IB Analysis and Approaches HL · Integral calculus · Section 10 of 13

Fundamental theorem of calculus and Barrow's rule

3 worked examples · By Álvaro López

Key idea

The fundamental theorem links the two big ideas of calculus: differentiation and integration are inverse operations. From it comes Barrow's rule, which calculates definite integrals with an antiderivative and avoids Riemann sums. With its properties, you solve most of the definite integrals in the exam in a few lines.

The accumulated area function

Take a continuous function $f$ and a fixed point $a$ . For each $x$ , the number $G (x) = \int_{a}^{x} f (t) d t$ is the (signed) area accumulated from $a$ up to $x$ . It is a function of $x$ ; the variable of integration is called $t$ only so as not to confuse it with the endpoint.

Accumulated area G(x) from a to x under y=f(t), and a thin strip of width h and height f(x)

How much does $G$ change when $x$ increases slightly, by $h$ ? A thin strip of width $h$ and height almost equal to $f (x)$ is added, that is, an area of approximately $h f (x)$ . So the quotient $\frac{G ( x + h ) - G ( x )}{h}$ is close to $f (x)$ and, as $h \to 0$ , it is exactly $f (x)$ .

If the upper limit is another function of $x$ , add the chain rule: $\frac{d}{d x} \int_{a}^{g (x)} f (t) d t = f (g (x)) g^{'} (x) .$

Barrow's rule

The fundamental theorem says that an antiderivative of $f$ exists. Any other antiderivative $F$ differs from $G$ by a constant, $G (x) = F (x) + C$ . Since $G (a) = 0$ , we get $C = - F (a)$ and, evaluating at $b$ , we obtain Barrow's formula.

To use it, find an antiderivative (without the constant: it would cancel out when you subtract), evaluate it at the upper limit and subtract its value at the lower limit. For example, $\int_{0}^{π /2} cos x d x = [sin x]_{0}^{π /2} = 1 - 0 = 1.$

With substitution you can change the limits at the same time as the variable; with parts, apply the formula between the limits: $\int_{a}^{b} u d v = [uv]_{a}^{b} - \int_{a}^{b} v d u$ .

Properties of the definite integral

If the function contains $x + k$ or $c x$ , you do not need to know the antiderivative: with the substitution $u = x + k$ or $u = c x$ (and changing the limits) you get $\int_{p}^{q} f (x + k) d x = \int_{p + k}^{q + k} f (u) d u, \int_{p}^{q} f (c x) d x = \frac{1}{c} \int_{c p}^{c q} f (u) d u .$

Even and odd functions

A function is even if $f (- x) = f (x)$ (its graph is symmetric about the $y$ -axis) and odd if $f (- x) = - f (x)$ (symmetric about the origin). On a symmