As $x \to - \infty$ , $e^{x} \to 0$ : $f \to \frac{2}{- 1} = - 2$ : horizontal asymptote $y = - 2$ (from the left). 4. There are three asymptotes.

c) 1. Quotient rule: $f^{'} (x) = \frac{( - e ^{x} ) ( 2 e ^{x} - 1 ) - ( 2 - e ^{x} ) ( 2 e ^{x} )}{( 2 e ^{x} - 1 ) ^{2}}$ . 2. Numerator: $- 2 e^{2 x} + e^{x} - 4 e^{x} + 2 e^{2 x} = - 3 e^{x}$ . 3. Therefore $f^{'} (x) = - \frac{3 e ^{x}}{( 2 e ^{x} - 1 ) ^{2}}$ .

d) 1. By c), $f^{'} (x) < 0$ on its whole domain, so $f$ is strictly decreasing on each of the two branches ( $x < - ln 2$ and $x > - ln 2$ ). 2. On the left branch, $f$ goes from $- 2$ (as $x \to - \infty$ ) down to $- \infty$ (as $x \to - ln 2^{-}$ ): it takes values in $] - \infty, - 2 [$ . 3. On the right branch, $f$ goes from $+ \infty$ (as $x \to - ln 2^{+}$ ) down to $- \frac{1}{2}$ : it takes values in $] - \frac{1}{2}, + \infty [$ . 4. The values of the two branches do not overlap, so $f$ is one-to-one on the whole of $D$ and has an inverse. 5. The domain of $f^{- 1}$ is the range of $f$ : $] - \infty, - 2 [\cup] - \frac{1}{2}, + \infty [$ .

e) 1. Write $y = \frac{2 - e ^{x}}{2 e ^{x} - 1}$ and solve for $e^{x}$ : $2 y e^{x} - y = 2 - e^{x}$ , so $e^{x} (2 y + 1) = 2 + y$ . 2. $e^{x} = \frac{2 + y}{2 y + 1}$ and $x = ln \frac{2 + y}{2 y + 1}$ . 3. Interchange the variables: $f^{- 1} (x) = ln (\frac{2 + x}{2 x + 1})$ .

f) 1. $f (0) = \frac{2 - 1}{2 - 1} = 1$ , and $f (x) = 0$ when $e^{x} = 2$ , i.e. $x = ln 2$ . The region $R$ is bounded by the curve, the $y$ -axis (from $y = 0$ to $y = 1$ ) and the $x$ -axis (from $x = 0$ to $x = ln 2$ ). 2. When rotating about the $y$ -axis, we integrate with respect to $y$ with $x = f^{- 1} (y)$ : $V = π \int_{0}^{1} [f^{- 1} (y)]^{2} d y = π \int_{0}^{1} [ln (\frac{2 + y}{2 y + 1})]^{2} d y$ . 3. Using the GDC: $V \approx 0.331$ .

Exercise 252 HardIB · P2 November 1997

(i) a) For the function $y = 2 x sin 2 x + cos 2 x$ , show that $\frac{d y}{d x} = 4 x cos 2 x$ .

b) Using the above, prove that the volume of the solid formed when the graph of $y = x cos x$ , $0 \leq x \leq \frac{π}{2}$ , is rotated $36 0^{\circ}$ about the $x$ -axis is $\frac{π}{16} (π^{2} - 4)$ .

(ii) a) The function $g$ is defined by $g (x) = x + 1 - e^{x}$ .

Calculate the maximum value of $g (x)$ , and then prove that $x + 1 \leq e^{x}$ , $x \in R$ .

b) The function $f (x)$ is defined by $f (x) = \frac{e ^{x} - 1}{x e ^{2 x}}$ , $x \neq = 0$ .

(i) Use the previous result to prove that, for $x > 0$ , $f (x) > e^{- 2 x}$ .

(ii) Show that $f (x) = \frac{1 - e ^{- x}}{x e ^{x}}$ for $x > 0$ .

(iii) Use the result of (ii)(a) to explain why $g (- x) \leq 0$ .

Next, show that $f (x) < e^{- x}$ for $x > 0$ .

(iv) Calculate $x \to 0 lim f (x)$ for positive values of $x$ .

Show answer

Answer: i) b) proven; ii) a) the maximum is $g (0) = 0$ , so $x + 1 \leq e^{x}$ ; b) (i) and (iii) proven; (iv) $x \to 0^{+} lim f (x) = 1$

Show working and answer

(i) a) 1. Differentiate using the product rule: $\frac{d y}{d x} = 2 sin 2 x + 2 x \cdot 2 cos 2 x - 2 sin 2 x = 4 x cos 2 x$ .

b) 1. When $y = x cos x$ is rotated about the $x$ -axis: $V = π \int_{0}^{π /2} x cos^{2} x d x$ . 2. Since $cos^{2} x = \frac{1 + cos 2 x}{2}$ : $V = \frac{π}{2} \int_{0}^{π /2} x d x + \frac{π}{2} \int_{0}^{π /2} x cos 2 x d x$ . 3. The first integral is $[\frac{x ^{2}}{2}]_{0}^{π /2} = \frac{π ^{2}}{8}$ . 4. By part a), $x cos 2 x = \frac{1}{4} \frac{d y}{d x}$ , so $\int_{0}^{π /2} x cos 2 x d x = \frac{1}{4} [2 x sin 2 x + cos 2 x]_{0}^{π /2} = \frac{1}{4} [(0 - 1) - (0 + 1)] = - \frac{1}{2}$ . 5. $V = \frac{π}{2} \cdot \frac{π ^{2}}{8} + \frac{π}{2} \cdot (- \frac{1}{2}) = \frac{π ^{3}}{16} - \frac{π}{4} = \frac{π}{16} (π^{2} - 4)$ .

(ii) a) 1. $g^{'} (x) = 1 - e^{x}$ , which is zero at $x = 0$ . Also $g^{''} (x) = - e^{x} < 0$ , so this is a maximum. 2. The maximum value is $g (0) = 0 + 1 - 1 = 0$ . 3. Since $g (x) \leq 0$ for all $x$ , we have $x + 1 - e^{x} \leq 0$ , i.e. $x + 1 \leq e^{x}$ . Equality holds only at $x = 0$ .

b) (i) 1. Split: $f (x) = \frac{e ^{x} - 1}{x} \cdot e^{- 2 x}$ . 2. For $x > 0$ , the result of a) gives $e^{x} - 1 > x$ (strict inequality, because $x \neq = 0$ ), so $\frac{e ^{x} - 1}{x} > 1$ . 3. Multiply by $e^{- 2 x} > 0$ : $f (x) > e^{- 2 x}$ .

(ii) 1. Divide the numerator and denominator of $f (x)$ by $e^{x}$ : $f (x) = \frac{1 - e ^{- x}}{x e ^{x}}$ .

(iii) 1. The result of ii) a) holds for all real $x$ , and in particular for $- x$ : $g (- x) = - x + 1 - e^{- x} \leq 0$ . 2. That is, $1 - e^{- x} \leq x$ , with strict inequality if $x > 0$ (the maximum of $g$ is reached only at $0$ ). 3. For $x > 0$ , using the form in ii): $f (x) = \frac{1 - e ^{- x}}{x e ^{x}} < \frac{x}{x e ^{x}} = e^{- x}$ .

(iv) 1. For $x > 0$ , by i) and iii): $e^{- 2 x} < f (x) < e^{- x}$ . 2. As $x \to 0^{+}$ , both $e^{- 2 x}$ and $e^{- x}$ tend to $1$ . By the squeeze theorem, $x \to 0^{+} lim f (x) = 1$ .

Answer: i) b) proven; ii) a) the maximum is $g (0) = 0$ , so $x + 1 \leq e^{x}$ ; b) (i) and (iii) proven; (iv) $x \to 0^{+} lim f (x) = 1$

Exercise 253 HardIB · P2 TZ0 November 2006

The function $f$ is defined by $f (x) = \frac{ln x}{x ^{3}}$ , $x \geq 1$ .

a) Find $f^{'} (x)$ and $f^{''} (x)$ , simplifying both answers.

b) i) Find the exact value of the $x$ -coordinate of the maximum point, and justify why it is a maximum.

(ii) Solve $f^{''} (x) = 0$ , and verify that for this value of $x$ , the graph of $f$ has a point of inflexion.

(iii) Sketch the graph of $f$ approximately, indicating the maximum point and the point of inflexion.

Let $R$ be the region bounded by the $x$ -axis, the graph of $f$ and the line $x = 3$ .

c) Find the volume of the solid of revolution formed when $R$ is rotated through $36 0^{\circ}$ about the $x$ -axis.

d) Verify that the area of $R$ is equal to $\frac{1}{18} (4 - ln 3)$ .

View solution

a) 1. Quotient rule with $u = ln x$ , $v = x^{3}$ : $f^{'} (x) = \frac{\frac{1}{x} \cdot x ^{3} - 3 x ^{2} ln x}{x ^{6}} = \frac{x ^{2} ( 1 - 3 ln x )}{x ^{6}} = \frac{1 - 3 ln x}{x ^{4}}$ . 2. Second derivative: $f^{''} (x) = \frac{- \frac{3}{x} \cdot x ^{4} - 4 x ^{3} ( 1 - 3 ln x )}{x ^{8}} = \frac{- 3 x ^{3} - 4 x ^{3} + 12 x ^{3} ln x}{x ^{8}} = \frac{- 7 + 12 ln x}{x ^{5}}$ .

b) i) 1. $f^{'} (x) = 0$ if $1 - 3 ln x = 0$ , i.e. $x = e^{1/3}$ . 2. $f^{''} (e^{1/3}) = \frac{- 7 + 4}{e ^{5/3}} < 0$ , so it is a maximum.

(ii) 1. $f^{''} (x) = 0$ if $ln x = \frac{7}{12}$ , that is, $x = e^{7/12}$ . 2. For $x < e^{7/12}$ , $- 7 + 12 ln x < 0$ and $f^{''} < 0$ ; for $x > e^{7/12}$ , $f^{''} > 0$ . It changes sign, so it is a point of inflexion.

(iii) Key features: $f (1) = 0$ ; it rises to the maximum at $x = e^{1/3} \approx 1.40$ , with $f \approx 0.13$ ; then it falls, with an inflexion at $x = e^{7/12} \approx 1.79$ , and tends to $0$ as $x \to \infty$ (the $x$ -axis is a horizontal asymptote).

Approximate graph of f(x) = ln x / x³ for x ≥ 1, with the maximum point and the point of inflexion

c) 1. Volume of revolution about the $x$ -axis between $x = 1$ and $x = 3$ : $V = π \int_{1}^{3} (\frac{ln x}{x ^{3}})^{2} d x = π \int_{1}^{3} \frac{( ln x ) ^{2}}{x ^{6}} d x$ . 2. Using the GDC: $V \approx 0.0458 \approx 0.046$ .

d) 1. Area $= \int_{1}^{3} \frac{ln x}{x ^{3}} d x$ ( $f \geq 0$ ). Integrating by parts with $u = ln x$ and $d v = x^{- 3} d x$ ( $d u = \frac{1}{x} d x$ , $v = - \frac{1}{2 x ^{2}}$ ). 2. $\int \frac{ln x}{x ^{3}} d x = - \frac{ln x}{2 x ^{2}} + \frac{1}{2} \int x^{- 3} d x = - \frac{ln x}{2 x ^{2}} - \frac{1}{4 x ^{2}}$ . 3. We evaluate: $(- \frac{ln 3}{18} - \frac{1}{36}) - (0 - \frac{1}{4}) = \frac{8}{36} - \frac{ln 3}{18} = \frac{4}{18} - \frac{ln 3}{18} = \frac{1}{18} (4 - ln 3)$ .

Answer: a) $f^{'} (x) = \frac{1 - 3 ln x}{x ^{4}}$ , $f^{''} (x) = \frac{- 7 + 12 ln x}{x ^{5}}$ ; b) i) $x = e^{1/3}$ , a maximum because $f^{''} (e^{1/3}) < 0$ ; ii) $x = e^{7/12}$ , point of inflexion; c) $V \approx 0.046$ ; d) shown.

Exercise 254 HardIB · P1 TZ1 May 2007

a) Using the formula for $cos (A + B)$ , show that $cos^{2} θ = \frac{cos 2 θ + 1}{2}$ .

b) Hence, find $\int cos^{2} x d x$ .

Let $f (x) = 4 cos x$ and $g (x) = sec x$ for $x \in] - \frac{π}{2}, \frac{π}{2} [$ .

Let $R$ be the region enclosed by the two functions.

c) Find the exact values of the $x$ -coordinates of the points of intersection.

d) Sketch the functions $f$ and $g$ , and clearly shade the region $R$ .

The region $R$ is rotated through $2 π$ about the $x$ -axis to generate a solid.

e) i) Write down an integral that represents the volume of this solid.

(ii) Hence, find the exact value of the volume.

View solution

a) 1. Using $cos (A + B) = cos A cos B - sin A sin B$ with $A = B = θ$ : $cos 2 θ = cos^{2} θ - sin^{2} θ$ . 2. As $sin^{2} θ = 1 - cos^{2} θ$ : $cos 2 θ = 2 cos^{2} θ - 1$ . 3. Rearranging: $cos^{2} θ = \frac{cos 2 θ + 1}{2}$ .

b) 1. By (a): $\int cos^{2} x d x = \frac{1}{2} \int (cos 2 x + 1) d x$ . 2. $= \frac{1}{2} (\frac{sin 2 x}{2} + x) + C$ .

c) 1. Equating: $4 cos x = sec x = \frac{1}{cos x}$ , so $4 cos^{2} x = 1$ and $cos x = \pm \frac{1}{2}$ . 2. On $] - \frac{π}{2}, \frac{π}{2} [$ cosine is positive, so $cos x = \frac{1}{2}$ and $x = \pm \frac{π}{3}$ .

d) Key features: $f$ is an arc with a maximum of $4$ at $x = 0$ that is zero at $\pm \frac{π}{2}$ ; $g$ has a minimum of $1$ at $x = 0$ and vertical asymptotes at $x = \pm \frac{π}{2}$ . They intersect at $x = \pm \frac{π}{3}$ (with $y = 2$ ), and $R$ is the region between both curves, with $f$ above $g$ , for $- \frac{π}{3} \leq x \leq \frac{π}{3}$ .

Graphs of f(x) = 4 cos x and g(x) = sec x on the interval (−π/2, π/2), with the region R between them shaded, lying between x = −π/3 and x = π/3

e) i) 1. Between $- \frac{π}{3}$ and $\frac{π}{3}$ , $f \geq g \geq 0$ , so the volume is the difference of the two solids: $V = π \int_{- π /3}^{π /3} (f^{2} - g^{2}) d x$ . 2. By symmetry (the integrand is even): $V = 2 π \int_{0}^{\frac{π}{3}} (16 cos^{2} x - sec^{2} x) d x$ .

(ii) 1. Using (b): $\int 16 cos^{2} x d x = 16 \cdot \frac{1}{2} (\frac{sin 2 x}{2} + x) = 4 sin 2 x + 8 x$ , and $\int sec^{2} x d x = tan x$ . 2. $V = 2 π [4 sin 2 x + 8 x - tan x]_{0}^{\frac{π}{3}}$ . 3. At $x = \frac{π}{3}$ : $4 \cdot \frac{3}{2} + \frac{8 π}{3} - 3 = 3 + \frac{8 π}{3}$ ; at $x = 0$ it equals $0$ . 4. $V = 2 π (3 + \frac{8 π}{3})$ .

Answer: a) shown; b) $\frac{1}{2} (\frac{sin 2 x}{2} + x) + C$ ; c) $x = \frac{π}{3}$ and $x = - \frac{π}{3}$ ; d) sketch of $f$ and $g$ with $R$ shaded; e) i) $2 π \int_{0}^{\frac{π}{3}} (16 cos^{2} x - sec^{2} x) d x$ ; ii) $2 π (3 + \frac{8 π}{3})$

Exercise 255 HardIB · P2 TZ1 May 2011

An open vessel is formed by rotating the curve $y = x^{2}$ , defined on the domain $x \in [0, 10]$ , through $2 π$ radians about the $y$ -axis. The units on the coordinate axes are defined in centimetres.

a) When the vessel contains water to a height of $h$ cm, find the volume $V$ of water in terms of $h$ .

b) If the water in the vessel evaporates at a rate of $3 c m^{3}$ per hour for each $c m^{2}$ of exposed surface area of the water, show that $\frac{d V}{d t} = - 3 2 π V$ , where $t$ is measured in hours.

c) If the vessel is completely full, how long will it take for all the water to evaporate?

Show answer

Answer: a) $V = \frac{π h ^{2}}{2}$ cm $^{3}$ ; b) shown; c) $\frac{100}{3} \approx 33.3$ hours

Show working and answer

a) 1. When $y = x^{2}$ is rotated about the $y$ -axis, the volume up to height $h$ is $V = π \int_{0}^{h} x^{2} d y$ , with $x^{2} = y$ . 2. $V = π \int_{0}^{h} y d y = π \frac{h ^{2}}{2}$ cm $^{3}$ .

b) 1. The surface of the water is a circle of radius $x$ with $x^{2} = h$ : area $= π h$ cm $^{2}$ . 2. The water evaporates at $3$ cm $^{3}$ per hour per cm $^{2}$ : $\frac{d V}{d t} = - 3 π h$ . 3. From $V = \frac{π h ^{2}}{2}$ : $h = \frac{2 V}{π}$ ( $h > 0$ ). 4. $\frac{d V}{d t} = - 3 π \frac{2 V}{π} = - 3 π^{2} \cdot \frac{2 V}{π} = - 3 2 π V$ .

c) 1. Full vessel: $h = 100$ cm (since $x = 10$ and $y = x^{2} = 100$ ), so $V_{0} = \frac{π \cdot 10 000}{2} = 5000 π$ cm $^{3}$ . 2. Separate the variables: $\frac{d V}{V} = - 3 2 π d t$ , and on integrating, $2 V = - 3 2 π t + C$ . 3. With $V = 5000 π$ at $t = 0$ : $C = 2 5000 π$ . 4. $V = 0$ when $3 2 π t = 2 5000 π$ , i.e. $t = \frac{2 2500}{3} = \frac{100}{3} \approx 33.3$ hours.

Answer: a) $V = \frac{π h ^{2}}{2}$ cm $^{3}$ ; b) shown; c) $\frac{100}{3} \approx 33.3$ hours

Exercise 256 HardIB · P2 TZ0 November 2010

Let $f (x) = \frac{a + b e ^{x}}{a e ^{x} + b}$ , where $0 < b < a$ .

a) Verify that $f^{'} (x) = \frac{( b ^{2} - a ^{2} ) e ^{x}}{( a e ^{x} + b ) ^{2}}$ .

b) Hence, justify why the graph of $f$ has no local maximum and no local minimum.

c) Given that the graph of $f$ has a point of inflexion, find its coordinates.

d) Verify that the graph of $f$ has exactly two asymptotes.

e) Let $a = 4$ and $b = 1$ . Consider the region $R$ bounded by the graph of $y = f (x)$ , the $y$ -axis and the line with equation $y = \frac{1}{2}$ . Find the volume $V$ of the solid obtained when $R$ is rotated $2 π$ about the $x$ -axis.

Show answer

Answer: a) shown; b) $f^{'} (x) < 0$ for all $x$ ; c) $(ln \frac{b}{a}, \frac{a ^{2} + b ^{2}}{2 ab})$ ; d) $y = \frac{b}{a}$ and $y = \frac{a}{b}$ , no vertical asymptotes; e) $V \approx 1.09$

Show working and answer

a) 1. Quotient rule with $u = a + b e^{x}$ and $v = a e^{x} + b$ : $f^{'} (x) = \frac{b e ^{x} ( a e ^{x} + b ) - ( a + b e ^{x} ) a e ^{x}}{( a e ^{x} + b ) ^{2}}$ . 2. Numerator: $e^{x} [ab e^{x} + b^{2} - a^{2} - ab e^{x}] = (b^{2} - a^{2}) e^{x}$ . 3. Therefore $f^{'} (x) = \frac{( b ^{2} - a ^{2} ) e ^{x}}{( a e ^{x} + b ) ^{2}}$ .

b) 1. As $0 < b < a$ , $b^{2} - a^{2} < 0$ ; moreover $e^{x} > 0$ and the denominator is positive. 2. Hence $f^{'} (x) < 0$ for all $x$ : $f$ is always strictly decreasing and $f^{'}$ is never zero, so there are no local maxima or minima.

c) 1. We differentiate $f^{'} (x) = (b^{2} - a^{2}) e^{x} (a e^{x} + b)^{- 2}$ : $f^{''} (x) = (b^{2} - a^{2}) [e^{x} (a e^{x} + b)^{- 2} - 2 a e^{2 x} (a e^{x} + b)^{- 3}]$ . 2. Common factor: $f^{''} (x) = \frac{( b ^{2} - a ^{2} ) e ^{x} ( b - a e ^{x} )}{( a e ^{x} + b ) ^{3}}$ . 3. $f^{''} (x) = 0$ if $a e^{x} = b$ , i.e. $x = ln \frac{b}{a}$ . The factor $b - a e^{x}$ changes sign there, so it is a point of inflexion. 4. $f (ln \frac{b}{a}) = \frac{a + b \cdot \frac{b}{a}}{a \cdot \frac{b}{a} + b} = \frac{\frac{a ^{2} + b ^{2}}{a}}{2 b} = \frac{a ^{2} + b ^{2}}{2 ab}$ . 5. Point of inflexion: $(ln \frac{b}{a}, \frac{a ^{2} + b ^{2}}{2 ab})$ .

d) 1. Vertical asymptotes: the denominator $a e^{x} + b > 0$ is never zero, so there are none. 2. As $x \to + \infty$ , dividing by $e^{x}$ : $f (x) = \frac{a e ^{- x} + b}{a + b e ^{- x}} \to \frac{b}{a}$ . 3. As $x \to - \infty$ , $e^{x} \to 0$ and $f (x) \to \frac{a}{b}$ . 4. There are exactly two horizontal asymptotes, $y = \frac{b}{a}$ and $y = \frac{a}{b}$ (distinct because $a \neq = b$ ).

e) 1. With $a = 4$ , $b = 1$ : $f (x) = \frac{4 + e ^{x}}{4 e ^{x} + 1}$ , with $f (0) = 1$ , decreasing towards $\frac{1}{4}$ . 2. $f (x) = \frac{1}{2}$ when $2 (4 + e^{x}) = 4 e^{x} + 1$ , i.e. $8 + 2 e^{x} = 4 e^{x} + 1$ , so $e^{x} = \frac{7}{2}$ and $x = ln 3.5 \approx 1.25$ . 3. $R$ lies between $y = f (x)$ (above) and $y = \frac{1}{2}$ (below), from $x = 0$ to $x = ln 3.5$ . 4. Using the GDC: $V = π \int_{0}^{l n 3.5} [(\frac{4 + e ^{x}}{4 e ^{x} + 1})^{2} - \frac{1}{4}] d x \approx 1.09$ .

Answer: a) shown; b) $f^{'} (x) < 0$ for all $x$ ; c) $(ln \frac{b}{a}, \frac{a ^{2} + b ^{2}}{2 ab})$ ; d) $y = \frac{b}{a}$ and $y = \frac{a}{b}$ , no vertical asymptotes; e) $V \approx 1.09$

Exercise 257 HardIB · P2 TZ0 November 2014

The following figure shows the vertical cross-section of a container.

Vertical cross-section of a container: sides in the form of parabolic arcs that widen upwards, base 16 wide on the x-axis and top edge at height 48

The sides of this cross-section are given by the equation $y = 0.25 x^{2} - 16$ . The horizontal cross-sections are circular. The container has a height of $48$ cm.

a) If the container is filled with water to a height of $h$ cm, show that the volume of water, $V$ $c m^{3}$ , is given by $V = 4 π (\frac{h ^{2}}{2} + 16 h)$ .

The container, which is initially full of water, starts to drip through a small hole at a rate given by $\frac{d V}{d t} = - \frac{250 h}{π ( h + 16 )}$ , where $t$ is measured in seconds.

b) (i) Show that $\frac{d h}{d t} = - \frac{250 h}{4 π ^{2} ( h + 16 ) ^{2}}$ .

(ii) State $\frac{d t}{d h}$ and, hence, show that $t = \frac{- 4 π ^{2}}{250} \int (h^{\frac{3}{2}} + 32 h^{\frac{1}{2}} + 256 h^{- \frac{1}{2}}) d h$ .

(iii) Find the time taken for the container to empty, correct to the nearest minute. ( $60$ seconds $= 1$ minute)

Once it is empty, water is poured back into the container at a rate of $8.5 c m^{3} s^{- 1}$ .

At the same time, the container continues to drip at a rate of $\frac{250 h}{π ( h + 16 )} c m^{3} s^{- 1}$ .

c) Using an appropriate approximate graph, determine the height at which the water level eventually stabilises in the container.

Show answer

Answer: b) (iii) $2688.8 s \approx 45$ min; c) $h = 5.06$ cm

Show working and answer

a) 1. The horizontal sections are circles. At height $y$ , the radius $x$ satisfies $y = 0.25 x^{2} - 16$ , so $x^{2} = 4 (y + 16)$ . 2. Volume by discs about the $y$ -axis: $V = π \int_{0}^{h} x^{2} d y = π \int_{0}^{h} 4 (y + 16) d y$ . 3. $V = 4 π [\frac{y ^{2}}{2} + 16 y]_{0}^{h} = 4 π (\frac{h ^{2}}{2} + 16 h)$ .

b) (i) 1. Chain rule: $\frac{d h}{d t} = \frac{d V}{d t} / \frac{d V}{d h}$ . 2. $\frac{d V}{d h} = 4 π (h + 16)$ . 3. $\frac{d h}{d t} = - \frac{250 h}{π ( h + 16 )} \cdot \frac{1}{4 π ( h + 16 )} = - \frac{250 h}{4 π ^{2} ( h + 16 ) ^{2}}$ .

(ii) 1. $\frac{d t}{d h} = \frac{1}{d h / d t} = - \frac{4 π ^{2} ( h + 16 ) ^{2}}{250 h}$ . 2. We expand $(h + 16)^{2} = h^{2} + 32 h + 256$ and divide by $h$ : $\frac{( h + 16 ) ^{2}}{h} = h^{3/2} + 32 h^{1/2} + 256 h^{- 1/2}$ . 3. Integrating with respect to $h$ : $t = - \frac{4 π ^{2}}{250} \int (h^{3/2} + 32 h^{1/2} + 256 h^{- 1/2}) d h$ .

(iii) 1. The container goes from $h = 48$ to $h = 0$ : $t = \frac{4 π ^{2}}{250} \int_{0}^{48} (h^{3/2} + 32 h^{1/2} + 256 h^{- 1/2}) d h$ (the minus sign cancels when the limits are reversed). 2. An antiderivative is $\frac{2}{5} h^{5/2} + \frac{64}{3} h^{3/2} + 512 h^{1/2}$ . Using the GDC (or by hand) between $0$ and $48$ , the integral multiplied by $\frac{4 π ^{2}}{250}$ gives $t \approx 2688.8$ s. 3. $\frac{2688.8}{60} \approx 44.8$ , which is $45$ minutes.

c) 1. The level stabilises when the inflow equals the outflow: $8.5 = \frac{250 h}{π ( h + 16 )}$ . 2. We draw on the GDC the curve $y = \frac{250 h}{π ( h + 16 )}$ and the line $y = 8.5$ for $0 \leq h \leq 48$ . They meet at a single point. 3. For $h$ less than this value the outflow is less than $8.5$ and the level rises; for $h$ greater, the outflow is greater and the level falls: this is the equilibrium level. 4. The intersection is $h \approx 5.06$ cm.

Answer: b) (iii) $2688.8 s \approx 45$ min; c) $h = 5.06$ cm

Exercise 258 HardIB · P1 TZ2 May 2016

The following graph shows the relation $x = 3 cos 2 y + 4$ , $0 \leq y \leq π$ .

Graph of x = 3cos2y + 4 for 0 ≤ y ≤ π, with y on the vertical axis: a sideways S-shaped curve running from (7, 0) to (7, π), passing through x = 1 at y = π/2

The curve is rotated through $36 0^{\circ}$ about the $y$ -axis to generate the volume of revolution, $V$ .

a) Calculate the value of $V$ .

A container is manufactured in this shape, with a volume equal to $V$ $c m^{3}$ and a solid base of diameter $14$ cm. The container is being filled with water at a rate of $2 c m^{3} mi n^{- 1}$ . The depth of the water is $h$ cm, $0 \leq h \leq π$ .

b) (i) Given that $\frac{d V}{d h} = π (3 cos 2 h + 4)^{2}$ , find an expression for $\frac{d h}{d t}$ .

(ii) Find the value of $\frac{d h}{d t}$ when $h = \frac{π}{4}$ .

c) (i) Find $\frac{d ^{2} h}{d t ^{2}}$ .

(ii) Find the values of $h$ for which $\frac{d ^{2} h}{d t ^{2}} = 0$ .

(iii) With explicit reference to the shape of the container, interpret $\frac{d h}{d t}$ at the values of $h$ you found in part (c)(ii).

Show answer

Answer: a) $V = \frac{41 π ^{2}}{2} c m^{3}$ ; b) (i) $\frac{d h}{d t} = \frac{2}{π ( 3 cos 2 h + 4 ) ^{2}}$ ; (ii) $\frac{1}{8 π}$ ; c) (i) $\frac{48 sin 2 h}{π ^{2} ( 3 cos 2 h + 4 ) ^{5}}$ ; (ii) $h = 0, \frac{π}{2}, π$

Show working and answer

a) 1. Rotation about the $y$ -axis: $V = π \int_{0}^{π} x^{2} d y = π \int_{0}^{π} (3 cos 2 y + 4)^{2} d y$ . 2. We expand: $(3 cos 2 y + 4)^{2} = 9 cos^{2} 2 y + 24 cos 2 y + 16 = \frac{9}{2} (1 + cos 4 y) + 24 cos 2 y + 16$ . 3. Between $0$ and $π$ , the integrals of $cos 4 y$ and $cos 2 y$ are $0$ . This leaves $\int_{0}^{π} (\frac{9}{2} + 16) d y = \frac{41}{2} π$ . 4. $V = π \cdot \frac{41 π}{2} = \frac{41 π ^{2}}{2} c m^{3} \approx 202.3 c m^{3}$ .

b) (i) 1. $\frac{d h}{d t} = \frac{d V}{d t} / \frac{d V}{d h} = \frac{2}{π ( 3 cos 2 h + 4 ) ^{2}}$ .

(ii) 1. At $h = \frac{π}{4}$ : $cos \frac{π}{2} = 0$ , so $\frac{d h}{d t} = \frac{2}{π \cdot 16} = \frac{1}{8 π}$ cm/min.

c) (i) 1. Differentiating with respect to $t$ using the chain rule: $\frac{d ^{2} h}{d t ^{2}} = \frac{d}{d h} (\frac{d h}{d t}) \cdot \frac{d h}{d t}$ . 2. $\frac{d}{d h} [\frac{2}{π} (3 cos 2 h + 4)^{- 2}] = \frac{2}{π} (- 2) (3 cos 2 h + 4)^{- 3} (- 6 sin 2 h) = \frac{24 sin 2 h}{π ( 3 cos 2 h + 4 ) ^{3}}$ . 3. Multiplying by $\frac{2}{π ( 3 cos 2 h + 4 ) ^{2}}$ : $\frac{d ^{2} h}{d t ^{2}} = \frac{48 sin 2 h}{π ^{2} ( 3 cos 2 h + 4 ) ^{5}}$ .

(ii) 1. The denominator is positive ( $3 cos 2 h + 4 \geq 1$ ), so it is zero when $sin 2 h = 0$ . 2. With $0 \leq h \leq π$ : $2 h = 0, π, 2 π$ , that is, $h = 0, \frac{π}{2}, π$ .

(iii) 1. The radius of the container at height $h$ is $x = 3 cos 2 h + 4$ : it equals $7$ at $h = 0$ and $h = π$ (the widest parts) and $1$ at $h = \frac{π}{2}$ (the narrowest part). 2. $\frac{d ^{2} h}{d t ^{2}}$ is positive between $0$ and $\frac{π}{2}$ and negative between $\frac{π}{2}$ and $π$ : the rate of rise $\frac{d h}{d t}$ increases while the container narrows and decreases while it widens. 3. At $h = \frac{π}{2}$ the rate of rise is maximum (the container is at its narrowest and the water rises fastest); at $h = 0$ and $h = π$ it is minimum (the widest parts, where the water rises slowest).

Exercise 259 HardIB · P2 TZ2 May 2021

The function $f$ is given by $f (x) = \frac{k e ^{\frac{x}{2}}}{1 + e ^{x}}$ , where $x \in R$ , $x \geq 0$ , $k \in R^{+}$ .

The region bounded by the graph of $y = f (x)$ , the $x$ -axis, the $y$ -axis and the line $x = ln 16$ is rotated through $36 0^{\circ}$ about the $x$ -axis to generate a solid of revolution.

a) Show that the volume of the solid formed is equal to $\frac{15 k ^{2} π}{34}$ cubic units.

Pedro wants to make a small bowl with a volume of $300 c m^{3}$ , based on the result in part (a). Pedro's design is shown in the following diagrams.

Graph of y = f(x) decreasing for 0 ≤ x ≤ ln 16: A is the point of the curve on the y-axis, C the right-hand end and B the point on the x-axis below C

Bowl, a solid of revolution narrowing downwards: circular top edge with centre O and A at the end of the radius, base with centre B and C at the end of the radius, height ln 16

The vertical height of the bowl ( $BO$ ) is measured along the $x$ -axis. The radius of the top of the bowl is $OA$ and the radius of the base of the bowl is $BC$ . All lengths are measured in cm.

b) Find the value of $k$ that satisfies the requirements of Pedro's design.

c) Find:

(i) $OA$

(ii) $BC$

For the design, Pedro decides to investigate how the radius of a cross-section of the bowl changes.

d) (i) By sketching an approximate graph of a suitable derivative of $f$ , find where the radius of the cross-section of the bowl decreases most rapidly.

(ii) State the radius of the cross-section of the bowl at that point.

Show answer

Answer: b) $k = 2 \frac{170}{π} = 14.7$ ; c) (i) $OA = 7.36$ cm; (ii) $BC = 3.46$ cm; d) (i) $x = 1.76$ cm, with $f^{'} (x) = - 1.84$ ; (ii) $f (1.76) = 5.20$ cm

Show working and answer

a) 1. Volume of revolution about the $x$ -axis: $V = π \int_{0}^{l n 16} [f (x)]^{2} d x = π k^{2} \int_{0}^{l n 16} \frac{e ^{x}}{( 1 + e ^{x} ) ^{2}} d x$ . 2. Technique: substitution $u = 1 + e^{x}$ , $d u = e^{x} d x$ . Limits: $x = 0 \Rightarrow u = 2$ ; $x = ln 16 \Rightarrow u = 17$ . 3. $\int_{2}^{17} \frac{d u}{u ^{2}} = [- \frac{1}{u}]_{2}^{17} = \frac{1}{2} - \frac{1}{17} = \frac{15}{34}$ . 4. $V = \frac{15 k ^{2} π}{34}$ .

b) 1. We impose $\frac{15 k ^{2} π}{34} = 300$ , so $k^{2} = \frac{10200}{15 π} = \frac{680}{π}$ . 2. Since $k > 0$ , $k = \frac{680}{π} = 2 \frac{170}{π} \approx 14.7$ .

c) (i) 1. $OA = f (0) = \frac{k}{1 + 1} = \frac{k}{2} \approx 7.36$ cm.

(ii) 1. $BC = f (ln 16) = \frac{k \cdot e ^{(l n 16) /2}}{1 + 16} = \frac{4 k}{17} \approx 3.46$ cm.

d) (i) 1. The radius of the cross-section is $f (x)$ ; it decreases most rapidly where $f^{'} (x)$ is as negative as possible. 2. We differentiate: $f^{'} (x) = \frac{k e ^{x /2} ( 1 - e ^{x} )}{2 ( 1 + e ^{x} ) ^{2}}$ . It is $0$ at $x = 0$ and negative for $x > 0$ . 3. We draw $y = f^{'} (x)$ using the GDC on $[0, ln 16]$ : it starts at $0$ , falls to a minimum and rises a little towards the end. 4. The minimum of $f^{'}$ is at $x \approx 1.76$ cm, with $f^{'} \approx - 1.84$ .

(ii) 1. Radius at that point: $f (1.76) \approx 5.20$ cm.

Answer: b) $k = 2 \frac{170}{π} = 14.7$ ; c) (i) $OA = 7.36$ cm; (ii) $BC = 3.46$ cm; d) (i) $x = 1.76$ cm, with $f^{'} (x) = - 1.84$ ; (ii) $f (1.76) = 5.20$ cm

Exercise 260 HardIB · P2 TZ1 May 2022

Consider the function $f (x) = x^{2} - 1$ , where $1 \leq x \leq 2$ .

a) Sketch the curve $y = f (x)$ , clearly indicating the coordinates of the end points.

b) (i) Show that the inverse function of $f$ is given by $f^{- 1} (x) = x^{2} + 1$ .

(ii) State the domain and range of $f^{- 1}$ .

The curve $y = f (x)$ is rotated through $2 π$ about the $y$ -axis to form a solid of revolution which is used to model a water container.

c) (i) Show that the volume, $V$ $m^{3}$ , of water in the container when it is filled to a height of $h$ metres is given by $V = π (\frac{1}{3} h^{3} + h)$ .

(ii) Using the above, determine the maximum volume of the container.

When $t = 0$ , the container is empty. Water is then added to the container at a constant rate of $0.4 m^{3} s^{- 1}$ .

d) Find the time taken to fill the container to its maximum volume.

e) Find the rate of change of the height of the water when the container is half full.

View solution

a) 1. End points: $f (1) = 0$ and $f (2) = 3 \approx 1.73$ . The function is increasing and concave, with a vertical tangent at $(1, 0)$ (because $f^{'} = \frac{x}{x ^{2} - 1} \to \infty$ ). 2. It is the arc of the curve from $(1, 0)$ to $(2, 3)$ .

Graph of y = √(x² − 1) for 1 ≤ x ≤ 2, with end points (1, 0) and (2; 1.73)

b) (i) 1. We write $y = x^{2} - 1$ and interchange $x$ and $y$ : $x = y^{2} - 1$ . 2. Squaring: $x^{2} = y^{2} - 1$ , so $y^{2} = x^{2} + 1$ . 3. As the range of $f^{- 1}$ is $[1, 2]$ (positive), $y = + x^{2} + 1$ , that is, $f^{- 1} (x) = x^{2} + 1$ .

(ii) 1. Domain of $f^{- 1}$ = range of $f$ = $[0, 3]$ . 2. Range of $f^{- 1}$ = domain of $f$ = $[1, 2]$ .

c) (i) 1. When $y = f (x)$ is rotated about the $y$ -axis, the radius at height $y$ is $x = f^{- 1} (y)$ , with $x^{2} = y^{2} + 1$ . 2. $V = π \int_{0}^{h} x^{2} d y = π \int_{0}^{h} (y^{2} + 1) d y = π (\frac{h ^{3}}{3} + h)$ .

(ii) 1. The maximum height is $3$ (the range of $f$ ). 2. $V_{max} = π (\frac{3 3}{3} + 3) = 2 π 3 \approx 10.9 m^{3}$ .

d) 1. $t = \frac{V _{max}}{0.4} = \frac{2 π 3}{0.4} \approx 27.2$ s.

e) 1. Half the volume is $π 3$ : $π (\frac{h ^{3}}{3} + h) = π 3$ , that is, $\frac{h ^{3}}{3} + h = 3$ . Using the GDC: $h \approx 1.18$ m. 2. Chain rule: $\frac{d V}{d t} = \frac{d V}{d h} \frac{d h}{d t}$ , with $\frac{d V}{d h} = π (h^{2} + 1)$ and $\frac{d V}{d t} = 0.4$ . 3. $\frac{d h}{d t} = \frac{0.4}{π ( h ^{2} + 1 )} = \frac{0.4}{π ( 1.181 8 ^{2} + 1 )} \approx 0.0531 m s^{- 1}$ .

Answer: b) (ii) $D_{f^{- 1}} = [0, 3]$ , $R_{f^{- 1}} = [1, 2]$ ; c) (ii) $V = 2 π 3 = 10.9 m^{3}$ ; d) $t = 27.2$ s; e) $0.0531 m s^{- 1}$

Exercise 261 HardIB · P2 TZ2 May 2022

The following figure shows the curve $\frac{x ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1$ , where $h \leq y \leq 4$ .

Fragment of the ellipse x²/36 + (y−4)²/16 = 1 running from Q to B, with the rectangle OPQR of height h below Q and the dashed segment from O to Q

The fragment of the curve from point $Q$ to point $B$ is rotated through $36 0^{\circ}$ about the $y$ -axis to generate the inner surface of a bowl. The rectangle $OPQR$ , of height $h$ cm, is rotated through $36 0^{\circ}$ about the $y$ -axis to generate a solid base. The bowl is assumed to have negligible thickness.

Given that the internal volume of the bowl must be equal to $285 c m^{3}$ , find the height of the base.

Show answer

Answer: $h = 0.793$ cm

Show working and answer

The ellipse is $\frac{x ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1$ . Solving for $x^{2}$ : $x^{2} = 36 (1 - \frac{( y - 4 ) ^{2}}{16}) = 36 - \frac{9}{4} (y - 4)^{2}$ .
The internal volume of the bowl (without the base) is obtained by rotating the region between the curve and the $y$ -axis from $y = h$ to $y = 4$ : $V = π \int_{h}^{4} x^{2} d y$ .
$V = π \int_{h}^{4} [36 - \frac{9}{4} (y - 4)^{2}] d y = π [36 y - \frac{3}{4} (y - 4)^{3}]_{h}^{4}$ .
$V = π [144 - (36 h - \frac{3}{4} (h - 4)^{3})] = π [36 (4 - h) - \frac{3}{4} (4 - h)^{3}]$ .
Setting $V = 285$ and solving with the GDC for $0 < h < 4$ : $π [36 (4 - h) - \frac{3}{4} (4 - h)^{3}] = 285$ gives $h \approx 0.793$ cm.

Answer: $h = 0.793$ cm

Improper integrals

Exercise 262 BasicIB · P3 TZ0 November 2012

a) Show that $H \to \infty lim \int_{a}^{H} \frac{1}{x ^{2}} d x$ exists and calculate its value in terms of $a$ (where $a \in R^{+}$ ).

Show answer

Answer: $\frac{1}{a}$

Show working and answer

We first calculate the integral with a finite upper limit $H > a$ : $\int_{a}^{H} \frac{1}{x ^{2}} d x = [- \frac{1}{x}]_{a}^{H} = - \frac{1}{H} + \frac{1}{a}$ .
As $H \to \infty$ , $\frac{1}{H} \to 0$ , so the limit exists.
$H \to \infty lim \int_{a}^{H} \frac{1}{x ^{2}} d x = \frac{1}{a}$ .

Answer: $\frac{1}{a}$

Exercise 263 MediumIB · P3 TZ2 May 2008

Find the exact value of $\int_{0}^{+ \infty} \frac{d x}{( x + 2 ) ( 2 x + 1 )}$ .

Show answer

Answer: $\frac{1}{3} ln 4$

Show working and answer

We decompose into partial fractions: $\frac{1}{( x + 2 ) ( 2 x + 1 )} = \frac{A}{x + 2} + \frac{B}{2 x + 1}$ , so $1 = A (2 x + 1) + B (x + 2)$ .
With $x = - 2$ : $1 = - 3 A$ , so $A = - \frac{1}{3}$ . With $x = - \frac{1}{2}$ : $1 = \frac{3}{2} B$ , so $B = \frac{2}{3}$ .
Antiderivative: $- \frac{1}{3} ln (x + 2) + \frac{2}{3} \cdot \frac{1}{2} ln (2 x + 1) = \frac{1}{3} ln \frac{2 x + 1}{x + 2}$ (for $x \geq 0$ ).
Improper integral, written with a limit: $\int_{0}^{+ \infty} = b \to \infty lim \frac{1}{3} [ln \frac{2 x + 1}{x + 2}]_{0}^{b}$ .
As $b \to \infty$ , $\frac{2 b + 1}{b + 2} \to 2$ . At $x = 0$ the quotient equals $\frac{1}{2}$ .
This gives $\frac{1}{3} (ln 2 - ln \frac{1}{2}) = \frac{2}{3} ln 2 = \frac{1}{3} ln 4$ .

Answer: $\frac{1}{3} ln 4$

Exercise 264 MediumIB · P3 TZ2 May 2007

Find the set of values of $p$ for which $\int_{1}^{\infty} \frac{1}{x ^{p}} d x$ converges.

Show answer

Answer: converges if and only if $p > 1$ (and then it equals $\frac{1}{p - 1}$ ); for $p \leq 1$ it diverges.

Show working and answer

An improper integral with an infinite upper limit is defined using a limit: $\int_{1}^{\infty} x^{- p} d x = b \to \infty lim \int_{1}^{b} x^{- p} d x$ . We consider two cases.
Case $p = 1$ : $\int_{1}^{b} \frac{d x}{x} = ln b \to + \infty$ . It diverges.
Case $p \neq = 1$ : $\int_{1}^{b} x^{- p} d x = [\frac{x ^{1 - p}}{1 - p}]_{1}^{b} = \frac{b ^{1 - p} - 1}{1 - p}$ .
If $p > 1$ , the exponent $1 - p < 0$ and $b^{1 - p} \to 0$ : the limit is $\frac{- 1}{1 - p} = \frac{1}{p - 1}$ . It converges.
If $p < 1$ , the exponent $1 - p > 0$ and $b^{1 - p} \to + \infty$ . It diverges.
There is convergence only for $p > 1$ , and then the integral equals $\frac{1}{p - 1}$ .

Answer: converges if and only if $p > 1$ (and then it equals $\frac{1}{p - 1}$ ); for $p \leq 1$ it diverges.

Exercise 265 MediumIB · P3 TZ0 May 2010

a) Using integration by parts, verify that $\int_{0}^{\infty} e^{- x} cos x d x = \int_{0}^{\infty} e^{- x} sin x d x$ .

b) Find the value of these two integrals.

Show answer

Answer: a) shown; b) $\int_{0}^{\infty} e^{- x} cos x d x = \int_{0}^{\infty} e^{- x} sin x d x = \frac{1}{2}$

Show working and answer

a) 1. Let $I = \int_{0}^{\infty} e^{- x} cos x d x$ . We integrate by parts with $u = e^{- x}$ and $d v = cos x d x$ : $d u = - e^{- x} d x$ , $v = sin x$ . 2. $I = [e^{- x} sin x]_{0}^{\infty} + \int_{0}^{\infty} e^{- x} sin x d x$ . 3. The bracketed term is $0$ : at $x = 0$ because of the sine and, as $x \to \infty$ , because $e^{- x} \to 0$ while the sine is bounded. 4. Hence $I = \int_{0}^{\infty} e^{- x} sin x d x = J$ , which is what was required.

b) 1. We integrate $J$ by parts with $u = e^{- x}$ and $d v = sin x d x$ : $d u = - e^{- x} d x$ , $v = - cos x$ . 2. $J = [- e^{- x} cos x]_{0}^{\infty} - \int_{0}^{\infty} e^{- x} cos x d x = (0 - (- 1)) - I = 1 - I$ . 3. Since $I = J$ by part (a): $I = 1 - I$ , so $I = \frac{1}{2}$ .

Answer: a) shown; b) $\int_{0}^{\infty} e^{- x} cos x d x = \int_{0}^{\infty} e^{- x} sin x d x = \frac{1}{2}$

Exercise 266 MediumIB · P3 TZ0 May 2015

a) Use L'Hôpital's rule to find $x \to \infty lim x^{2} e^{- x}$ .

b) Show that the improper integral $\int_{0}^{\infty} x^{2} e^{- x} d x$ is convergent, and calculate its value.

Show answer

Answer: a) $0$ ; b) the integral converges and equals $2$

Show working and answer

a) 1. As $x \to \infty$ , $x^{2} e^{- x} = \frac{x ^{2}}{e ^{x}}$ is of the form $\frac{\infty}{\infty}$ , so we apply L'Hôpital's rule. 2. Differentiating the numerator and the denominator: $x \to \infty lim \frac{x ^{2}}{e ^{x}} = x \to \infty lim \frac{2 x}{e ^{x}}$ , which is still $\frac{\infty}{\infty}$ . 3. Applying the rule again: $x \to \infty lim \frac{2 x}{e ^{x}} = x \to \infty lim \frac{2}{e ^{x}} = 0$ .

b) 1. It is improper at the upper limit, so we write it as a limit: $\int_{0}^{\infty} x^{2} e^{- x} d x = b \to \infty lim \int_{0}^{b} x^{2} e^{- x} d x$ . 2. First integration by parts with $u = x^{2}$ and $d v = e^{- x} d x$ ( $d u = 2 x d x$ , $v = - e^{- x}$ ): $\int x^{2} e^{- x} d x = - x^{2} e^{- x} + 2 \int x e^{- x} d x$ . 3. Second by parts with $u = x$ and $d v = e^{- x} d x$ : $\int x e^{- x} d x = - x e^{- x} + \int e^{- x} d x = - x e^{- x} - e^{- x}$ . 4. An antiderivative is $F (x) = - e^{- x} (x^{2} + 2 x + 2)$ . 5. Hence $\int_{0}^{b} x^{2} e^{- x} d x = [F (x)]_{0}^{b} = - e^{- b} (b^{2} + 2 b + 2) + 2$ . 6. As $b \to \infty$ : $b^{2} e^{- b} \to 0$ by part a), and $b e^{- b} \to 0$ and $e^{- b} \to 0$ . The limit exists and is finite, so the integral converges and equals $2$ .

Answer: a) $0$ ; b) the integral converges and equals $2$

Exercise 267 MediumIB · P3 TZ0 May 2012

Find the set of values of $k$ for which the improper integral $\int_{2}^{\infty} \frac{d x}{x ( ln x ) ^{k}}$ converges.

Show answer

Answer: the integral converges if $k > 1$

Show working and answer

This is an improper integral at the upper limit: $\int_{2}^{\infty} \frac{d x}{x ( ln x ) ^{k}} = b \to \infty lim \int_{2}^{b} \frac{d x}{x ( ln x ) ^{k}}$ .
Substitute $u = ln x$ , $d u = \frac{d x}{x}$ ; the limits become $ln 2$ and $ln b$ : $\int_{l n 2}^{l n b} u^{- k} d u$ .
If $k = 1$ : $[ln u]_{l n 2}^{l n b} = ln (ln b) - ln (ln 2) \to \infty$ as $b \to \infty$ . It diverges.
If $k \neq = 1$ : $[\frac{u ^{1 - k}}{1 - k}]_{l n 2}^{l n b} = \frac{( ln b ) ^{1 - k} - ( ln 2 ) ^{1 - k}}{1 - k}$ .
As $b \to \infty$ , $ln b \to \infty$ . If $k > 1$ , then $1 - k < 0$ and $(ln b)^{1 - k} \to 0$ : the limit exists and equals $\frac{( ln 2 ) ^{1 - k}}{k - 1}$ . If $k < 1$ , then $1 - k > 0$ and $(ln b)^{1 - k} \to \infty$ : it diverges.
The integral converges only for $k > 1$ .

Answer: the integral converges if $k > 1$

Exercise 268 MediumIB · P3 TZ1 May 2008

a) Using L'Hôpital's rule, prove that $x \to \infty lim x e^{- x} = 0$ .

b) Calculate $\int_{0}^{a} x e^{- x} d x$ .

c) Show that the integral $\int_{0}^{\infty} x e^{- x} d x$ is convergent, and calculate its value.

Show answer

Answer: a) shown; b) $- a e^{- a} - e^{- a} + 1$ ; c) convergent, with value $1$ .

Show working and answer

a) 1. $x e^{- x} = \frac{x}{e ^{x}}$ is of the form $\frac{\infty}{\infty}$ as $x \to \infty$ . 2. L'Hôpital: $x \to \infty lim \frac{x}{e ^{x}} = x \to \infty lim \frac{1}{e ^{x}} = 0$ .

b) 1. By parts with $u = x$ and $d v = e^{- x} d x$ ( $d u = d x$ , $v = - e^{- x}$ ): $\int x e^{- x} d x = - x e^{- x} + \int e^{- x} d x = - x e^{- x} - e^{- x}$ . 2. $\int_{0}^{a} x e^{- x} d x = [- x e^{- x} - e^{- x}]_{0}^{a} = - a e^{- a} - e^{- a} + 1$ .

c) 1. The improper integral is $\int_{0}^{\infty} x e^{- x} d x = a \to \infty lim (- a e^{- a} - e^{- a} + 1)$ . 2. By (a), $a e^{- a} \to 0$ , and also $e^{- a} \to 0$ . The limit exists and equals $1$ . 3. Hence the integral is convergent and its value is $1$ .

Answer: a) shown; b) $- a e^{- a} - e^{- a} + 1$ ; c) convergent, with value $1$ .

Exercise 269 MediumIB · P3 TZ0 May 2018

a) Find the value of $\int_{4}^{\infty} \frac{1}{x ^{3}} d x$ .

b) Illustrate graphically the inequality $n = 5 \sum \infty \frac{1}{n ^{3}} < \int_{4}^{\infty} \frac{1}{x ^{3}} d x < n = 4 \sum \infty \frac{1}{n ^{3}}$ .

c) Hence, write down a lower bound for $n = 4 \sum \infty \frac{1}{n ^{3}}$ .

d) Find an upper bound for $n = 4 \sum \infty \frac{1}{n ^{3}}$ .

View solution

a) 1. $\int_{4}^{\infty} \frac{1}{x ^{3}} d x = H \to \infty lim [- \frac{1}{2 x ^{2}}]_{4}^{H} = 0 + \frac{1}{32} = \frac{1}{32}$ .

b) 1. $f (x) = \frac{1}{x ^{3}}$ is decreasing. On each interval $[n, n + 1]$ with $n \geq 4$ , the rectangle of height $f (n + 1)$ lies below the curve and the one of height $f (n)$ lies above it. 2. Adding the areas of the rectangles of base $1$ : $n = 5 \sum \infty \frac{1}{n ^{3}} < \int_{4}^{\infty} \frac{1}{x ^{3}} d x < n = 4 \sum \infty \frac{1}{n ^{3}}$ .

Graph of y = 1/x³ for x ≥ 4 with rectangles of base 1: the upper ones of height f(n) and the lower ones of height f(n+1)

c) 1. From the second inequality: $n = 4 \sum \infty \frac{1}{n ^{3}} > \frac{1}{32} = 0.03125$ .

d) 1. From the first inequality: $n = 5 \sum \infty \frac{1}{n ^{3}} < \frac{1}{32}$ . 2. We add the term $n = 4$ , which is $\frac{1}{64}$ : $n = 4 \sum \infty \frac{1}{n ^{3}} < \frac{1}{64} + \frac{1}{32} = \frac{3}{64} \approx 0.0469$ .

Answer: a) $\frac{1}{32}$ ; c) $n = 4 \sum \infty \frac{1}{n ^{3}} > \frac{1}{32} = 0.03125$ ; d) $n = 4 \sum \infty \frac{1}{n ^{3}} < \frac{3}{64} \approx 0.0469$

Exercise 270 HardIB · P3 TZ0 May 2014

Consider the functions $f$ and $g$ given by $f (x) = \frac{e ^{x} + e ^{- x}}{2}$ and $g (x) = \frac{e ^{x} - e ^{- x}}{2}$ .

a) Show that $f^{'} (x) = g (x)$ and $g^{'} (x) = f (x)$ .

b) Find the first three non-zero terms of the Maclaurin series expansion of $f (x)$ .

c) Hence, find the value of $x \to 0 lim \frac{1 - f ( x )}{x ^{2}}$ .

d) Find the value of the improper integral $\int_{0}^{\infty} \frac{g ( x )}{[ f ( x ) ] ^{2}} d x$ .

Show answer

Answer: a) proved; b) $1 + \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ ; c) $- \frac{1}{2}$ ; d) $1$

Show working and answer

a) 1. $f (x) = \frac{e ^{x} + e ^{- x}}{2}$ , so $f^{'} (x) = \frac{e ^{x} - e ^{- x}}{2} = g (x)$ . 2. $g (x) = \frac{e ^{x} - e ^{- x}}{2}$ , so $g^{'} (x) = \frac{e ^{x} + e ^{- x}}{2} = f (x)$ .

b) 1. By (a), the derivatives alternate: $f^{'} = g$ , $f^{''} = g^{'} = f$ , $f^{'''} = g$ , $f^{(4)} = f$ . 2. At $x = 0$ : $f (0) = 1$ , $g (0) = 0$ , so $f (0) = 1$ , $f^{'} (0) = 0$ , $f^{''} (0) = 1$ , $f^{'''} (0) = 0$ , $f^{(4)} (0) = 1$ . 3. Maclaurin: $f (x) = 1 + 0 \cdot x + \frac{1}{2 !} x^{2} + 0 \cdot x^{3} + \frac{1}{4 !} x^{4} + \dots$ . 4. The first three non-zero terms: $1 + \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ .

c) 1. $1 - f (x) \approx - \frac{x ^{2}}{2} - \frac{x ^{4}}{24}$ . 2. $\frac{1 - f ( x )}{x ^{2}} \approx - \frac{1}{2} - \frac{x ^{2}}{24} \to - \frac{1}{2}$ as $x \to 0$ .

d) 1. As $f^{'} (x) = g (x)$ , substitute $u = f (x)$ , $d u = g (x) d x$ . Limits: $x = 0 \to u = f (0) = 1$ ; $x = b \to u = f (b)$ . 2. $\int_{0}^{b} \frac{g ( x )}{[ f ( x ) ] ^{2}} d x = \int_{1}^{f (b)} \frac{d u}{u ^{2}} = [- \frac{1}{u}]_{1}^{f (b)} = 1 - \frac{1}{f ( b )}$ . 3. As $b \to \infty$ , $f (b) = \frac{e ^{b} + e ^{- b}}{2} \to \infty$ , so $\frac{1}{f ( b )} \to 0$ . 4. The integral converges and equals $1$ .

Answer: a) proved; b) $1 + \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ ; c) $- \frac{1}{2}$ ; d) $1$

Exercise 271 HardIB · P3 TZ0 May 2011

The integral $I_{n}$ is defined as follows: $I_{n} = \int_{nπ}^{(n + 1) π} e^{- x} ∣ sin x ∣ d x$ , for $n \in N$ .

a) Verify that $I_{0} = \frac{1}{2} (1 + e^{- π})$ .

b) Verify that $I_{n} = e^{- nπ} \cdot I_{0}$ , by taking $y = x - nπ$ .

c) Hence, determine the exact value of $\int_{0}^{\infty} e^{- x} ∣ sin x ∣ d x$ .

Show answer

Answer: a) verified; b) verified; c) $\frac{e ^{π} + 1}{2 ( e ^{π} - 1 )}$

Show working and answer

a) 1. On $[0, π]$ we have $sin x \geq 0$ , so $∣ sin x ∣ = sin x$ and $I_{0} = \int_{0}^{π} e^{- x} sin x d x$ . 2. An antiderivative is $F (x) = - \frac{1}{2} e^{- x} (sin x + cos x)$ . We check by differentiating: $F^{'} (x) = - \frac{1}{2} [- e^{- x} (sin x + cos x) + e^{- x} (cos x - sin x)] = e^{- x} sin x$ . 3. $I_{0} = F (π) - F (0) = - \frac{1}{2} e^{- π} (0 - 1) + \frac{1}{2} (0 + 1) = \frac{1}{2} e^{- π} + \frac{1}{2} = \frac{1}{2} (1 + e^{- π})$ .

b) 1. With $y = x - nπ$ : $x = y + nπ$ , $d x = d y$ , and the limits $x = nπ$ and $x = (n + 1) π$ become $y = 0$ and $y = π$ . 2. $e^{- x} = e^{- y} e^{- nπ}$ and $∣ sin (y + nπ) ∣ = ∣ sin y ∣$ (the sine only changes sign when $nπ$ is added). 3. $I_{n} = e^{- nπ} \int_{0}^{π} e^{- y} ∣ sin y ∣ d y = e^{- nπ} I_{0}$ .

c) 1. This is an improper integral: $\int_{0}^{\infty} e^{- x} ∣ sin x ∣ d x = N \to \infty lim \int_{0}^{N π} e^{- x} ∣ sin x ∣ d x = N \to \infty lim n = 0 \sum N - 1 I_{n}$ . 2. By b), $I_{n} = I_{0} (e^{- π})^{n}$ : this is a geometric series with ratio $e^{- π}$ , where $∣ e^{- π} ∣ < 1$ , so it converges and its sum is $\frac{I _{0}}{1 - e ^{- π}}$ . 3. $\frac{\frac{1}{2} ( 1 + e ^{- π} )}{1 - e ^{- π}}$ ; multiplying numerator and denominator by $e^{π}$ : $\frac{e ^{π} + 1}{2 ( e ^{π} - 1 )}$ .

Answer: a) verified; b) verified; c) $\frac{e ^{π} + 1}{2 ( e ^{π} - 1 )}$

Exercise 272 HardIB · P3 TZ1 May 2008

The diagram shows part of the graph of $y = \frac{1}{x ^{3}}$ together with line segments parallel to the coordinate axes.

Graph of y = 1/x³ for x ≥ 3 with rectangles of base 1 over the intervals (3,4), (4,5) and (5,6), above and below the curve

a) Using the diagram, show that $\frac{1}{4 ^{3}} + \frac{1}{5 ^{3}} + \frac{1}{6 ^{3}} + \dots < \int_{3}^{\infty} \frac{1}{x ^{3}} d x < \frac{1}{3 ^{3}} + \frac{1}{4 ^{3}} + \frac{1}{5 ^{3}} + \dots$ .

b) Hence, find upper and lower bounds for $n = 1 \sum \infty \frac{1}{n ^{3}}$ .

Show answer

Answer: b) $1.18 < n = 1 \sum \infty \frac{1}{n ^{3}} < 1.22$

Show working and answer

a) 1. $f (x) = \frac{1}{x ^{3}}$ is decreasing. Each rectangle has base $1$ . 2. The upper rectangles, with heights $f (3), f (4), f (5), \dots$ , stick out above the curve: their sum $\frac{1}{3 ^{3}} + \frac{1}{4 ^{3}} + \dots$ is greater than $\int_{3}^{\infty} \frac{1}{x ^{3}} d x$ . 3. The lower rectangles, with heights $f (4), f (5), f (6), \dots$ , lie below the curve: their sum $\frac{1}{4 ^{3}} + \frac{1}{5 ^{3}} + \dots$ is less than the integral. 4. Hence, $\frac{1}{4 ^{3}} + \frac{1}{5 ^{3}} + \frac{1}{6 ^{3}} + \dots < \int_{3}^{\infty} \frac{1}{x ^{3}} d x < \frac{1}{3 ^{3}} + \frac{1}{4 ^{3}} + \frac{1}{5 ^{3}} + \dots$ .

b) 1. $\int_{3}^{\infty} \frac{1}{x ^{3}} d x = H \to \infty lim [- \frac{1}{2 x ^{2}}]_{3}^{H} = \frac{1}{18}$ . 2. From the left-hand side of the inequality: $n = 4 \sum \infty \frac{1}{n ^{3}} < \frac{1}{18}$ . 3. From the right-hand side: $n = 3 \sum \infty \frac{1}{n ^{3}} > \frac{1}{18}$ , so $n = 4 \sum \infty \frac{1}{n ^{3}} > \frac{1}{18} - \frac{1}{27} = \frac{1}{54}$ . 4. Add the first three terms, $1 + \frac{1}{8} + \frac{1}{27}$ : $1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{54} < n = 1 \sum \infty \frac{1}{n ^{3}} < 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{18}$ . 5. Numerically: $1.1806 < n = 1 \sum \infty \frac{1}{n ^{3}} < 1.2176$ , i.e. $1.18 < S < 1.22$ .

Answer: b) $1.18 < n = 1 \sum \infty \frac{1}{n ^{3}} < 1.22$

Applications and contextual problems

Exercise 273 BasicIB · P1 May 2000

The velocity, $v$ , of an object at an instant $t$ is given by $v = k e^{- \frac{t}{2}}$ , where $t$ is in seconds and $v$ is in $m s^{- 1}$ . Calculate the distance travelled between $t = 0$ and $t = a$ .

Show answer

Answer: distance $= 2 k (1 - e^{- \frac{a}{2}})$

Show working and answer

We assume $k > 0$ (otherwise the velocity is not positive). Then $v = k e^{- t /2} > 0$ and the distance is equal to the displacement.
$d = \int_{0}^{a} k e^{- \frac{t}{2}} d t = [- 2 k e^{- \frac{t}{2}}]_{0}^{a}$ .
$d = - 2 k e^{- \frac{a}{2}} + 2 k = 2 k (1 - e^{- \frac{a}{2}})$ .

Answer: distance $= 2 k (1 - e^{- \frac{a}{2}})$

Exercise 274 MediumIB · P1 November 2002

A particle moves in a straight line with velocity, in metres per second, at time $t$ seconds given by $v (t) = 6 t^{2} - 6 t$ , $t \geq 0$ . Calculate the total distance travelled by the particle in the first two seconds of motion.

Show answer

Answer: distance $= 6$ m

Show working and answer

Factorising: $v (t) = 6 t (t - 1)$ . It is negative on $] 0, 1 [$ (the particle moves backwards) and positive on $] 1, 2]$ .
As the total distance is not cancelled out, we integrate the absolute value: $d = \int_{0}^{2} ∣ v ∣ d t = - \int_{0}^{1} v d t + \int_{1}^{2} v d t$ .
An antiderivative of $v$ is $2 t^{3} - 3 t^{2}$ . At $t = 0$ it equals $0$ , at $t = 1$ it equals $- 1$ and at $t = 2$ it equals $4$ .
$\int_{0}^{1} v d t = - 1 - 0 = - 1$ and $\int_{1}^{2} v d t = 4 - (- 1) = 5$ .
$d = 1 + 5 = 6$ m.

Answer: distance $= 6$ m

Exercise 275 MediumIB · P1 May 2003

A particle moves in a straight line. Its velocity $v m s^{- 1}$ , $t$ seconds later, is given by $v = e^{- t} sin t$ . Find the total distance travelled in the time interval $[0, 2 π]$ .

Show answer

Answer: distance $\approx 0.852$ m

Show working and answer

$v = e^{- t} sin t$ changes sign at $t = π$ : it is positive on $] 0, π [$ and negative on $] π, 2 π [$ .
The total distance is $d = \int_{0}^{2 π} ∣ v ∣ d t = \int_{0}^{π} v d t - \int_{π}^{2 π} v d t$ .
There is no simple antiderivative, so using the GDC: $\int_{0}^{π} v d t \approx 0.620$ and $\int_{π}^{2 π} v d t \approx - 0.232$ .
$d \approx 0.620 + 0.232 = 0.852$ m.

Answer: distance $\approx 0.852$ m

Exercise 276 MediumIB · P1 TZ0 November 2017

A particle moves in a straight line such that, at time $t$ seconds, $t \geq 0$ , its velocity $v$ , in $m s^{- 1}$ , is given by $v = 10 t e^{- 2 t}$ . Calculate the exact distance travelled by the particle in the first half second.

Show answer

Answer: distance $= \frac{5 ( e - 2 )}{2 e}$ m

Show working and answer

Since $v = 10 t e^{- 2 t} \geq 0$ for $t \geq 0$ , the distance is $d = \int_{0}^{0.5} 10 t e^{- 2 t} d t$ .
By parts: $u = t$ , $d v = e^{- 2 t} d t$ ; then $d u = d t$ and $v = - \frac{1}{2} e^{- 2 t}$ .
$\int t e^{- 2 t} d t = - \frac{t}{2} e^{- 2 t} + \frac{1}{2} \int e^{- 2 t} d t = - \frac{t}{2} e^{- 2 t} - \frac{1}{4} e^{- 2 t}$ .
Between $0$ and $\frac{1}{2}$ : at $t = \frac{1}{2}$ it equals $- \frac{1}{4 e} - \frac{1}{4 e} = - \frac{1}{2 e}$ ; at $t = 0$ it equals $- \frac{1}{4}$ . The difference is $\frac{1}{4} - \frac{1}{2 e}$ .
$d = 10 (\frac{1}{4} - \frac{1}{2 e}) = \frac{5}{2} - \frac{5}{e} = \frac{5 ( e - 2 )}{2 e}$ m.

Answer: distance $= \frac{5 ( e - 2 )}{2 e}$ m

Exercise 277 MediumIB · P1 November 1999

The acceleration, $a (t)$ , of an express train during the first $80$ seconds of motion is given by $a (t) = - \frac{1}{20} t + 2$ , where $t$ is the time in seconds. If the train starts from rest at $t = 0$ , find the distance travelled by the train in the first minute.

Show answer

Answer: distance $= 1800$ m

Show working and answer

Since $v^{'} (t) = a (t)$ , we integrate: $v (t) = \int (- \frac{t}{20} + 2) d t = - \frac{t ^{2}}{40} + 2 t + C$ .
It starts from rest: $v (0) = 0$ , so $C = 0$ and $v (t) = 2 t - \frac{t ^{2}}{40}$ .
This velocity is positive for $0 < t < 80$ , so in the first minute ( $0 \leq t \leq 60$ ) there is no change of direction and the distance is the integral of $v$ .
$d = \int_{0}^{60} (2 t - \frac{t ^{2}}{40}) d t = [t^{2} - \frac{t ^{3}}{120}]_{0}^{60} = 3600 - 1800 = 1800$ m.

Answer: distance $= 1800$ m

Exercise 278 MediumIB · P2 TZ2 May 2011

A stunt parachutist jumps from a stationary hot-air balloon at a height of $2000$ m above the ground. Her velocity, $v m s^{- 1}$ , $t$ seconds after she jumps, is given by $v = 50 (1 - e^{- 0.2 t})$ .

a) Find her acceleration $10$ seconds after she jumps.

b) How far above the ground is she $10$ seconds after she jumps?

Show answer

Answer: a) $a (10) = \frac{10}{e ^{2}} \approx 1.35 m s^{- 2}$ ; b) $\approx 1716$ m above the ground

Show working and answer

a) 1. The acceleration is the derivative of the velocity: $a (t) = v^{'} (t) = 50 \cdot 0.2 e^{- 0.2 t} = 10 e^{- 0.2 t}$ . 2. At $t = 10$ : $a (10) = 10 e^{- 2} = \frac{10}{e ^{2}} \approx 1.35 m s^{- 2}$ .

b) 1. The distance fallen in $10$ s is $\int_{0}^{10} v d t = \int_{0}^{10} 50 (1 - e^{- 0.2 t}) d t$ (the velocity is positive). 2. $= 50 [t + 5 e^{- 0.2 t}]_{0}^{10} = 50 (10 + 5 e^{- 2} - 5) = 250 (1 + e^{- 2}) \approx 283.8$ m. 3. Height above the ground: $2000 - 283.8 \approx 1716$ m.

Answer: a) $a (10) = \frac{10}{e ^{2}} \approx 1.35 m s^{- 2}$ ; b) $\approx 1716$ m above the ground

Exercise 279 MediumIB · P3 TZ0 November 2010

a) Using the Maclaurin series for the function $e^{x}$ , write down the first four terms of the Maclaurin series for $e^{- \frac{x ^{2}}{2}}$ .

b) Hence, find the first four terms of the series for $\int_{0}^{x} e^{- \frac{u ^{2}}{2}} d u$ .

c) Use the result of part (b) to find an approximate value of $\frac{1}{2 π} \int_{0}^{1} e^{- \frac{x ^{2}}{2}} d x$ .

Show answer

Answer: a) $1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{8} - \frac{x ^{6}}{48}$ ; b) $x - \frac{x ^{3}}{6} + \frac{x ^{5}}{40} - \frac{x ^{7}}{336}$ ; c) $\approx 0.341$

Show working and answer

a) 1. Recall $e^{u} = 1 + u + \frac{u ^{2}}{2 !} + \frac{u ^{3}}{3 !} + \dots$ . 2. Substituting $u = - \frac{x ^{2}}{2}$ : $1 - \frac{x ^{2}}{2} + \frac{1}{2} \cdot \frac{x ^{4}}{4} - \frac{1}{6} \cdot \frac{x ^{6}}{8}$ . 3. Simplifying: $e^{- \frac{x ^{2}}{2}} \approx 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{8} - \frac{x ^{6}}{48}$ .

b) 1. We integrate term by term between $0$ and $x$ (the variable is $u$ ): $\int_{0}^{x} (1 - \frac{u ^{2}}{2} + \frac{u ^{4}}{8} - \frac{u ^{6}}{48}) d u$ . 2. $= x - \frac{x ^{3}}{6} + \frac{x ^{5}}{40} - \frac{x ^{7}}{336}$ .

c) 1. We evaluate the series from part (b) at $x = 1$ : $1 - \frac{1}{6} + \frac{1}{40} - \frac{1}{336} \approx 0.8556$ . 2. We divide by $2 π \approx 2.5066$ : $\frac{0.8556}{2.5066} \approx 0.341$ .

Answer: a) $1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{8} - \frac{x ^{6}}{48}$ ; b) $x - \frac{x ^{3}}{6} + \frac{x ^{5}}{40} - \frac{x ^{7}}{336}$ ; c) $\approx 0.341$

Exercise 280 MediumIB · P1 May 2002

A particle is launched along a straight path. After $t$ seconds, its velocity $v$ in metres per second is given by $v = \frac{1}{2 + t ^{2}}$ .

a) Find the distance travelled during the first second.

b) Find an expression for the acceleration at time $t$ .

Show answer

Answer: a) $s (1) = \frac{2}{2} arctan (\frac{1}{2}) \approx 0.435$ m; b) $a (t) = - \frac{2 t}{( 2 + t ^{2} ) ^{2}}$

Show working and answer

a) 1. The velocity is positive, so the distance is $\int_{0}^{1} \frac{d t}{2 + t ^{2}}$ . 2. We use $\int \frac{d t}{t ^{2} + a ^{2}} = \frac{1}{a} arctan \frac{t}{a}$ with $a = 2$ . 3. $d = \frac{1}{2} [arctan \frac{t}{2}]_{0}^{1} = \frac{2}{2} arctan \frac{1}{2} \approx 0.435$ m.

b) 1. $a = \frac{d v}{d t}$ with $v = (2 + t^{2})^{- 1}$ . By the chain rule: $a = - (2 + t^{2})^{- 2} \cdot 2 t$ . 2. $a (t) = - \frac{2 t}{( 2 + t ^{2} ) ^{2}}$ .

Answer: a) $s (1) = \frac{2}{2} arctan (\frac{1}{2}) \approx 0.435$ m; b) $a (t) = - \frac{2 t}{( 2 + t ^{2} ) ^{2}}$

Exercise 281 MediumIB · P2 TZ0 November 2014

A particle moves in a straight line, such that its velocity, $v m s^{- 1}$ , at time $t$ seconds is given by $v (t) = ⎩ ⎨ ⎧ 5 - (t - 2)^{2}, 3 - \frac{t}{2}, 0 \leq t \leq 4 t > 4$ .

a) Find the value of $t$ at which the particle is momentarily at rest.

The particle returns to its initial position at $t = T$ .

b) Find the value of $T$ .

Show answer

Answer: a) $t = 6$ s; b) $T \approx 13.9$ s

Show working and answer

a) 1. On $0 \leq t \leq 4$ : $5 - (t - 2)^{2} = 0 \Rightarrow t = 2 \pm 5$ , but $2 + 5 \approx 4.24 > 4$ and $2 - 5 < 0$ . Neither lies in the interval (in fact $v \geq 1$ there). 2. For $t > 4$ : $3 - \frac{t}{2} = 0 \Rightarrow t = 6$ , which does satisfy $t > 4$ .

b) 1. The particle is back at its initial position when the total displacement is $0$ : $\int_{0}^{T} v d t = 0$ . 2. As $v > 0$ up to $t = 6$ , the particle moves away until $t = 6$ and then returns; hence $T > 6$ . 3. Interval $0 \leq t \leq 4$ : $\int_{0}^{4} [5 - (t - 2)^{2}] d t = [5 t - \frac{( t - 2 ) ^{3}}{3}]_{0}^{4} = (20 - \frac{8}{3}) - \frac{8}{3} = \frac{44}{3}$ . 4. Interval $4 \leq t \leq T$ : $\int_{4}^{T} (3 - \frac{t}{2}) d t = [3 t - \frac{t ^{2}}{4}]_{4}^{T} = 3 T - \frac{T ^{2}}{4} - 8$ . 5. Set equal to zero: $\frac{44}{3} + 3 T - \frac{T ^{2}}{4} - 8 = 0$ , i.e. $3 T^{2} - 36 T - 80 = 0$ . 6. $T = \frac{36 \pm 1296 + 960}{6}$ ; we take the positive root: $T = 6 + \frac{2 141}{3} \approx 13.9$ s.

Answer: a) $t = 6$ s; b) $T \approx 13.9$ s

Exercise 282 MediumIB · P1 TZ2 May 2007

A particle moves in a straight line. After a time of $t$ seconds, its displacement from a fixed point $O$ is $s$ metres, and its velocity $v$ in metres per second is given by $v = 3 t^{2} - 4 t + 2$ , with $t \geq 0$ . When $t = 0$ , $s = - 3$ .

Find the value of $t$ at which the particle is at $O$ .

Show answer

Answer: $t \approx 1.81$ s

Show working and answer

We integrate the velocity: $s = \int (3 t^{2} - 4 t + 2) d t = t^{3} - 2 t^{2} + 2 t + C$ .
At $t = 0$ , $s = - 3$ , so $C = - 3$ and $s = t^{3} - 2 t^{2} + 2 t - 3$ .
The particle is at $O$ when $s = 0$ : $t^{3} - 2 t^{2} + 2 t - 3 = 0$ .
Using the GDC (or by trial; $v = 3 (t - \frac{2}{3})^{2} + \frac{2}{3} > 0$ , so $s$ is strictly increasing and there is only one real root) we obtain $t \approx 1.81$ s.

Answer: $t \approx 1.81$ s

Exercise 283 MediumIB · P2 TZ2 May 2019

A particle moves along a horizontal line such that, at time $t$ seconds, $t \geq 0$ , its acceleration $a$ is given by $a = 2 t - 1$ . When $t = 6$ , its displacement $s$ from a fixed origin $O$ is $18.25$ m. When $t = 15$ , the displacement from $O$ is $922.75$ m. Find an expression for $s$ in terms of $t$ .

Show answer

Answer: $s = \frac{t ^{3}}{3} - \frac{t ^{2}}{2} - 6 t + 0.25$

Show working and answer

Integrate the acceleration: $v = \int (2 t - 1) d t = t^{2} - t + C_{1}$ .
Integrate the velocity: $s = \frac{t ^{3}}{3} - \frac{t ^{2}}{2} + C_{1} t + C_{2}$ .
At $t = 6$ : $\frac{216}{3} - \frac{36}{2} + 6 C_{1} + C_{2} = 54 + 6 C_{1} + C_{2} = 18.25$ .
At $t = 15$ : $\frac{3375}{3} - \frac{225}{2} + 15 C_{1} + C_{2} = 1012.5 + 15 C_{1} + C_{2} = 922.75$ .
Subtract the first equation from the second: $958.5 + 9 C_{1} = 904.5$ , so $C_{1} = - 6$ . Then $C_{2} = 18.25 - 54 + 36 = 0.25$ .
$s = \frac{t ^{3}}{3} - \frac{t ^{2}}{2} - 6 t + 0.25$ .

Answer: $s = \frac{t ^{3}}{3} - \frac{t ^{2}}{2} - 6 t + 0.25$

Exercise 284 MediumIB · P3 TZ0 May 2019

a) Show that $1 - x^{2} + x^{4} - x^{6} + \dots = \frac{1}{1 + x ^{2}}$ , where $∣ x ∣ < 1$ .

b) Hence, write down the first four non-zero terms of the power series for $f (x) = \frac{1}{1 + 4 x ^{2}}$ .

c) Using the result of b), find the first four non-zero terms of the power series for $f (x) = arctan 2 x$ .

Show answer

Answer: a) shown; b) $f (x) = 1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + \dots$ ; c) $arctan 2 x = 2 x - \frac{8 x ^{3}}{3} + \frac{32 x ^{5}}{5} - \frac{128 x ^{7}}{7} + \dots$

Show working and answer

a) 1. The sum is a geometric series with first term $1$ and common ratio $- x^{2}$ (each term is the previous one multiplied by $- x^{2}$ ). 2. It converges if $- x^{2} < 1$ , that is, $∣ x ∣ < 1$ , and its sum is $\frac{1}{1 - ( - x ^{2} )} = \frac{1}{1 + x ^{2}}$ .

b) Substituting $2 x$ for $x$ in the expansion above (valid if $∣2 x ∣ < 1$ ): $f (x) = \frac{1}{1 + 4 x ^{2}} = 1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + \dots$

c) 1. As $\frac{d}{d x} arctan 2 x = \frac{2}{1 + 4 x ^{2}} = 2 f (x)$ , we have $arctan 2 x = \int_{0}^{x} 2 f (t) d t$ (since $arctan 0 = 0$ ). 2. Integrating term by term: $2 \int_{0}^{x} (1 - 4 t^{2} + 16 t^{4} - 64 t^{6} + \dots) d t = 2 x - \frac{8 x ^{3}}{3} + \frac{32 x ^{5}}{5} - \frac{128 x ^{7}}{7} + \dots$

Answer: a) shown; b) $f (x) = 1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + \dots$ ; c) $arctan 2 x = 2 x - \frac{8 x ^{3}}{3} + \frac{32 x ^{5}}{5} - \frac{128 x ^{7}}{7} + \dots$

Exercise 285 MediumIB · P2 November 2000

A uniform rod of length $l$ metres is placed with its ends on two supports $A$ , $B$ at the same horizontal level.

Rod of length l metres resting on the ends A and B, at the same height, sagging in the middle; the length l, the distance x measured from A and the sag y(x) below the segment AB are indicated

If $y (x)$ metres is the amount of sag (that is, the distance below $[A B]$ at a distance $x$ metres from the support $A$ , then it is known that $\frac{d ^{2} y}{d x ^{2}} = \frac{1}{125 l ^{3}} (x^{2} - l x)$ .

a) i) Let $z = \frac{1}{125 l ^{3}} (\frac{x ^{3}}{3} - \frac{l x ^{2}}{2}) + \frac{1}{1500}$ . Show that $\frac{d z}{d x} = \frac{1}{125 l ^{3}} (x^{2} - l x)$ .

(ii) Given that $\frac{d w}{d x} = z$ and $w (0) = 0$ , calculate $w (x)$ .

(iii) Show that $w$ satisfies $\frac{d ^{2} w}{d x ^{2}} = \frac{1}{125 l ^{3}} (x^{2} - l x)$ , and that $w (l) = w (0) = 0$ .

b) Calculate the sag at the centre of a rod of $2.4$ metres in length.

Show answer

Answer: a) ii) $w (x) = \frac{1}{125 l ^{3}} (\frac{x ^{4}}{12} - \frac{l x ^{3}}{6}) + \frac{1}{1500} x$ ; b) $0.0005$ m

Show working and answer

a) i) 1. We differentiate $z$ term by term: $\frac{d z}{d x} = \frac{1}{125 l ^{3}} (\frac{3 x ^{2}}{3} - \frac{2 l x}{2}) + 0$ . 2. Simplifying, $\frac{d z}{d x} = \frac{1}{125 l ^{3}} (x^{2} - l x)$ .

(ii) 1. We integrate $\frac{d w}{d x} = z$ : $w = \frac{1}{125 l ^{3}} (\frac{x ^{4}}{12} - \frac{l x ^{3}}{6}) + \frac{x}{1500} + C$ . 2. $w (0) = 0$ gives $C = 0$ . 3. $w (x) = \frac{1}{125 l ^{3}} (\frac{x ^{4}}{12} - \frac{l x ^{3}}{6}) + \frac{1}{1500} x$ .

(iii) 1. $\frac{d ^{2} w}{d x ^{2}} = \frac{d z}{d x} = \frac{1}{125 l ^{3}} (x^{2} - l x)$ by (i). 2. $w (0) = 0$ by construction. 3. $w (l) = \frac{1}{125 l ^{3}} (\frac{l ^{4}}{12} - \frac{l ^{4}}{6}) + \frac{l}{1500} = - \frac{l}{1500} + \frac{l}{1500} = 0$ .

b) 1. The maximum sag is at the centre, $x = \frac{l}{2} = 1.2$ , with $l = 2.4$ ; $y = w$ satisfies the same equation and the same conditions at the ends. 2. We calculate: $125 l^{3} = 125 \cdot 13.824 = 1728$ ; $\frac{x ^{4}}{12} = \frac{2.0736}{12} = 0.1728$ ; $\frac{l x ^{3}}{6} = \frac{2.4 \cdot 1.728}{6} = 0.6912$ . 3. $y = \frac{0.1728 - 0.6912}{1728} + \frac{1.2}{1500} = - 0.0003 + 0.0008 = 0.0005$ m.

Answer: a) ii) $w (x) = \frac{1}{125 l ^{3}} (\frac{x ^{4}}{12} - \frac{l x ^{3}}{6}) + \frac{1}{1500} x$ ; b) $0.0005$ m

Exercise 286 MediumIB · P2 TZ0 November 2007

The acceleration (in $m s^{- 2}$ ) at time $t$ seconds, (with $t > 0$ ), of a particle moving in a straight line is given by the formula $a = - \frac{1}{( 1 + t ) ^{2}}$ . When $t = 1$ , the velocity is $8 m s^{- 1}$ .

a) Find the velocity when $t = 3$ .

b) Find the limit of the velocity as $t \to \infty$ .

c) Find the distance travelled between $t = 1$ and $t = 3$ .

Show answer

Answer: a) $\frac{31}{4}$ m s $^{- 1}$ ; b) $\frac{15}{2}$ ; c) $ln 2 + 15$ m

Show working and answer

a) 1. Integrating the acceleration: $v = \int - \frac{1}{( 1 + t ) ^{2}} d t = \frac{1}{1 + t} + C$ . 2. With $v (1) = 8$ : $\frac{1}{2} + C = 8$ , so $C = \frac{15}{2}$ and $v (t) = \frac{1}{1 + t} + \frac{15}{2}$ . 3. $v (3) = \frac{1}{4} + \frac{15}{2} = \frac{31}{4}$ m s $^{- 1}$ .

b) 1. As $t \to \infty$ , $\frac{1}{1 + t} \to 0$ . 2. $t \to \infty lim v (t) = \frac{15}{2}$ m s $^{- 1}$ .

c) 1. $v > 0$ , so the distance is the displacement: $\int_{1}^{3} (\frac{1}{1 + t} + \frac{15}{2}) d t$ . 2. $= [ln (1 + t) + \frac{15}{2} t]_{1}^{3} = (ln 4 + \frac{45}{2}) - (ln 2 + \frac{15}{2}) = ln 2 + 15$ m.

Answer: a) $\frac{31}{4}$ m s $^{- 1}$ ; b) $\frac{15}{2}$ ; c) $ln 2 + 15$ m

Exercise 287 MediumIB · P2 TZ1 May 2021

A particle moves in a straight line. The velocity, $v$ $m s^{- 1}$ , of the particle at time $t$ seconds is given by $v (t) = t sin t - 3$ , for $0 \leq t \leq 10$ .

The following diagram shows the graph of $v$ .

Graph of the velocity v(t) = t·sin t − 3 for 0 ≤ t ≤ 10, with two points marked where it crosses the t-axis

a) Calculate the smallest value of $t$ for which the particle is at rest.

b) Calculate the total distance travelled by the particle.

c) Calculate the acceleration of the particle when $t = 7$ .

Show answer

Answer: a) $t = 6.74$ s; b) $37.1$ m; c) $a (7) = 5.93 m s^{- 2}$

Show working and answer

Paper 2: we use the GDC.

a) 1. The particle is at rest when $v = 0$ : $t sin t - 3 = 0$ . 2. Drawing $v (t)$ on the GDC, the first crossing with the $t$ -axis is at $t \approx 6.74$ s (before that instant $v < 0$ : the maximum of $t sin t$ does not reach $3$ ).

b) 1. The total distance is $\int_{0}^{10} ∣ v (t) ∣ d t$ (the absolute value stops the parts with negative and positive velocity from cancelling). 2. With the GDC: $\int_{0}^{10} ∣ t sin t - 3∣ d t \approx 37.1$ m.

c) 1. $a (t) = v^{'} (t) = sin t + t cos t$ . 2. $a (7) = sin 7 + 7 cos 7 \approx 0.657 + 5.277 = 5.93 m s^{- 2}$ (with the calculator in radian mode).

Answer: a) $t = 6.74$ s; b) $37.1$ m; c) $a (7) = 5.93 m s^{- 2}$

Exercise 288 MediumIB · P1 TZ0 November 2021

A particle $P$ moves along the $x$ -axis. The velocity of $P$ is $v$ $m s^{- 1}$ at time $t$ seconds, where $v (t) = 4 + 4 t - 3 t^{2}$ , $0 \leq t \leq 3$ .

When $t = 0$ , $P$ is at the origin $O$ .

a) (i) Find the value of $t$ when $P$ reaches its maximum velocity.

(ii) Show that the distance of $P$ from $O$ at this instant is $\frac{88}{27}$ m.

b) Sketch a graph of $v$ against $t$ , showing any points of intersection with the axes.

c) Calculate the total distance travelled by $P$ .

View solution

a) (i) 1. The velocity is maximum where $v^{'} (t) = 4 - 6 t = 0$ , that is, $t = \frac{2}{3}$ . It is a maximum because $v^{''} = - 6 < 0$ .

(ii) 1. The displacement is $s (t) = \int_{0}^{t} (4 + 4 u - 3 u^{2}) d u = 4 t + 2 t^{2} - t^{3}$ (with $s (0) = 0$ ). 2. $s (\frac{2}{3}) = \frac{8}{3} + \frac{8}{9} - \frac{8}{27} = \frac{72 + 24 - 8}{27} = \frac{88}{27}$ m. As it is positive, it is also the distance from $O$ .

b) 1. $v$ is a parabola opening downwards. Intercept with the $v$ -axis: $(0, 4)$ . Maximum at $(\frac{2}{3}, \frac{16}{3})$ . 2. Intercepts with the $t$ -axis: $3 t^{2} - 4 t - 4 = 0$ gives $(3 t + 2) (t - 2) = 0$ ; in $[0, 3]$ only $t = 2$ : $(2, 0)$ . 3. End point: $v (3) = 4 + 12 - 27 = - 11$ , that is, $(3, - 11)$ .

Graph of the velocity v = 4 + 4t − 3t² for 0 ≤ t ≤ 3

c) 1. The particle changes direction at $t = 2$ ( $v > 0$ before, $v < 0$ after). We calculate the positions: $s (2) = 8 + 8 - 8 = 8$ and $s (3) = 12 + 18 - 27 = 3$ . 2. Total distance: from $0$ to $8$ (outward) and from $8$ to $3$ (return): $8 + 5 = 13$ m.

Answer: a) (i) $t = \frac{2}{3}$ ; (ii) shown ( $\frac{88}{27}$ m); c) $13$ m

Exercise 289 MediumIB · P2 TZ1 May 2022

A particle moves along a straight line such that its velocity, $v$ $m s^{- 1}$ , after $t$ seconds is given by $v (t) = e^{s i n t} + 4 sin t$ for $0 \leq t \leq 6$ .

a) Find the value of $t$ when the particle is at rest.

b) Calculate the acceleration of the particle when it changes direction.

c) Calculate the total distance travelled by the particle.

Show answer

Answer: a) $t = 3.35$ s; b) $a = - 4.71 m s^{- 2}$ ; c) $20.8$ m

Show working and answer

Paper 2: we use the GDC.

a) 1. At rest: $v = 0$ , that is, $e^{s i n t} + 4 sin t = 0$ . 2. For $0 \leq t \leq π$ both terms are positive or zero, so the solution can only occur where $sin t < 0$ . Using the GDC, on $[0, 6]$ : $t \approx 3.35$ s.

b) 1. The particle changes direction where $v$ changes sign, at $t \approx 3.35$ . 2. $a (t) = v^{'} (t) = cos t e^{s i n t} + 4 cos t = cos t (e^{s i n t} + 4)$ . 3. $a (3.35) \approx - 4.71 m s^{- 2}$ .

c) 1. Total distance $= \int_{0}^{6} ∣ v (t) ∣ d t$ . 2. Using the GDC: $\approx 20.8$ m.

Answer: a) $t = 3.35$ s; b) $a = - 4.71 m s^{- 2}$ ; c) $20.8$ m

Exercise 290 MediumIB · P2 TZ2 May 2022

A particle moves in a straight line such that its velocity ( $m s^{- 1}$ ) at time $t$ seconds is given by $v = \frac{( t ^{2} + 1 ) cos t}{4}$ , $0 \leq t \leq 3$ .

a) Determine when the particle changes direction of motion.

b) Find the times at which the acceleration of the particle is equal to $- 1.9 m s^{- 2}$ .

c) Find the acceleration of the particle when its speed is at a maximum.

Show answer

Answer: a) $t = 1.57$ s; b) $t = 2.26$ s and $t = 2.96$ s; c) $a = - 1.84 m s^{- 2}$

Show working and answer

a) 1. The particle changes direction when $v = 0$ with a change of sign: $(t^{2} + 1) cos t = 0$ . As $t^{2} + 1 > 0$ , we need $cos t = 0$ . 2. In $[0, 3]$ : $t = \frac{π}{2} \approx 1.57$ s ( $\frac{3 π}{2} \approx 4.71$ lies outside the interval). There is a change of sign (from positive to negative).

b) 1. $a = v^{'} = \frac{2 t cos t - ( t ^{2} + 1 ) sin t}{4}$ . 2. Using the GDC, we solve $a (t) = - 1.9$ on $[0, 3]$ (intersections of $y = a (t)$ with $y = - 1.9$ ): $t \approx 2.26$ s and $t \approx 2.96$ s.

c) 1. The speed is $∣ v ∣$ . We study $v$ on $[0, 3]$ : $v (0) = 0.25$ , the local maximum of $v$ is small (around $t \approx 0.8$ , where $v \approx 0.3$ ), and then $v$ decreases to $v (3) = \frac{10 cos 3}{4} \approx - 2.47$ . 2. The maximum speed is at the end point $t = 3$ , where $∣ v ∣ \approx 2.47$ is greater than at any other point. (At the local maximum of $v$ , at $t \approx 0.71$ , the acceleration is $a \approx 0$ , but there the speed is only $\approx 0.29$ , so it is not the maximum speed.) 3. $a (3) = \frac{6 cos 3 - 10 sin 3}{4} \approx - 1.84 m s^{- 2}$ .

Answer: a) $t = 1.57$ s; b) $t = 2.26$ s and $t = 2.96$ s; c) $a = - 1.84 m s^{- 2}$

Exercise 291 HardIB · P3 TZ0 May 2010

a) Using the Maclaurin series for $(1 + x)^{n}$ , write down and simplify the approximation given by the Maclaurin series for $(1 - x^{2})^{- \frac{1}{2}}$ , up to and including the $x^{4}$ term.

b) Use this result to verify that a series approximation for $arccos x$ is $\frac{π}{2} - x - \frac{1}{6} x^{3} - \frac{3}{40} x^{5}$ .

c) Evaluate $x \to 0 lim \frac{\frac{π}{2} - arccos ( x ^{2} ) - x ^{2}}{x ^{6}}$ .

Show answer

Answer: a) $1 + \frac{x ^{2}}{2} + \frac{3 x ^{4}}{8}$ ; b) verified; c) $\frac{1}{6}$

Show working and answer

a) 1. Maclaurin series of the binomial: $(1 + u)^{n} \approx 1 + n u + \frac{n ( n - 1 )}{2} u^{2}$ . 2. With $n = - \frac{1}{2}$ and $u = - x^{2}$ : $\frac{n ( n - 1 )}{2} = \frac{( - \frac{1}{2} ) ( - \frac{3}{2} )}{2} = \frac{3}{8}$ . 3. $(1 - x^{2})^{- \frac{1}{2}} \approx 1 + (- \frac{1}{2}) (- x^{2}) + \frac{3}{8} x^{4} = 1 + \frac{x ^{2}}{2} + \frac{3 x ^{4}}{8}$ .

b) 1. Since $\frac{d}{d x} arccos x = - \frac{1}{1 - x ^{2}}$ , we have $arccos x = - \int \frac{d x}{1 - x ^{2}}$ . 2. Integrating the approximation from (a): $arccos x \approx - (x + \frac{x ^{3}}{6} + \frac{3 x ^{5}}{40}) + C$ . 3. At $x = 0$ : $arccos 0 = \frac{π}{2} = C$ . 4. Hence $arccos x \approx \frac{π}{2} - x - \frac{1}{6} x^{3} - \frac{3}{40} x^{5}$ , as required.

c) 1. We substitute $x^{2}$ for $x$ in the series from (b): $arccos (x^{2}) \approx \frac{π}{2} - x^{2} - \frac{x ^{6}}{6} - \frac{3 x ^{10}}{40}$ . 2. The numerator is $\frac{π}{2} - arccos (x^{2}) - x^{2} \approx \frac{x ^{6}}{6} + \frac{3 x ^{10}}{40}$ . 3. Dividing by $x^{6}$ : $\frac{1}{6} + \frac{3 x ^{4}}{40} \to \frac{1}{6}$ as $x \to 0$ .

Answer: a) $1 + \frac{x ^{2}}{2} + \frac{3 x ^{4}}{8}$ ; b) verified; c) $\frac{1}{6}$

Exercise 292 HardIB · P1 TZ2 May 2009

The complex number $z$ is defined as $z = cos θ + i sin θ$ .

a) State De Moivre's theorem.

b) Verify that $z^{n} - \frac{1}{z ^{n}} = 2 i sin (n θ)$ .

c) Use the binomial theorem to expand $(z - \frac{1}{z})^{5}$ , giving your answer in simplified form.

d) Hence, show that $16 sin^{5} θ = sin 5 θ - 5 sin 3 θ + 10 sin θ$ .

e) Verify that the result obtained in part (d) is valid for $θ = \frac{π}{4}$ .

f) Find $\int_{0}^{\frac{π}{2}} sin^{5} θ d θ$ .

g) Hence, with reference to graphs of circular functions, find $\int_{0}^{\frac{π}{2}} cos^{5} θ d θ$ , and explain your reasoning.

Show answer

Answer: a) $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ ; b) shown; c) $z^{5} - 5 z^{3} + 10 z - \frac{10}{z} + \frac{5}{z ^{3}} - \frac{1}{z ^{5}}$ ; d) shown; e) both sides equal $22$ ; f) $\frac{8}{15}$ ; g) $\frac{8}{15}$

Show working and answer

a) 1. De Moivre's theorem: $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ , that is $(r cis θ)^{n} = r^{n} cis (n θ)$ .

b) 1. By De Moivre, $z^{n} = cos n θ + i sin n θ$ . 2. And $\frac{1}{z ^{n}} = z^{- n} = cos (- n θ) + i sin (- n θ) = cos n θ - i sin n θ$ . 3. Subtracting: $z^{n} - \frac{1}{z ^{n}} = 2 i sin (n θ)$ .

c) 1. Binomial theorem with coefficients $1, 5, 10, 10, 5, 1$ : $(z - \frac{1}{z})^{5} = z^{5} - 5 z^{4} \cdot \frac{1}{z} + 10 z^{3} \cdot \frac{1}{z ^{2}} - 10 z^{2} \cdot \frac{1}{z ^{3}} + 5 z \cdot \frac{1}{z ^{4}} - \frac{1}{z ^{5}}$ . 2. Simplifying: $z^{5} - 5 z^{3} + 10 z - \frac{10}{z} + \frac{5}{z ^{3}} - \frac{1}{z ^{5}}$ .

d) 1. Grouping (c) in pairs: $(z^{5} - \frac{1}{z ^{5}}) - 5 (z^{3} - \frac{1}{z ^{3}}) + 10 (z - \frac{1}{z})$ . 2. By (b), this is $2 i sin 5 θ - 10 i sin 3 θ + 20 i sin θ$ . 3. On the other hand, by (b) with $n = 1$ : $(z - \frac{1}{z})^{5} = (2 i sin θ)^{5} = 32 i^{5} sin^{5} θ = 32 i sin^{5} θ$ . 4. Equating and dividing by $2 i$ : $16 sin^{5} θ = sin 5 θ - 5 sin 3 θ + 10 sin θ$ .

e) 1. With $θ = \frac{π}{4}$ : $sin \frac{π}{4} = \frac{2}{2}$ , so the left-hand side is $16 (\frac{2}{2})^{5} = 16 \cdot \frac{4 2}{32} = 22$ . 2. Right-hand side: $sin \frac{5 π}{4} = - \frac{2}{2}$ , $sin \frac{3 π}{4} = \frac{2}{2}$ , $sin \frac{π}{4} = \frac{2}{2}$ . 3. $- \frac{2}{2} - 5 \cdot \frac{2}{2} + 10 \cdot \frac{2}{2} = 4 \cdot \frac{2}{2} = 22$ . 4. Both sides equal $22$ ✓.

f) 1. By (d): $sin^{5} θ = \frac{1}{16} (sin 5 θ - 5 sin 3 θ + 10 sin θ)$ . 2. $\int_{0}^{\frac{π}{2}} sin^{5} θ d θ = \frac{1}{16} [- \frac{cos 5 θ}{5} + \frac{5 cos 3 θ}{3} - 10 cos θ]_{0}^{\frac{π}{2}}$ . 3. At $θ = \frac{π}{2}$ the three cosines ( $cos \frac{5 π}{2}$ , $cos \frac{3 π}{2}$ , $cos \frac{π}{2}$ ) are all $0$ . At $θ = 0$ : $- \frac{1}{5} + \frac{5}{3} - 10 = - \frac{128}{15}$ . 4. This gives $\frac{1}{16} (0 + \frac{128}{15}) = \frac{8}{15}$ .

g) 1. Since $cos θ = sin (\frac{π}{2} - θ)$ , the graph of $y = cos θ$ on $[0, \frac{π}{2}]$ is the graph of $y = sin θ$ reflected in the line $θ = \frac{π}{4}$ . 2. The same holds for $y = cos^{5} θ$ and $y = sin^{5} θ$ , so the areas under both curves on $[0, \frac{π}{2}]$ are equal. 3. Therefore $\int_{0}^{\frac{π}{2}} cos^{5} θ d θ = \frac{8}{15}$ .

Exercise 293 HardIB · P2 TZ2 May 2010

The functions $f$ , $g$ and $h$ are defined as follows.

$f (x) = 1 + e^{x}$ , for $x \in R$

$g (x) = \frac{1}{x}$ , for $x \in R - {0}$

$h (x) = sec x$ , for $x \in R - {\frac{2 n + 1}{2} π, n \in Z}$

a) Find the range of the composite function $g \circ f$ .

b) Find the inverse of the function $g \circ f$ , clearly stating its domain.

c) (i) Verify that the function $y = (f \circ g \circ h) (x)$ satisfies the differential equation $\frac{d y}{d x} = (1 - y) sin x$ .

(ii) Hence, or otherwise, find $\int y sin x d x$ , in terms of $x$ .

(iii) It is said that the domain of $y = (f \circ g \circ h) (x)$ can be extended to the whole real line. The part of the graph of $y = (f \circ g \circ h) (x)$ between the maximum at $x = 0$ and the first minimum with $x$ positive is rotated through $2 π$ about the $y$ -axis. Calculate the volume of the solid of revolution generated.

Show answer

Answer: a) $] 0, 1 [$ ; b) $(g \circ f)^{- 1} (x) = ln \frac{1 - x}{x}$ , domain $] 0, 1 [$ ; c) (i) verified, $y = 1 + e^{c o s x}$ ; (ii) $- cos x - e^{c o s x} + C$ ; (iii) $V \approx 13.6$

Show working and answer

a) 1. $(g \circ f) (x) = g (f (x)) = \frac{1}{1 + e ^{x}}$ . 2. Since $e^{x} > 0$ , the denominator $1 + e^{x}$ takes values in $] 1, + \infty [$ , and the quotient takes values in $] 0, 1 [$ . 3. Range of $g \circ f$ : $] 0, 1 [$ .

b) 1. Let $y = \frac{1}{1 + e ^{x}}$ . Solving for $x$ : $1 + e^{x} = \frac{1}{y}$ , so $e^{x} = \frac{1 - y}{y}$ and $x = ln \frac{1 - y}{y}$ . 2. The inverse is $(g \circ f)^{- 1} (x) = ln \frac{1 - x}{x}$ , with domain equal to the range of $g \circ f$ : $] 0, 1 [$ .

c) (i) 1. $(f \circ g \circ h) (x) = f (g (sec x)) = 1 + e^{\frac{1}{s e c x}} = 1 + e^{c o s x}$ . That is, $y = 1 + e^{c o s x}$ . 2. $\frac{d y}{d x} = - sin x e^{c o s x}$ . 3. On the other hand, $1 - y = - e^{c o s x}$ , so $(1 - y) sin x = - e^{c o s x} sin x$ ✓. They agree.

(ii) 1. Rearranging (i): $y sin x = sin x - (1 - y) sin x = sin x - \frac{d y}{d x}$ . 2. Integrating: $\int y sin x d x = - cos x - y + C = - cos x - 1 - e^{c o s x} + C$ . 3. Absorbing the constant: $- cos x - e^{c o s x} + C$ .

(iii) 1. $y = 1 + e^{c o s x}$ has its maximum at $x = 0$ ( $y = 1 + e$ ) and its first minimum for $x > 0$ at $x = π$ ( $y = 1 + e^{- 1}$ ), and it is decreasing between the two. 2. When rotated about the $y$ -axis, the solid lies between the $y$ -axis and the curve, from $y = 1 + e^{- 1}$ to $y = 1 + e$ . Solving for $x$ : $e^{c o s x} = y - 1$ , so $cos x = ln (y - 1)$ and $x = arccos (ln (y - 1))$ . 3. $V = π \int_{1 + e^{- 1}}^{1 + e} x^{2} d y = π \int_{1 + e^{- 1}}^{1 + e} [arccos (ln (y - 1))]^{2} d y$ . 4. Using the GDC: $V \approx 13.6$ .

Answer: a) $] 0, 1 [$ ; b) $(g \circ f)^{- 1} (x) = ln \frac{1 - x}{x}$ , domain $] 0, 1 [$ ; c) (i) verified, $y = 1 + e^{c o s x}$ ; (ii) $- cos x - e^{c o s x} + C$ ; (iii) $V \approx 13.6$

Exercise 294 HardIB · P2 TZ0 November 2008

The function $f$ is defined by $f (x) = x 9 - x^{2} + 2 arcsin (\frac{x}{3})$ .

a) Write down the largest possible domain for each of the two terms of the function $f$ and, hence, state the largest possible domain, $D$ , for $f$ .

b) Find the volume generated when the region bounded by the curve $y = x 9 - x^{2}$ , the $x$ -axis, the $y$ -axis and the line $x = 2$ is rotated through $2 π$ radians about the $x$ -axis.

c) Find $f^{'} (x)$ , giving the result in simplified form.

d) Hence, verify that: $\int_{- p}^{p} \frac{11 - 2 x ^{2}}{9 - x ^{2}} d x = 2 p 9 - p^{2} + 4 arcsin (\frac{p}{3})$ , $p \in D$ .

e) Find the value of $p$ for which the value of the integral in (d) is a maximum.

f) (i) Verify that $f^{''} (x) = \frac{x ( 2 x ^{2} - 25 )}{( 9 - x ^{2} ) ^{\frac{3}{2}}}$ .

(ii) Hence, justify why $f (x)$ has a point of inflexion at $x = 0$ , but not at $x = \pm \frac{25}{2}$ .

Show answer

Answer: a) $D = [- 3, 3]$ ; b) $V = \frac{88 π}{5} \approx 55.3$ ; c) $f^{'} (x) = \frac{11 - 2 x ^{2}}{9 - x ^{2}}$ ; d) proved; e) $p = \frac{11}{2}$ ; f) proved: inflexion at $x = 0$ and not at $\pm \frac{25}{2}$ .

Show working and answer

a) 1. First term, $x 9 - x^{2}$ : requires $9 - x^{2} \geq 0$ , i.e. $- 3 \leq x \leq 3$ . 2. Second term, $2 arcsin \frac{x}{3}$ : requires $- 1 \leq \frac{x}{3} \leq 1$ , i.e. $- 3 \leq x \leq 3$ . 3. The domain of $f$ is the intersection: $D = [- 3, 3]$ .

b) 1. Volume about the $x$ -axis between $x = 0$ and $x = 2$ (the curve is positive there): $V = π \int_{0}^{2} x^{2} (9 - x^{2}) d x$ . 2. $V = π \int_{0}^{2} (9 x^{2} - x^{4}) d x = π [3 x^{3} - \frac{x ^{5}}{5}]_{0}^{2} = π (24 - \frac{32}{5}) = \frac{88 π}{5} \approx 55.3$ .

c) 1. Differentiate the first term using the product rule: $9 - x^{2} + x \cdot \frac{- x}{9 - x ^{2}} = 9 - x^{2} - \frac{x ^{2}}{9 - x ^{2}}$ . 2. Differentiate the second term: $2 \cdot \frac{1}{1 - \frac{x ^{2}}{9}} \cdot \frac{1}{3} = \frac{2}{9 - x ^{2}}$ . 3. Using a common denominator: $f^{'} (x) = \frac{9 - x ^{2} - x ^{2} + 2}{9 - x ^{2}} = \frac{11 - 2 x ^{2}}{9 - x ^{2}}$ .

d) 1. By the fundamental theorem of calculus, for $p \in D$ : $\int_{- p}^{p} f^{'} (x) d x = f (p) - f (- p)$ . 2. $f$ is odd (sum of two odd functions), so $f (- p) = - f (p)$ and the integral equals $2 f (p)$ . 3. $2 f (p) = 2 p 9 - p^{2} + 4 arcsin \frac{p}{3}$ ✓.

e) 1. Let $I (p) = 2 f (p)$ . Then $I^{'} (p) = 2 f^{'} (p) = \frac{2 ( 11 - 2 p ^{2} )}{9 - p ^{2}}$ , for $0 \leq p < 3$ . 2. $I^{'} (p) = 0$ when $11 - 2 p^{2} = 0$ , i.e. $p = \frac{11}{2} \approx 2.35$ (which lies in $] 0, 3 [$ ). 3. For $p < \frac{11}{2}$ , $I^{'} > 0$ (the integral is increasing); for $p > \frac{11}{2}$ , $I^{'} < 0$ (it is decreasing). It is a maximum.

f) (i) 1. Write $f^{'} (x) = (11 - 2 x^{2}) (9 - x^{2})^{- \frac{1}{2}}$ and differentiate using the product rule. 2. $f^{''} (x) = - 4 x (9 - x^{2})^{- \frac{1}{2}} + (11 - 2 x^{2}) \cdot x (9 - x^{2})^{- \frac{3}{2}}$ . 3. Taking out the common factor $x (9 - x^{2})^{- \frac{3}{2}}$ : $f^{''} (x) = \frac{x [ - 4 ( 9 - x ^{2} ) + 11 - 2 x ^{2} ]}{( 9 - x ^{2} ) ^{\frac{3}{2}}} = \frac{x ( 2 x ^{2} - 25 )}{( 9 - x ^{2} ) ^{\frac{3}{2}}}$ ✓.

(ii) 1. $f^{''} (x) = 0$ when $x = 0$ or $2 x^{2} - 25 = 0$ , i.e. $x = \pm \frac{25}{2} \approx \pm 3.54$ . 2. At $x = 0$ : inside $D$ , the factor $2 x^{2} - 25 < 0$ and the denominator is positive; so $f^{''}$ has the opposite sign to $x$ : $f^{''} > 0$ for $x < 0$ and $f^{''} < 0$ for $x > 0$ . $f^{''}$ changes sign, so there is a point of inflexion. 3. At $x = \pm \frac{25}{2} \approx \pm 3.54$ : these lie outside the domain $D = [- 3, 3]$ (and $f^{''}$ is not defined there, since $9 - x^{2} < 0$ ). They cannot be points of inflexion of $f$ .

Exercise 295 HardIB · P2 TZ2 May 2014

A particle $A$ moves so that its velocity $v m s^{- 1}$ , at time $t$ seconds, is given by $v (t) = \frac{t}{12 + t ^{4}}$ , $t \geq 0$ .

a) Sketch the graph of $y = v (t)$ approximately. Clearly indicate the local maximum and write down its coordinates.

b) Use the substitution $u = t^{2}$ to find $\int \frac{t}{12 + t ^{4}} d t$ .

c) Find the exact distance travelled by particle $A$ between $t = 0$ and $t = 6$ seconds. Give your answer in the form $k arctan (b)$ , $k, b \in R$ .

Particle $B$ moves in such a way that its velocity $v m s^{- 1}$ and its displacement $s$ m are related by the equation $v (s) = arcsin (s)$ .

d) Find the acceleration of particle $B$ when $s = 0.1$ m.

View solution

a) 1. We differentiate: $v^{'} (t) = \frac{( 12 + t ^{4} ) - t \cdot 4 t ^{3}}{( 12 + t ^{4} ) ^{2}} = \frac{12 - 3 t ^{4}}{( 12 + t ^{4} ) ^{2}}$ . 2. $v^{'} (t) = 0 \Rightarrow t^{4} = 4 \Rightarrow t = 2 \approx 1.4142$ (with $t \geq 0$ ). Before this value $v^{'} > 0$ and after it $v^{'} < 0$ , so it is a maximum. 3. $v (2) = \frac{2}{16} \approx 0.088$ . 4. The graph starts at $(0, 0)$ , rises to the local maximum and then decreases slowly towards $0$ (the $t$ -axis is a horizontal asymptote).

Graph of the velocity v(t) = t/(12 + t⁴) for t between 0 and 3.5: starts at the origin, rises to a local maximum and then decreases slowly towards 0

Local maximum: $(1.4142; 0.088)$ .

b) Substitution $u = t^{2}$ .

$d u = 2 t d t$ , i.e. $t d t = \frac{1}{2} d u$ . Also $t^{4} = u^{2}$ .
$\int \frac{t}{12 + t ^{4}} d t = \frac{1}{2} \int \frac{d u}{12 + u ^{2}} = \frac{1}{2} \cdot \frac{1}{12} arctan \frac{u}{12} + C$ .
As $\frac{1}{2 12} = \frac{1}{4 3} = \frac{3}{12}$ : $\int \frac{t}{12 + t ^{4}} d t = \frac{3}{12} arctan (\frac{t ^{2}}{12}) + C$ .

c) 1. As $v (t) \geq 0$ for $t \geq 0$ , the distance is $\int_{0}^{6} v (t) d t$ . 2. $= \frac{3}{12} [arctan \frac{t ^{2}}{12}]_{0}^{6} = \frac{3}{12} arctan \frac{36}{2 3} - 0$ . 3. As $\frac{36}{2 3} = \frac{18}{3} = 63$ , the distance is $\frac{3}{12} arctan (63)$ m, with $k = \frac{3}{12}$ and $b = 63$ .

d) 1. The acceleration is $a = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ (chain rule). 2. $\frac{d v}{d s} = \frac{1}{1 - s} \cdot \frac{1}{2 s}$ . 3. At $s = 0.1$ : $v = arcsin 0.1 \approx 0.32175$ and $\frac{d v}{d s} = \frac{1}{0.9 \cdot 2 0.1} \approx 1.6667$ . 4. $a \approx 0.32175 \times 1.6667 \approx 0.5362 m s^{- 2}$ .

Answer: a) local maximum $(1.4142; 0.088)$ ; b) $\frac{3}{12} arctan (\frac{t ^{2}}{12}) + C$ ; c) $\frac{3}{12} arctan (63)$ m; d) $0.5362 m s^{- 2}$

Exercise 296 HardIB · P2 TZ0 November 2013

a) A particle $P$ moves in a straight line with velocity $v m s^{- 1}$ . At the instant $t = 0$ , $P$ is at the point $Q$ and has a velocity of $12 m s^{- 1}$ . Its acceleration at time $t$ seconds is given by $\frac{d v}{d t} = 3 cos \frac{t}{4} m s^{- 2}$ , $(t \geq 0)$ .

(i) Find an expression for the velocity of the particle, $v$ , in terms of $t$ .

(ii) Sketch the graph of the velocity of the particle against time for $0 \leq t \leq 8 π$ , clearly showing the points where the curve meets the axes and all maximum and minimum points.

(iii) Find the distance travelled by the particle before it first comes to rest.

b) Another particle $Q$ moves in a straight line, its displacement being $s$ metres and its velocity $v m s^{- 1}$ . Its acceleration is given by $a = - (v^{2} + 4) m s^{- 2}$ , $(0 \leq t \leq 1)$ . At $t = 0$ , $Q$ is at the point $O$ and has a velocity of $2 m s^{- 1}$ .

(i) Verify that the velocity $v$ at time $t$ is given by $v = 2 tan (\frac{π - 8 t}{4})$ .

(ii) Verify that $\frac{d v}{d s} = - \frac{( v ^{2} + 4 )}{v}$ .

(iii) Find the distance travelled by the particle before it comes to rest.

View solution

a) (i) 1. Integrating the acceleration: $v = \int 3 cos \frac{t}{4} d t = 12 sin \frac{t}{4} + C$ . 2. At $t = 0$ , $v = 12$ , so $C = 12$ and $v = 12 sin (\frac{t}{4}) + 12$ .

(ii) 1. It is a sinusoid with amplitude $12$ , period $8 π$ , shifted up by $12$ : its value oscillates between $0$ and $24$ . 2. Intercept with the $v$ -axis: $(0, 12)$ . Maximum: $sin \frac{t}{4} = 1 \Rightarrow t = 2 π$ , $v = 24$ . Minimum: $sin \frac{t}{4} = - 1 \Rightarrow t = 6 π$ , $v = 0$ (the curve touches the $t$ -axis without crossing it). At $t = 8 π$ , $v = 12$ .

Graph of the velocity v against time t for t between 0 and 8π: a sinusoidal curve of period 8π starting at the point (0, 12), with a maximum, a minimum touching the t-axis, and the points labelled

Intercepts: $(0, 12)$ and $(6 π, 0)$ . Maximum: $(2 π, 24)$ . Minimum: $(6 π, 0)$ .

(iii) 1. The particle comes to rest when $v = 0$ : $sin \frac{t}{4} = - 1$ , first at $\frac{t}{4} = \frac{3 π}{2}$ , that is, $t = 6 π$ . 2. Up to that instant $v \geq 0$ , so the distance is $\int_{0}^{6 π} (12 sin \frac{t}{4} + 12) d t = [- 48 cos \frac{t}{4} + 12 t]_{0}^{6 π}$ . 3. $= (- 48 cos \frac{3 π}{2} + 72 π) - (- 48) = 72 π + 48 \approx 274.2$ m.

b) (i) 1. Separating variables in $\frac{d v}{d t} = - (v^{2} + 4)$ : $\int \frac{d v}{v ^{2} + 4} = - \int d t$ . 2. $\frac{1}{2} arctan \frac{v}{2} = - t + C$ . 3. At $t = 0$ , $v = 2$ : $\frac{1}{2} \cdot \frac{π}{4} = C$ , so $arctan \frac{v}{2} = \frac{π}{4} - 2 t = \frac{π - 8 t}{4}$ . 4. Solving for $v$ : $v = 2 tan (\frac{π - 8 t}{4})$ .

(ii) By the chain rule, $\frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ . As $\frac{d v}{d t} = - (v^{2} + 4)$ , it follows that $\frac{d v}{d s} = - \frac{v ^{2} + 4}{v}$ .

(iii) 1. The particle comes to rest when $v = 0$ , which occurs when $\frac{π - 8 t}{4} = 0$ , that is, $t = \frac{π}{8} \approx 0.39$ s (within $0 \leq t \leq 1$ ); before that $v > 0$ . 2. Separating variables in the result of (ii): $\int \frac{v}{v ^{2} + 4} d v = - \int d s$ , that is, $\frac{1}{2} ln (v^{2} + 4) = - s + K$ . 3. At $s = 0$ , $v = 2$ : $K = \frac{1}{2} ln 8$ . 4. At $v = 0$ : $s = \frac{1}{2} ln 8 - \frac{1}{2} ln 4 = \frac{1}{2} ln 2 \approx 0.347$ m.

Answer: a) (i) $v = 12 sin (\frac{t}{4}) + 12$ ; (ii) intercepts $(0, 12)$ and $(6 π, 0)$ , maximum $(2 π, 24)$ , minimum $(6 π, 0)$ ; (iii) $72 π + 48 \approx 274.2$ m; b) (i) and (ii) verified; (iii) $\frac{ln 2}{2} \approx 0.347$ m

Exercise 297 HardIB · P2 TZ1 May 2013

A particle $A$ moves in a straight line from a point $O_{A}$ , such that its velocity, in metres per second, for $0 \leq t \leq 9$ is given by $v_{A} = - \frac{1}{2} t^{2} + 3 t + \frac{3}{2}$ . A particle $B$ moves in a straight line from a point $O_{B}$ , such that its velocity, in metres per second, for $0 \leq t \leq 9$ is given by $v_{B} = e^{0.2 t}$ .

a) Calculate the maximum value of $v_{A}$ .

b) Calculate the acceleration of $B$ when $t = 4$ .

The displacements of $A$ and $B$ from $O_{A}$ and $O_{B}$ respectively are $s_{A}$ metres and $s_{B}$ metres at time $t$ . When $t = 0$ , $s_{A} = 0$ and $s_{B} = 5$ .

c) Calculate expressions for $s_{A}$ and for $s_{B}$ , giving your answers in terms of $t$ .

d) i) Draw the graphs of $s_{A}$ and $s_{B}$ on a diagram.

(ii) Calculate the values of $t$ for which $s_{A} = s_{B}$ .

View solution

a) $v_{A} = - \frac{1}{2} t^{2} + 3 t + \frac{3}{2}$ is a parabola opening downwards; its vertex is at $t = - \frac{3}{2 \cdot ( - 1/2 )} = 3$ , which lies within $[0, 9]$ . Maximum value: $v_{A} (3) = - \frac{9}{2} + 9 + \frac{3}{2} = 6$ . (At the endpoints: $v_{A} (0) = 1.5$ and $v_{A} (9) = - 12$ .)

b) 1. $a_{B} = \frac{d v _{B}}{d t} = 0.2 e^{0.2 t}$ . 2. At $t = 4$ : $a_{B} = 0.2 e^{0.8} \approx 0.445 m s^{- 2}$ .

c) 1. $s_{A} = \int v_{A} d t = - \frac{t ^{3}}{6} + \frac{3 t ^{2}}{2} + \frac{3 t}{2} + C$ , and with $s_{A} (0) = 0$ we get $C = 0$ . 2. $s_{B} = \int e^{0.2 t} d t = 5 e^{0.2 t} + C$ , and with $s_{B} (0) = 5$ we get $C = 0$ : $s_{B} = 5 e^{0.2 t}$ .

d) (i) The curve $s_{A}$ starts at $(0, 0)$ , rises to a maximum near $t = 6.5$ (a value close to $27$ ) and falls to $s_{A} (9) = 13.5$ . The curve $s_{B}$ starts at $(0, 5)$ and increases exponentially to $s_{B} (9) \approx 30$ . They intersect twice.

Graphs of s_A and s_B against t for t between 0 and 9: s_A starts at 0, reaches a maximum close to 27 and falls; s_B starts at 5 and increases; they intersect at two points

(ii) Using the GDC, we solve $s_{A} = s_{B}$ on $[0, 9]$ : $t \approx 1.95$ and $t \approx 7.81$ .

Answer: a) $v_{A} = 6$ for $t = 3$ ; b) $a_{B} (4) = 0.2 e^{0.8} \approx 0.445 m s^{- 2}$ ; c) $s_{A} = - \frac{t ^{3}}{6} + \frac{3 t ^{2}}{2} + \frac{3 t}{2}$ , $s_{B} = 5 e^{0.2 t}$ ; d) (i) graphs; (ii) $t \approx 1.95$ and $t \approx 7.81$

Exercise 298 HardIB · P2 TZ1 May 2019

The voltage $v$ in a circuit is given by the equation $v (t) = 3 sin (100 π t)$ , $t \geq 0$ , where $t$ is measured in seconds.

a) Write down the maximum and minimum values of $v$ .

The current $i$ in this circuit is given by the equation $i (t) = 2 sin (100 π (t + 0.003))$ .

b) Write down two transformations that will transform the graph of $y = v (t)$ into the graph of $y = i (t)$ .

The power $p$ in this circuit is given by $p (t) = v (t) \times i (t)$ .

c) Sketch the graph of $y = p (t)$ for $0 \leq t \leq 0.02$ , clearly showing the coordinates of the first maximum and the first minimum.

d) Find the total time in the interval $0 \leq t \leq 0.02$ for which $p (t) \geq 3$ .

The average power $p_{a v}$ in this circuit between $t = 0$ and $t = T$ is given by the equation $p_{a v} (T) = \frac{1}{T} \int_{0}^{T} p (t) d t$ , where $T > 0$ .

e) Calculate $p_{a v} (0.007)$ .

f) With reference to your graph of $y = p (t)$ , explain why $p_{a v} (T) > 0$ for all $T > 0$ .

g) Given that $p (t)$ can be written as $p (t) = a sin (b (t - c)) + d$ where $a, b, c, d > 0$ , use your graph to find the values of $a$ , $b$ , $c$ and $d$ .

View solution

a) $v (t) = 3 sin (100 π t)$ oscillates between $- 3$ and $3$ : maximum $3$ , minimum $- 3$ .

b) 1. $i (t) = 2 sin (100 π (t + 0.003))$ : the factor $t + 0.003$ is a horizontal translation of $0.003$ units to the left, with vector $(- 0.003 0)$ . 2. The amplitude changes from $3$ to $2$ : a vertical stretch with scale factor $\frac{2}{3}$ .

c) 1. Using the GDC, we plot $p (t) = 3 sin (100 π t) \times 2 sin (100 π (t + 0.003))$ for $0 \leq t \leq 0.02$ . 2. The graph oscillates with period $0.01$ about a positive level (see part g), $p (0) = 0$ , it is positive until $t = 0.007$ , negative between $0.007$ and $0.01$ , and then repeats. 3. The first maximum is at $(0.0035; 4.76)$ and the first minimum at $(0.0085; - 1.24)$ .

Graph of the power p(t) for t between 0 and 0.02: upward-shifted sinusoidal curve starting at (0, 0), with first maximum at (0.0035; 4.76), first minimum at (0.0085; −1.24) and repeating a second time within the interval

d) 1. Using the GDC, we solve $p (t) = 3$ (the intersection of the graph with the line $y = 3$ ) for $0 \leq t \leq 0.02$ . 2. We obtain $p (t) \geq 3$ for $0.00168 \leq t \leq 0.00532$ and for $0.01168 \leq t \leq 0.01532$ . 3. Total time: $2 \times 0.00364 \approx 0.0073$ s.

e) $p_{a v} (0.007) = \frac{1}{0.007} \int_{0}^{0.007} p (t) d t \approx 2.87$ (integral calculated using the GDC).

f) The graph of $p$ oscillates between $- 1.24$ and $4.76$ , that is, it has more height above the axis than below it. The negative sections are short ( $0.007 < t < 0.01$ , etc.) and each one is preceded by a larger positive section. Thus, for any $T > 0$ , the area above the $t$ -axis is greater than the area below it, the integral $\int_{0}^{T} p d t$ is positive and $p_{a v} (T) > 0$ .

g) 1. Amplitude: $a = \frac{4.76 - ( - 1.24 )}{2} = 3$ . Central level: $d = \frac{4.76 + ( - 1.24 )}{2} = 1.76$ . 2. Period $0.01$ : $b = \frac{2 π}{0.01} = 200 π$ . 3. The curve passes through the central level going upwards at $t = 0.001$ (a quarter of a period, $0.0025$ , before the first maximum at $0.0035$ ): $c = 0.001$ . 4. (This can be checked with the identity $2 sin A sin B = cos (A - B) - cos (A + B)$ : $p (t) = 3 cos (0.3 π) - 3 cos (200 π t + 0.3 π)$ .)

Answer: a) maximum $v = 3$ ; minimum $v = - 3$ ; b) translation with vector $(- 0.003 0)$ and vertical stretch with scale factor $\frac{2}{3}$ ; c) first maximum $(0.0035; 4.76)$ and first minimum $(0.0085; - 1.24)$ ; d) $0.0017 \leq t \leq 0.0053$ and $0.0117 \leq t \leq 0.0153$ , total $\approx 0.0073$ s; e) $p_{a v} (0.007) \approx 2.87$ ; f) explained; g) $a = 3$ , $b = 200 π$ , $c = 0.001$ , $d = 1.76$

Exercise 299 HardIB · P2 TZ1 May 2017

Xavier, the parachutist, jumps from an aeroplane at a height of $h$ metres above the ground. After free fall for $10$ seconds he opens his parachute. His velocity, $v m s^{- 1}$ , $t$ seconds after jumping from the aeroplane, can be modelled by the function

$v (t) = ⎩ ⎨ ⎧ 9.8 t, \frac{98}{1 + ( t - 10 ) ^{2}}, 0 \leq t \leq 10 t > 10$

a) Find his velocity when $t = 15$ .

b) Calculate the vertical distance that Xavier travelled in the first $10$ seconds.

His velocity when he reaches the ground is $2.8 m s^{- 1}$ .

c) Determine the value of $h$ .

Show answer

Answer: a) $v (15) = 19.2 m s^{- 1}$ ; b) distance $= 490$ m; c) $h = 906.3$ m

Show working and answer

a) 1. As $15 > 10$ , we use the second branch: $v (15) = \frac{98}{1 + 5 ^{2}} = \frac{98}{26} \approx 19.2$ .

b) 1. The distance is the integral of the velocity: $\int_{0}^{10} 9.8 t d t = [4.9 t^{2}]_{0}^{10} = 490$ .

c) 1. Find the instant $T$ at which the velocity is $2.8$ : $\frac{98}{1 + ( T - 10 ) ^{2}} = 2.8$ , so $1 + (T - 10)^{2} = 35$ and $(T - 10)^{2} = 1224$ . As $T > 10$ , $T = 10 + 1224 \approx 44.99$ . 2. The total distance is $h = 490 + \int_{10}^{T} \frac{98}{1 + ( t - 10 ) ^{2}} d t$ . 3. Using the GDC (or the antiderivative $98 arsinh (t - 10)$ ): the integral is $\approx 416.3$ . 4. $h \approx 490 + 416.3 = 906.3$ m.

Answer: a) $v (15) = 19.2 m s^{- 1}$ ; b) distance $= 490$ m; c) $h = 906.3$ m

Exercise 300 HardIB · P2 November 2001

(i) Let $y = sin (k x) - k x cos (k x)$ , where $k$ is a constant.

Show that $\frac{d y}{d x} = k^{2} x sin (k x)$ .

(ii) A particle moves in a straight line so that $t$ seconds after passing a fixed point $O$ its velocity $v (t) m s^{- 1}$ is given by $v (t) = t sin (\frac{π}{3} t)$ .

a) Calculate the values of $t$ for which $v (t) = 0$ , where $0 \leq t \leq 6$ .

b) i) Write down a mathematical expression for the total distance travelled by the particle in the first six seconds after passing $O$ .

(ii) Calculate the distance.

Show answer

Answer: ii) a) $t = 3$ and $t = 6$ ; b) i) $\int_{0}^{3} t sin (\frac{π}{3} t) d t + \int_{3}^{6} - t sin (\frac{π}{3} t) d t$ ; ii) $\frac{36}{π}$ m

Show working and answer

(i) 1. We differentiate each term with respect to $x$ ; the second is a product: $\frac{d y}{d x} = k cos (k x) - [k cos (k x) + k x \cdot (- k sin (k x))]$ . 2. Simplifying, $\frac{d y}{d x} = k cos (k x) - k cos (k x) + k^{2} x sin (k x) = k^{2} x sin (k x)$ .

(ii) a) 1. We set $v (t) = t sin (\frac{π}{3} t) = 0$ , that is, $t = 0$ or $sin (\frac{π}{3} t) = 0$ . 2. Since $0 \leq t \leq 6$ , $\frac{π}{3} t \in [0, 2 π]$ and the sine vanishes at $0, π, 2 π$ , that is, $t = 0$ , $t = 3$ and $t = 6$ . Leaving aside the start at $O$ ( $t = 0$ ), the required instants are $t = 3$ and $t = 6$ .

b) i) 1. The sign of $v$ on $[0, 6]$ : for $0 < t < 3$ , $sin (\frac{π}{3} t) > 0$ and $v > 0$ ; for $3 < t < 6$ , $v < 0$ . 2. The total distance is the integral of the absolute value of the velocity, so we change the sign on the second interval: $\int_{0}^{3} t sin (\frac{π}{3} t) d t + \int_{3}^{6} - t sin (\frac{π}{3} t) d t$

(ii) 1. Using part (i) with $k = \frac{π}{3}$ we have $\int t sin (k t) d t = \frac{sin ( k t ) - k t cos ( k t )}{k ^{2}} =: F (t)$ , that is, $F (t) = \frac{9}{π ^{2}} [sin (\frac{π t}{3}) - \frac{π t}{3} cos (\frac{π t}{3})]$ . 2. $F (0) = 0$ , $F (3) = \frac{9}{π ^{2}} [0 - π \cdot (- 1)] = \frac{9}{π}$ and $F (6) = \frac{9}{π ^{2}} [0 - 2 π \cdot 1] = - \frac{18}{π}$ . 3. Distance $= (F (3) - F (0)) - (F (6) - F (3)) = \frac{9}{π} - (- \frac{18}{π} - \frac{9}{π}) = \frac{9}{π} + \frac{27}{π} = \frac{36}{π}$ m.

Answer: ii) a) $t = 3$ and $t = 6$ ; b) i) $\int_{0}^{3} t sin (\frac{π}{3} t) d t + \int_{3}^{6} - t sin (\frac{π}{3} t) d t$ ; ii) $\frac{36}{π}$ m

Exercise 301 HardIB · P2 TZ0 May 2006

Particle $A$ moves in a straight line from the point $O_{A}$ , such that its velocity in metres per second for $0 \leq t \leq 9$ is given by $v_{A} = - \frac{1}{2} t^{2} + 3 t + \frac{3}{2}$ .

Particle $B$ moves in a straight line from the point $O_{B}$ , such that its velocity in metres per second for $0 \leq t \leq 9$ is given by $v_{B} = e^{0.2 t}$ .

a) Find the maximum value of $v_{A}$ , justifying that it is a maximum.

b) Find the acceleration of $B$ when $t = 4$ .

The displacements of $A$ and $B$ from $O_{A}$ and $O_{B}$ respectively, at time $t$ , are $s_{A}$ and $s_{B}$ metres. When $t = 0$ , $s_{A} = 0$ and $s_{B} = 5$ .

c) Find an expression for $s_{A}$ and for $s_{B}$ , giving your answers in terms of $t$ .

d) i) Sketch, on the same diagram, the graphs of $s_{A}$ and $s_{B}$ .

(ii) Find the values of $t$ for which $s_{A} = s_{B}$ .

View solution

a) 1. $v_{A}^{'} (t) = - t + 3 = 0$ when $t = 3$ . 2. $v_{A}^{''} (t) = - 1 < 0$ , so it is a maximum. 3. $v_{A} (3) = - \frac{9}{2} + 9 + \frac{3}{2} = 6$ m s $^{- 1}$ .

b) 1. $a_{B} = v_{B}^{'} = 0.2 e^{0.2 t}$ . 2. When $t = 4$ : $a_{B} = 0.2 e^{0.8} \approx 0.445$ m s $^{- 2}$ .

c) 1. $s_{A} = \int v_{A} d t = - \frac{t ^{3}}{6} + \frac{3 t ^{2}}{2} + \frac{3 t}{2} + C$ , with $s_{A} (0) = 0$ : $C = 0$ . 2. $s_{B} = \int e^{0.2 t} d t = 5 e^{0.2 t} + C$ , with $s_{B} (0) = 5$ : $5 + C = 5$ , so $C = 0$ .

d) i) $s_{A}$ starts at $(0, 0)$ , rises to a maximum (close to $27$ , at $t \approx 6.5$ , where $v_{A} = 0$ ) and then falls to $13.5$ at $t = 9$ ; $s_{B}$ starts at $(0, 5)$ and increases to about $30$ at $t = 9$ .

Graphs of the positions s_A and s_B as functions of t, for 0 ≤ t ≤ 9, on the same set of axes: s_A starts at the origin and s_B at 5, and they intersect at two points

(ii) 1. Using the GDC, solve $- \frac{t ^{3}}{6} + \frac{3 t ^{2}}{2} + \frac{3 t}{2} = 5 e^{0.2 t}$ in $[0, 9]$ . 2. The two solutions are $t \approx 1.95$ s and $t \approx 7.81$ s.

Answer: a) $6$ m s $^{- 1}$ ; b) $0.2 e^{0.8}$ m s $^{- 2}$ ; c) $s_{A} = - \frac{t ^{3}}{6} + \frac{3 t ^{2}}{2} + \frac{3 t}{2}$ m, $s_{B} = 5 e^{0.2 t}$ m; d) ii) $t \approx 1.95$ s and $t \approx 7.81$ s

Exercise 302 HardIB · P3 TZ0 November 2006

a) Show that the polynomial approximation for $ln x$ on the interval $[0.5, 1.5]$ , obtained by taking the first three non-zero terms of the Taylor series for $x = 1$ , is given by

$ln x \approx \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + 3 x - \frac{11}{6}$ .

b) Given that $\int ln x d x = x ln x - x + C$ , show, by integrating the above series, that another approximation to $ln x$ is given by $ln x \approx \frac{x ^{3}}{12} - \frac{x ^{2}}{2} + \frac{3 x}{2} - \frac{5}{6} - \frac{1}{4 x}$ .

c) Which is the better approximation when $x = 1.5$ ?

Show answer

Answer: a) shown; b) shown; c) with (a) $ln 1.5 \approx 0.41667$ and with (b) $ln 1.5 \approx 0.40625$ ; the better one is (b), since $ln 1.5 \approx 0.4055$ .

Show working and answer

a) 1. Derivatives of $ln x$ at $x = 1$ : $f = ln x \to 0$ ; $f^{'} = \frac{1}{x} \to 1$ ; $f^{''} = - \frac{1}{x ^{2}} \to - 1$ ; $f^{'''} = \frac{2}{x ^{3}} \to 2$ . 2. The first term ( $ln 1$ ) is zero; the first three non-zero terms are: $(x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{2 ( x - 1 ) ^{3}}{3 !} = (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3}$ . 3. Expanding: $(x - 1) - \frac{x ^{2} - 2 x + 1}{2} + \frac{x ^{3} - 3 x ^{2} + 3 x - 1}{3} = \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + 3 x - \frac{11}{6}$ .

b) 1. We integrate both sides of the approximation in (a) between $1$ and $x$ (so that the constant is fixed): $\int_{1}^{x} ln t d t \approx \int_{1}^{x} (\frac{t ^{3}}{3} - \frac{3 t ^{2}}{2} + 3 t - \frac{11}{6}) d t$ . 2. Left side: $[t ln t - t]_{1}^{x} = x ln x - x + 1$ . 3. Right side: $[\frac{t ^{4}}{12} - \frac{t ^{3}}{2} + \frac{3 t ^{2}}{2} - \frac{11 t}{6}]_{1}^{x} = \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} - \frac{11 x}{6} + \frac{3}{4}$ , because at $t = 1$ it equals $\frac{1}{12} - \frac{1}{2} + \frac{3}{2} - \frac{11}{6} = - \frac{3}{4}$ . 4. Hence $x ln x \approx \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} - \frac{11 x}{6} + x - 1 + \frac{3}{4} = \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} - \frac{5 x}{6} - \frac{1}{4}$ . 5. Dividing by $x$ : $ln x \approx \frac{x ^{3}}{12} - \frac{x ^{2}}{2} + \frac{3 x}{2} - \frac{5}{6} - \frac{1}{4 x}$ .

c) 1. Using (a) with $x = 1.5$ : $\frac{3.375}{3} - \frac{3 \cdot 2.25}{2} + 4.5 - \frac{11}{6} \approx 0.41667$ . 2. Using (b): $\frac{3.375}{12} - \frac{2.25}{2} + 2.25 - \frac{5}{6} - \frac{1}{6} = 0.40625$ . 3. The true value is $ln 1.5 \approx 0.40547$ , which is closer to the value from (b). The better approximation is (b).

Answer: a) shown; b) shown; c) with (a) $ln 1.5 \approx 0.41667$ and with (b) $ln 1.5 \approx 0.40625$ ; the better one is (b), since $ln 1.5 \approx 0.4055$ .

Exercise 303 HardIB · P2 TZ0 November 2015

A particle can move along a straight line starting from a point $O$ .

The velocity $v$ , in $m s^{- 1}$ , is given by the function $v (t) = 1 - e^{- s i n t^{2}}$ , where the time $t \geq 0$ is measured in seconds.

a) Write down the first two times $t_{1}, t_{2} > 0$ at which the particle changes direction.

b) i) Find the time $t < t_{2}$ at which the particle has a maximum velocity.

(ii) Find the time $t < t_{2}$ at which the particle has a minimum velocity.

c) Find the distance travelled by the particle between the times $t = t_{1}$ and $t = t_{2}$ .

Show answer

Answer: a) $t_{1} \approx 1.77$ s, $t_{2} \approx 2.51$ s; b) i) $t \approx 1.25$ s; ii) $t \approx 2.17$ s; c) $0.711$ m

Show working and answer

a) 1. The particle changes direction when $v$ is zero and changes sign. $v = 0$ if $sin t^{2} = 0$ , i.e. $t^{2} = nπ$ . 2. If $sin t^{2} > 0$ , $e^{- s i n t^{2}} < 1$ and $v > 0$ ; if $sin t^{2} < 0$ , $v < 0$ . The sign changes at $t^{2} = π$ and at $t^{2} = 2 π$ (at $t = 0$ the motion only starts). 3. $t_{1} = π \approx 1.77$ s and $t_{2} = 2 π \approx 2.51$ s.

b) i) 1. $v$ is maximum when $sin t^{2} = 1$ (so that $e^{- s i n t^{2}}$ is minimum): $t^{2} = \frac{π}{2}$ . 2. $t = \frac{π}{2} \approx 1.25$ s.

(ii) 1. $v$ is minimum when $sin t^{2} = - 1$ : $t^{2} = \frac{3 π}{2}$ . 2. $t = \frac{3 π}{2} \approx 2.17$ s (less than $t_{2}$ ).

c) 1. Between $t_{1}$ and $t_{2}$ the velocity is negative, so the distance is $- \int_{t_{1}}^{t_{2}} v d t$ . 2. Using the GDC: $\int_{1.77}^{2.51} 1 - e^{- s i n t^{2}} d t \approx 0.711$ m.

Answer: a) $t_{1} \approx 1.77$ s, $t_{2} \approx 2.51$ s; b) i) $t \approx 1.25$ s; ii) $t \approx 2.17$ s; c) $0.711$ m

Exercise 304 HardIB · P3 TZ0 November 2014Proof

In this question you may assume that $arctan x$ is continuous and differentiable for all $x \in R$ .

a) Consider the infinite geometric series $1 - x^{2} + x^{4} - x^{6} + \dots$ $∣ x ∣ < 1$ .

Show that the sum of the series is equal to $\frac{1}{1 + x ^{2}}$ .

b) Hence, show that an expansion of $arctan x$ is $arctan x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + \dots$

c) $f$ is a continuous function defined on $[a, b]$ and differentiable on $] a, b [$ , with $f^{'} (x) > 0$ on $] a, b [$ . Use the mean value theorem to show that for all $x, y \in [a, b]$ , if $y > x$ then $f (y) > f (x)$ .

d) i) Given that $g (x) = x - arctan x$ , show that $g^{'} (x) > 0$ for $x > 0$ .

(ii) Use the result of part (c) to show that $arctan x < x$ for $x > 0$ .

e) Use the result of part (c) to show that $arctan x > x - \frac{x ^{3}}{3}$ for $x > 0$ .

f) Hence, show that $\frac{16}{3 3} < π < \frac{6}{3}$ .

Show answer

Answer: shown.

Show working and answer

a) 1. It is a geometric series with first term $1$ and common ratio $r = - x^{2}$ , with $∣ r ∣ = x^{2} < 1$ because $∣ x ∣ < 1$ . 2. Sum $= \frac{1}{1 - r} = \frac{1}{1 + x ^{2}}$ .

b) 1. We know that $\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}} = 1 - x^{2} + x^{4} - x^{6} + \dots$ . 2. Integrating term by term: $arctan x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + \dots + C$ . 3. With $x = 0$ : $arctan 0 = 0$ , so $C = 0$ .

c) 1. Let $x, y \in [a, b]$ with $y > x$ . $f$ is continuous on $[x, y]$ and differentiable on $] x, y [$ . 2. By the mean value theorem there exists $c \in] x, y [$ with $f (y) - f (x) = f^{'} (c) (y - x)$ . 3. As $c \in] a, b [$ , $f^{'} (c) > 0$ , and $y - x > 0$ . Hence $f (y) - f (x) > 0$ , i.e. $f (y) > f (x)$ .

d) i) 1. $g (x) = x - arctan x$ , so $g^{'} (x) = 1 - \frac{1}{1 + x ^{2}} = \frac{x ^{2}}{1 + x ^{2}}$ . 2. For $x > 0$ , $x^{2} > 0$ and $1 + x^{2} > 0$ , so $g^{'} (x) > 0$ .

(ii) 1. $g$ is continuous and differentiable, with $g^{'} > 0$ on $] 0, x [$ for any $x > 0$ . By (c) with $a = 0$ : $g (x) > g (0) = 0$ . 2. That is, $x - arctan x > 0$ , i.e. $arctan x < x$ for $x > 0$ .

e) 1. Let $h (x) = arctan x - x + \frac{x ^{3}}{3}$ . Then $h (0) = 0$ and $h^{'} (x) = \frac{1}{1 + x ^{2}} - 1 + x^{2} = \frac{1 - ( 1 + x ^{2} ) + x ^{2} ( 1 + x ^{2} )}{1 + x ^{2}} = \frac{x ^{4}}{1 + x ^{2}}$ . 2. For $x > 0$ , $h^{'} (x) > 0$ . By (c): $h (x) > h (0) = 0$ . 3. Hence $arctan x > x - \frac{x ^{3}}{3}$ for $x > 0$ .

f) 1. Take $x = \frac{1}{3}$ , where $arctan \frac{1}{3} = \frac{π}{6}$ . 2. From (d)(ii): $\frac{π}{6} < \frac{1}{3}$ , i.e. $π < \frac{6}{3}$ . 3. From (e): $\frac{π}{6} > \frac{1}{3} - \frac{1}{3} \cdot \frac{1}{3 3} = \frac{1}{3} - \frac{1}{9 3} = \frac{8}{9 3}$ , because $(\frac{1}{3})^{3} = \frac{1}{3 3}$ . 4. Multiplying by $6$ : $π > \frac{48}{9 3} = \frac{16}{3 3}$ . 5. Combining: $\frac{16}{3 3} < π < \frac{6}{3}$ .

Answer: shown.

Exercise 305 HardIB · P2 TZ1 May 2010

A body moves through a liquid so that its acceleration can be expressed as $(- \frac{v ^{2}}{200} - 32) m s^{- 2}$ , where $v m s^{- 1}$ is the velocity of the body at time $t$ seconds.

It is known that the initial velocity of the body was $40 m s^{- 1}$ .

a) Show that the time taken, $T$ seconds, for the velocity of the body to decrease to $V m s^{- 1}$ is given by $T = 200 \int_{V}^{40} \frac{1}{v ^{2} + 8 0 ^{2}} d v$ .

b) i) Explain why the acceleration can be expressed as $v \frac{d v}{d s}$ , where $s$ is the displacement, in metres, of the body at time $t$ seconds.

(ii) Hence, find an integral similar to the one shown in part (a) for the distance, $S$ metres, travelled by the body until it slows down to $V m s^{- 1}$ .

c) Using parts (a) and (b), find the distance travelled and the time until the body comes to rest.

Show answer

Answer: a) shown; b) ii) $S = 200 \int_{V}^{40} \frac{v}{v ^{2} + 8 0 ^{2}} d v$ ; c) $T \approx 1.16$ s, $S \approx 22.3$ m

Show working and answer

a) 1. $a = \frac{d v}{d t} = - \frac{v ^{2}}{200} - 32 = - \frac{v ^{2} + 6400}{200} = - \frac{v ^{2} + 8 0 ^{2}}{200}$ . 2. We separate the variables: $d t = - 200 \frac{d v}{v ^{2} + 8 0 ^{2}}$ . 3. Between $t = 0$ ( $v = 40$ ) and $t = T$ ( $v = V$ ): $T = - 200 \int_{40}^{V} \frac{d v}{v ^{2} + 8 0 ^{2}} = 200 \int_{V}^{40} \frac{1}{v ^{2} + 8 0 ^{2}} d v$ .

b) i) 1. As $v = \frac{d s}{d t}$ , by the chain rule $a = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$ .

(ii) 1. We equate: $v \frac{d v}{d s} = - \frac{v ^{2} + 8 0 ^{2}}{200}$ , and separate: $d s = - 200 \frac{v}{v ^{2} + 8 0 ^{2}} d v$ . 2. Between $v = 40$ and $v = V$ : $S = 200 \int_{V}^{40} \frac{v}{v ^{2} + 8 0 ^{2}} d v$ .

c) 1. The body stops when $V = 0$ . 2. Time: $T = 200 \int_{0}^{40} \frac{d v}{v ^{2} + 8 0 ^{2}} = 200 \cdot \frac{1}{80} [arctan \frac{v}{80}]_{0}^{40} = 2.5 arctan \frac{1}{2} \approx 1.16$ s. 3. Distance: $S = 200 \int_{0}^{40} \frac{v}{v ^{2} + 8 0 ^{2}} d v = 100 [ln (v^{2} + 8 0^{2})]_{0}^{40} = 100 ln \frac{8000}{6400} = 100 ln 1.25 \approx 22.3$ m.

Answer: a) shown; b) ii) $S = 200 \int_{V}^{40} \frac{v}{v ^{2} + 8 0 ^{2}} d v$ ; c) $T \approx 1.16$ s, $S \approx 22.3$ m

Exercise 306 HardIB · P3 TZ0 May 2010

a) Using the Maclaurin series for $(1 + x)^{n}$ , write down and simplify the Maclaurin series approximation for $(1 - x^{2})^{- \frac{1}{2}}$ , up to and including the term in $x^{4}$ .

b) Use this result to verify that a series approximation for $arccos x$ is $\frac{π}{2} - x - \frac{1}{6} x^{3} - \frac{3}{40} x^{5}$ .

c) Evaluate $x \to 0 lim \frac{\frac{π}{2} - arccos ( x ^{2} ) - x ^{2}}{x ^{6}}$ .

d) Use the series approximation for $arccos x$ to find an approximate value of $\int_{0}^{0.2} arccos (x) d x$ .

Give your answer correct to $5$ decimal places.

Does your answer give the true value of the integral correct to $5$ decimal places?

Show answer

Answer: a) $1 + \frac{x ^{2}}{2} + \frac{3}{8} x^{4} + \dots$ ; b) shown; c) $\frac{1}{6}$ ; d) $\approx 0.25326$ ; does not match the true value ( $0.25325$ ) to $5$ decimal places.

Show working and answer

a) 1. Maclaurin series of the binomial: $(1 + u)^{n} = 1 + n u + \frac{n ( n - 1 )}{2 !} u^{2} + \dots$ , with $n = - \frac{1}{2}$ and $u = - x^{2}$ . 2. $1 + (- \frac{1}{2}) (- x^{2}) + \frac{( - \frac{1}{2} ) ( - \frac{3}{2} )}{2} x^{4} = 1 + \frac{x ^{2}}{2} + \frac{3}{8} x^{4}$ . 3. $(1 - x^{2})^{- 1/2} \approx 1 + \frac{x ^{2}}{2} + \frac{3}{8} x^{4}$ .

b) 1. $\frac{d}{d x} arccos x = - \frac{1}{1 - x ^{2}} = - (1 - x^{2})^{- 1/2} \approx - 1 - \frac{x ^{2}}{2} - \frac{3}{8} x^{4}$ . 2. We integrate: $arccos x \approx C - x - \frac{x ^{3}}{6} - \frac{3}{40} x^{5}$ . 3. With $arccos 0 = \frac{π}{2}$ : $C = \frac{π}{2}$ . 4. $arccos x \approx \frac{π}{2} - x - \frac{1}{6} x^{3} - \frac{3}{40} x^{5}$ .

c) 1. We substitute $x \to x^{2}$ in (b): $\frac{π}{2} - arccos (x^{2}) \approx x^{2} + \frac{x ^{6}}{6} + \frac{3}{40} x^{10}$ . 2. The numerator is $\frac{x ^{6}}{6} + \frac{3}{40} x^{10} + \dots$ , and dividing by $x^{6}$ gives $\frac{1}{6} + \frac{3}{40} x^{4} + \dots$ . 3. As $x \to 0$ the limit is $\frac{1}{6}$ .

d) 1. We substitute $x \to x$ in (b): $arccos x \approx \frac{π}{2} - x^{1/2} - \frac{1}{6} x^{3/2} - \frac{3}{40} x^{5/2}$ . 2. Integrating from $0$ to $0.2$ : $[\frac{π}{2} x - \frac{2}{3} x^{3/2} - \frac{1}{15} x^{5/2} - \frac{3}{140} x^{7/2}]_{0}^{0.2}$ . 3. Using the GDC the value is $0.25326$ . 4. The exact integral (using the GDC) is $0.25325$ , which differs in the fifth decimal place: the approximation does not give the true value correct to $5$ decimal places.

Answer: a) $1 + \frac{x ^{2}}{2} + \frac{3}{8} x^{4} + \dots$ ; b) shown; c) $\frac{1}{6}$ ; d) $\approx 0.25326$ ; does not match the true value ( $0.25325$ ) to $5$ decimal places.

Exercise 307 HardIB · P2 TZ1 May 2011

A jet aircraft travels horizontally along a straight path for one minute, starting at time $t = 0$ , where $t$ is measured in seconds. The acceleration, $a$ , measured in $m s^{- 2}$ , of the aircraft is given by the straight-line graph below.

Graph of the acceleration a (in m/s²) of the aircraft against time t (in s), for 0 ≤ t ≤ 60: a decreasing straight line from 15 at t = 0 to −10 at t = 60

a) Find an expression for the acceleration of the aircraft during this time, in terms of $t$ .

b) Given that at $t = 0$ the aircraft is travelling at $125 m s^{- 1}$ , find its maximum velocity in $m s^{- 1}$ .

c) Given that the aircraft breaks the sound barrier at $295 m s^{- 1}$ , calculate how long the aircraft travels at a speed greater than this.

Show answer

Answer: a) $a = - \frac{5}{12} t + 15$ ; b) $395$ m s $^{- 1}$ ; c) $43.8$ s

Show working and answer

a) 1. The line passes through $(0, 15)$ and $(60, - 10)$ : gradient $\frac{- 10 - 15}{60} = - \frac{5}{12}$ . 2. $a = - \frac{5}{12} t + 15$ .

b) 1. We integrate: $v = \int a d t = - \frac{5}{24} t^{2} + 15 t + C$ , and with $v (0) = 125$ : $C = 125$ . 2. The velocity is maximum when $a = 0$ (the acceleration changes from positive to negative): $- \frac{5}{12} t + 15 = 0$ , so $t = 36$ s. 3. $v (36) = - \frac{5}{24} \cdot 1296 + 540 + 125 = - 270 + 540 + 125 = 395$ m s $^{- 1}$ .

c) 1. We solve $v = 295$ : $- \frac{5}{24} t^{2} + 15 t + 125 = 295$ , i.e. $5 t^{2} - 360 t + 4080 = 0$ , or $t^{2} - 72 t + 816 = 0$ . 2. $t = 36 \pm 1296 - 816 = 36 \pm 480$ , i.e. $t \approx 14.09$ s and $t \approx 57.91$ s (both in $[0, 60]$ ). 3. The velocity exceeds $295$ between these two instants: $2480 \approx 43.8$ s.

Answer: a) $a = - \frac{5}{12} t + 15$ ; b) $395$ m s $^{- 1}$ ; c) $43.8$ s

Exercise 308 HardIB · P1 TZ0 November 2012

Consider a function $f$ , defined by $f (x) = \frac{x}{2 - x}$ , for $0 \leq x \leq 1$ .

a) Find an expression for $(f \circ f) (x)$ .

Let $F_{n} (x) = \frac{x}{2 ^{n} - ( 2 ^{n} - 1 ) x}$ , where $0 \leq x \leq 1$ .

b) Use mathematical induction to show that for all $n \in Z^{+}$ , $n times (f \circ f \circ \dots \circ f) (x) = F_{n} (x)$ .

c) Show that $F_{- n} (x)$ is an expression for the inverse of $F_{n}$ .

d) (i) Calculate $F_{n} (0)$ and $F_{n} (1)$ .

(ii) Show that $F_{n} (x) < x$ , given that $0 < x < 1$ , $n \in Z^{+}$ .

(iii) For $n \in Z^{+}$ , let $A_{n}$ be the area of the region enclosed by the graph of $F_{n}^{- 1}$ , the $x$ -axis and the line $x = 1$ . Find the area $B_{n}$ of the region enclosed by $F_{n}$ and $F_{n}^{- 1}$ in terms of $A_{n}$ .

Show answer

Answer: a) $\frac{x}{4 - 3 x}$ ; c) $F_{n}^{- 1} (x) = \frac{2 ^{n} x}{1 + ( 2 ^{n} - 1 ) x} = F_{- n} (x)$ ; d) (i) $F_{n} (0) = 0$ , $F_{n} (1) = 1$ ; (iii) $B_{n} = 2 A_{n} - 1$

Show working and answer

a) 1. $(f \circ f) (x) = f (\frac{x}{2 - x}) = \frac{\frac{x}{2 - x}}{2 - \frac{x}{2 - x}}$ . 2. Multiplying numerator and denominator by $2 - x$ : $\frac{x}{2 ( 2 - x ) - x} = \frac{x}{4 - 3 x}$ .

b) Induction on $n$ .

Case $n = 1$ : $F_{1} (x) = \frac{x}{2 - ( 2 - 1 ) x} = \frac{x}{2 - x} = f (x)$ . It holds.
Hypothesis: for $n = k$ , $k times (f \circ \dots \circ f) (x) = F_{k} (x) = \frac{x}{2 ^{k} - ( 2 ^{k} - 1 ) x}$ .
For $n = k + 1$ we apply $f$ once more: $f (F_{k} (x)) = \frac{F _{k} ( x )}{2 - F _{k} ( x )} = \frac{\frac{x}{2 ^{k} - ( 2 ^{k} - 1 ) x}}{2 - \frac{x}{2 ^{k} - ( 2 ^{k} - 1 ) x}}$ .
Multiplying top and bottom by $2^{k} - (2^{k} - 1) x$ : $\frac{x}{2 ( 2 ^{k} - ( 2 ^{k} - 1 ) x ) - x} = \frac{x}{2 ^{k + 1} - ( 2 ^{k + 1} - 2 + 1 ) x} = \frac{x}{2 ^{k + 1} - ( 2 ^{k + 1} - 1 ) x} = F_{k + 1} (x)$ .
It holds for $n = 1$ and, if it holds for $k$ , it holds for $k + 1$ . By induction it is true for all $n \in Z^{+}$ .

c) 1. To find $F_{n}^{- 1}$ we write $y = \frac{x}{2 ^{n} - ( 2 ^{n} - 1 ) x}$ and solve for $x$ : $2^{n} y - (2^{n} - 1) x y = x$ . 2. Grouping: $2^{n} y = x (1 + (2^{n} - 1) y)$ , so $x = \frac{2 ^{n} y}{1 + ( 2 ^{n} - 1 ) y}$ and, therefore, $F_{n}^{- 1} (x) = \frac{2 ^{n} x}{1 + ( 2 ^{n} - 1 ) x}$ . 3. On the other hand, substituting $n$ by $- n$ : $F_{- n} (x) = \frac{x}{2 ^{- n} - ( 2 ^{- n} - 1 ) x}$ . 4. Multiplying top and bottom by $2^{n}$ : $F_{- n} (x) = \frac{2 ^{n} x}{1 - ( 1 - 2 ^{n} ) x} = \frac{2 ^{n} x}{1 + ( 2 ^{n} - 1 ) x} = F_{n}^{- 1} (x)$ .

d) (i) 1. $F_{n} (0) = \frac{0}{2 ^{n}} = 0$ . 2. $F_{n} (1) = \frac{1}{2 ^{n} - ( 2 ^{n} - 1 )} = 1$ .

(ii) 1. Let $D = 2^{n} - (2^{n} - 1) x = 1 + (2^{n} - 1) (1 - x)$ . For $0 < x < 1$ and $n \geq 1$ we have $D > 0$ . 2. $x - F_{n} (x) = x (1 - \frac{1}{D}) = \frac{x ( D - 1 )}{D} = \frac{x ( 2 ^{n} - 1 ) ( 1 - x )}{D}$ . 3. All the factors are positive, so $x - F_{n} (x) > 0$ , i.e. $F_{n} (x) < x$ .

(iii) 1. By (i) and (ii), $F_{n}$ goes from $(0, 0)$ to $(1, 1)$ below the line $y = x$ ; its inverse is its reflection in $y = x$ , and lies above the line. 2. The area $B_{n}$ between the two curves is $B_{n} = \int_{0}^{1} (F_{n}^{- 1} - F_{n}) d x = A_{n} - \int_{0}^{1} F_{n} d x$ . 3. The area under $F_{n}$ and the area under $F_{n}^{- 1}$ (between $x = 0$ and $x = 1$ ) together fill the unit square, because they are reflections of each other in $y = x$ : $\int_{0}^{1} F_{n} d x = 1 - A_{n}$ . 4. Hence $B_{n} = A_{n} - (1 - A_{n}) = 2 A_{n} - 1$ .

Answer: a) $\frac{x}{4 - 3 x}$ ; c) $F_{n}^{- 1} (x) = \frac{2 ^{n} x}{1 + ( 2 ^{n} - 1 ) x} = F_{- n} (x)$ ; d) (i) $F_{n} (0) = 0$ , $F_{n} (1) = 1$ ; (iii) $B_{n} = 2 A_{n} - 1$

Exercise 309 HardIB · P2 TZ1 May 2013

A particle $A$ moves in a straight line. Its velocity $v_{A}$ $m s^{- 1}$ , at $t$ seconds after it starts moving, is given by $v_{A} = t^{3} - 5 t^{2} + 6 t$ .

a) Sketch the graph of $v_{A} = t^{3} - 5 t^{2} + 6 t$ for $t \geq 0$ , with $v_{A}$ on the vertical axis and $t$ on the horizontal axis. Show on your graph the local maximum and minimum points, and the $t$ -intercepts.

b) Write down the times for which the velocity of the particle is increasing.

c) Write down the times for which the magnitude of the velocity of the particle is increasing.

At $t = 0$ the particle is at the point $O$ .

d) Find an expression for the displacement, $x_{A}$ m, of the particle from $O$ at time $t$ .

A second particle, $B$ , moving along the same line, has position $x_{B}$ m, velocity $v_{B}$ $m s^{- 1}$ and acceleration $a_{B}$ $m s^{- 2}$ , where $a_{B} = - 2 v_{B}$ for $t \geq 0$ . At $t = 0$ , $x_{B} = 20$ and $v_{B} = - 20$ .

e) Calculate an expression for $v_{B}$ in terms of $t$ .

f) Calculate the value of $t$ at which the two particles meet.

View solution

a) 1. $v_{A} = t^{3} - 5 t^{2} + 6 t = t (t - 2) (t - 3)$ : it crosses the $t$ -axis at $t = 0$ , $t = 2$ and $t = 3$ . 2. Turning points: $\frac{d v _{A}}{d t} = 3 t^{2} - 10 t + 6 = 0$ , so $t = \frac{10 \pm 28}{6} = \frac{5 \pm 7}{3}$ , i.e. $t \approx 0.785$ and $t \approx 2.55$ . 3. It is a cubic with positive leading coefficient: the first turning point is a maximum and the second a minimum. $v_{A} (0.785) \approx 2.11$ and $v_{A} (2.55) \approx - 0.631$ . 4. Local maximum $(0.785; 2.11)$ , local minimum $(2.55; - 0.631)$ and intercepts $(0, 0)$ , $(2, 0)$ , $(3, 0)$ .

Graph of the velocity v_A = t³ − 5t² + 6t for t ≥ 0

b) 1. The velocity increases when the acceleration $a = \frac{d v _{A}}{d t}$ is positive, i.e. when the graph rises. 2. $3 t^{2} - 10 t + 6 > 0$ for $0 \leq t < 0.785$ and for $t > 2.55$ .

c) 1. The magnitude of the velocity (the modulus $∣ v ∣$ ) increases when $v$ and $a$ have the same sign. 2. $v > 0$ and $a > 0$ : $v > 0$ on $] 0, 2 [$ and on $] 3, \infty [$ , and $a > 0$ if $t < 0.785$ or $t > 2.55$ . Both hold together for $0 \leq t < 0.785$ and for $t > 3$ . 3. $v < 0$ and $a < 0$ : $v < 0$ on $] 2, 3 [$ , and $a < 0$ on $] 0.785; 2.55 [$ . Both hold together for $2 < t < 2.55$ .

d) 1. The displacement is the integral of the velocity, with $x_{A} = 0$ at $t = 0$ . 2. $x_{A} = \int_{0}^{t} (s^{3} - 5 s^{2} + 6 s) d s = \frac{t ^{4}}{4} - \frac{5 t ^{3}}{3} + 3 t^{2}$ .

e) 1. $a_{B} = \frac{d v _{B}}{d t} = - 2 v_{B}$ is a separable differential equation: $\int \frac{d v _{B}}{v _{B}} = - 2 \int d t$ , so $ln ∣ v_{B} ∣ = - 2 t + C$ . 2. Hence $v_{B} = K e^{- 2 t}$ . With $v_{B} (0) = - 20$ : $K = - 20$ . 3. $v_{B} = - 20 e^{- 2 t}$ .

f) 1. $x_{B} = \int - 20 e^{- 2 t} d t = 10 e^{- 2 t} + C$ . With $x_{B} (0) = 20$ : $C = 10$ , and $x_{B} = 10 + 10 e^{- 2 t}$ . 2. They meet when $x_{A} = x_{B}$ : $\frac{t ^{4}}{4} - \frac{5 t ^{3}}{3} + 3 t^{2} = 10 + 10 e^{- 2 t}$ . 3. Using the GDC (Paper 2) the solution with $t > 0$ is $t \approx 4.41$ s.

Answer: b) $0 \leq t < 0.785$ , $t > 2.55$ ; c) $0 \leq t < 0.78$ , $2 < t < 2.55$ , $t > 3$ ; d) $x_{A} = \frac{t ^{4}}{4} - \frac{5 t ^{3}}{3} + 3 t^{2}$ ; e) $v_{B} = - 20 e^{- 2 t}$ ; f) $t = 4.41$ s

Exercise 310 HardIB · P2 TZ2 May 2015

A particle moves in a straight line. Its velocity $v$ $m s^{- 1}$ at time $t$ seconds is given by $v = 9 t - 3 t^{2}$ , $0 \leq t \leq 5$ . At time $t = 0$ , the displacement $s$ of the particle from the origin $O$ is $3$ m.

a) Find the displacement of the particle when $t = 4$ .

b) Sketch the displacement/time graph for this particle, $0 \leq t \leq 5$ , clearly showing where the curve meets the axes and the coordinates of the points at which the displacement reaches its maximum and minimum values.

For $t > 5$ , the displacement of the particle is given by $s = a + b cos \frac{2 π t}{5}$ , such that $s$ is continuous for all $t \geq 0$ .

c) Given also that $s = 16.5$ when $t = 7.5$ , find the values of $a$ and $b$ .

d) Find the values $t_{1}$ and $t_{2}$ ( $0 < t_{1} < t_{2} < 8$ ) at which the particle returns to its starting point.

View solution

a) 1. Displacement is the integral of velocity: $s (t) = 3 + \int_{0}^{t} (9 u - 3 u^{2}) d u = 3 + \frac{9 t ^{2}}{2} - t^{3}$ (with $s (0) = 3$ ). 2. $s (4) = 3 + 72 - 64 = 11$ m.

b) 1. The turning points of $s$ occur where $v = 9 t - 3 t^{2} = 3 t (3 - t) = 0$ : $t = 0$ and $t = 3$ . We must also check the endpoint of the interval, $t = 5$ . 2. $s (0) = 3$ , $s (3) = 3 + 40.5 - 27 = 16.5$ , $s (5) = 3 + 112.5 - 125 = - 9.5$ . 3. Between $0$ and $3$ the velocity is positive (the graph rises); between $3$ and $5$ it is negative (the graph falls). Maximum at $(3; 16.5)$ and minimum at the endpoint $(5; - 9.5)$ . 4. Intercept with the $s$ -axis: $(0, 3)$ . Intercept with the $t$ -axis: using the GDC, $3 + \frac{9 t ^{2}}{2} - t^{3} = 0$ gives $t \approx 4.64$ , i.e. $(4.64; 0)$ .

Displacement–time graph of the particle for 0 ≤ t ≤ 5

c) 1. Continuity at $t = 5$ : $s (5) = - 9.5$ must equal $a + b cos (2 π) = a + b$ . So $a + b = - 9.5$ . 2. At $t = 7.5$ : $\frac{2 π \cdot 7.5}{5} = 3 π$ and $cos 3 π = - 1$ , so $a - b = 16.5$ . 3. Adding both equations: $2 a = 7$ , so $a = 3.5$ ; and $b = - 9.5 - 3.5 = - 13$ .

d) 1. The particle returns to its initial position when $s = 3$ . 2. For $t \leq 5$ : $\frac{9 t ^{2}}{2} - t^{3} = 0$ , so $t^{2} (\frac{9}{2} - t) = 0$ and $t_{1} = 4.5$ (the solution $t = 0$ is the start). 3. For $t > 5$ : $3.5 - 13 cos \frac{2 π t}{5} = 3$ , i.e. $cos \frac{2 π t}{5} = \frac{1}{26}$ . 4. With $5 < t < 8$ , the angle $\frac{2 π t}{5}$ lies between approximately $2 π$ and $3.2 π$ , and the first solution is $\frac{2 π t}{5} = 2 π + arccos \frac{1}{26} \approx 7.8155$ , so $t_{2} \approx 6.22$ (the next solution, $4 π - arccos \frac{1}{26}$ , gives $t > 8$ ).

Answer: a) $s = 11$ m; b) max. $(3; 16.5)$ , min. $(5; - 9.5)$ ; c) $a = 3.5$ , $b = - 13$ ; d) $t_{1} = 4.5$ s, $t_{2} = 6.22$ s

Exercise 311 HardIB · P2 TZ1 May 2015

Richard, a marine, jumps from a stationary helicopter, $1000$ m above the ground, at time $t = 0$ . His height above the ground, in metres, is given by $s (t)$ .

During the first $10$ seconds his velocity, $v (t)$ $m s^{- 1}$ , is given by $v (t) = - 10 t$ .

a) (i) Find his acceleration $a (t)$ for $t < 10$ .

(ii) Calculate $v (10)$ .

(iii) Show that $s (10) = 500$ .

At $t = 10$ his parachute opens and his acceleration $a (t)$ is then given by $a (t) = - 10 - 5 v$ , $t \geq 10$ .

b) Given that $\frac{d t}{d v} = \frac{1}{\frac{d v}{d t}}$ , write $\frac{d t}{d v}$ in terms of $v$ .

It is known that Richard's acceleration, $a (t) = - 10 - 5 v$ , is always positive, for $t \geq 10$ .

c) Hence show that $t = 10 + \frac{1}{5} ln (\frac{98}{- 2 - v})$ .

d) Using the above, find an expression for the velocity, $v$ , for $t \geq 10$ .

e) Calculate an expression for his height, $s$ , above the ground for $t \geq 10$ .

f) Calculate the value of $t$ when Richard lands on the ground.

Show answer

Answer: a) (i) $- 10 m s^{- 2}$ ; (ii) $- 100 m s^{- 1}$ ; b) $\frac{d t}{d v} = \frac{1}{- 10 - 5 v}$ ; d) $v = - 2 - 98 e^{- 5 (t - 10)}$ ; e) $s (t) = - 2 t + \frac{98}{5} e^{- 5 (t - 10)} + 500.4$ ; f) $t = 250$ s

Show working and answer

a) (i) 1. $a (t) = v^{'} (t) = - 10 m s^{- 2}$ .

(ii) 1. $v (10) = - 10 \cdot 10 = - 100 m s^{- 1}$ .

(iii) 1. The height is $s (t) = 1000 + \int_{0}^{t} - 10 u d u = 1000 - 5 t^{2}$ . 2. $s (10) = 1000 - 500 = 500$ m.

b) 1. As $\frac{d v}{d t} = a = - 10 - 5 v$ , the reciprocal is $\frac{d t}{d v} = \frac{1}{- 10 - 5 v}$ .

c) 1. Technique: separation of variables. $t = \int \frac{d v}{- 10 - 5 v} = - \frac{1}{5} ln ∣ 10 + 5 v ∣ + C_{1}$ , and since $ln ∣10 + 5 v ∣ = ln 5 + ln ∣2 + v ∣$ , the constant $ln 5$ can be absorbed into the constant: $t = - \frac{1}{5} ln ∣2 + v ∣ + C$ . 2. Since the acceleration is positive, $- 10 - 5 v > 0$ , i.e. $- 2 - v > 0$ , and $∣2 + v ∣ = - 2 - v$ : $t = - \frac{1}{5} ln (- 2 - v) + C$ . 3. At $t = 10$ , $v = - 100$ and $- 2 - v = 98$ : $10 = - \frac{1}{5} ln 98 + C$ , so $C = 10 + \frac{1}{5} ln 98$ . 4. $t = 10 + \frac{1}{5} ln 98 - \frac{1}{5} ln (- 2 - v) = 10 + \frac{1}{5} ln (\frac{98}{- 2 - v})$ .

d) 1. Rearranging for $v$ : $5 (t - 10) = ln \frac{98}{- 2 - v}$ , so $\frac{98}{- 2 - v} = e^{5 (t - 10)}$ . 2. $- 2 - v = 98 e^{- 5 (t - 10)}$ , and therefore $v = - 2 - 98 e^{- 5 (t - 10)}$ .

e) 1. $s (t) = s (10) + \int_{10}^{t} v (u) d u = 500 + [- 2 u + \frac{98}{5} e^{- 5 (u - 10)}]_{10}^{t}$ . 2. $= 500 - 2 t + \frac{98}{5} e^{- 5 (t - 10)} + 20 - \frac{98}{5}$ . 3. $20 - 19.6 = 0.4$ , so $s (t) = - 2 t + \frac{98}{5} e^{- 5 (t - 10)} + 500.4$ .

f) 1. He lands when $s = 0$ . Using the GDC, $- 2 t + \frac{98}{5} e^{- 5 (t - 10)} + 500.4 = 0$ gives $t \approx 250$ s (the exponential term is already negligible: $t \approx 250.2$ ).

Exercise 312 HardIB · P2 TZ0 November 2015

The following figure shows the vertical cross-section of a building.

Cross-section of a building: bell-shaped roof f(x) = 30e^(−x²/400) for −20 ≤ x ≤ 20 and rectangle CDEF resting on the x-axis, with E and F on the curve

The cross-section of the roof of the building can be modelled by the curve $f (x) = 30 e^{- \frac{x ^{2}}{400}}$ , where $- 20 \leq x \leq 20$ . The $x$ -axis represents street level.

a) Find $f^{''} (x)$ .

b) Show that the gradient of the roof function is maximum for $x = - 200$ .

The cross-section of the habitable space below the roof is modelled by the rectangle $CDEF$ , with points $C (- a, 0)$ and $D (a, 0)$ , where $0 < a \leq 20$ .

c) Show that the maximum area ( $A$ ) of the rectangle $CDEF$ is $6002 e^{- \frac{1}{2}}$ .

d) The function $I$ is known as the insulation factor of $CDEF$ .

The function is defined as $I (a) = \frac{P ( a )}{A ( a )}$ , where $P =$ perimeter and $A =$ area of the rectangle.

(i) Find an expression for $P$ in terms of $a$ .

(ii) Find the value of $a$ that minimises $I$ .

(iii) Using the value of $a$ found in part (ii), calculate the percentage of the area of the cross-section under the whole roof that is not included in the cross-section of the habitable space.

Show answer

Answer: a) $f^{''} (x) = - \frac{3}{20} e^{- x^{2} /400} (1 - \frac{x ^{2}}{200})$ ; d) (i) $P (a) = 4 a + 60 e^{- a^{2} /400}$ ; (ii) $a = 12.6$ ; (iii) $43.3 %$

Show working and answer

a) 1. Using the chain rule, $f^{'} (x) = 30 e^{- x^{2} /400} \cdot (- \frac{2 x}{400}) = - \frac{3 x}{20} e^{- x^{2} /400}$ . 2. Product rule: $f^{''} (x) = - \frac{3}{20} e^{- x^{2} /400} - \frac{3 x}{20} \cdot (- \frac{x}{200}) e^{- x^{2} /400}$ . 3. $f^{''} (x) = - \frac{3}{20} e^{- x^{2} /400} (1 - \frac{x ^{2}}{200})$ .

b) 1. The gradient of the roof is $f^{'}$ , and it is maximum when $f^{''} = 0$ (with $f^{'}$ changing from increasing to decreasing). 2. $f^{''} (x) = 0 ⟺ 1 - \frac{x ^{2}}{200} = 0 ⟺ x = \pm 200$ . 3. For $x < - 200$ , $1 - \frac{x ^{2}}{200} < 0$ and $f^{''} > 0$ ; for $- 200 < x < 0$ , $f^{''} < 0$ . So $f^{'}$ increases and then decreases: at $x = - 200$ the gradient is maximum. 4. We check that it is larger than the value at the endpoint: $f^{'} (- 200) = \frac{3 200}{20} e^{- 1/2} \approx 1.29$ and $f^{'} (- 20) = 3 e^{- 1} \approx 1.10$ . 5. For $x > 0$ the gradient is negative, so the maximum is at $x = - 200$ .

c) 1. The rectangle has base $2 a$ and height $f (a)$ : $A (a) = 2 a \cdot 30 e^{- a^{2} /400} = 60 a e^{- a^{2} /400}$ . 2. $A^{'} (a) = 60 e^{- a^{2} /400} (1 - \frac{a ^{2}}{200}) = 0$ when $a = 200 = 102 \approx 14.1$ , which lies in $] 0, 20]$ . 3. $A^{'} > 0$ before this value and $A^{'} < 0$ after it, so it is a maximum. 4. $A_{max} = 60 \cdot 102 e^{- 200/400} = 6002 e^{- 1/2}$ .

d) (i) 1. Perimeter: two horizontal sides of length $2 a$ and two vertical sides of height $f (a)$ . 2. $P (a) = 4 a + 2 \cdot 30 e^{- a^{2} /400} = 4 a + 60 e^{- a^{2} /400}$ .

(ii) 1. $I (a) = \frac{4 a + 60 e ^{- a^{2} /400}}{60 a e ^{- a^{2} /400}}$ . 2. Using the GDC (Paper 2), we look for the minimum of $I$ in $] 0, 20]$ : $a \approx 12.6$ .

(iii) 1. Area under the whole roof: $\int_{- 20}^{20} 30 e^{- x^{2} /400} d x \approx 896.19$ . 2. Area of the rectangle with $a = 12.6$ : $A = 60 \cdot 12.6 e^{- 12. 6^{2} /400} \approx 508.6$ . 3. Area not included: $896.19 - 508.6 \approx 387.6$ . 4. Percentage: $\frac{387.6}{896.19} \cdot 100 \approx 43.3 %$ .

Answer: a) $f^{''} (x) = - \frac{3}{20} e^{- x^{2} /400} (1 - \frac{x ^{2}}{200})$ ; d) (i) $P (a) = 4 a + 60 e^{- a^{2} /400}$ ; (ii) $a = 12.6$ ; (iii) $43.3 %$

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp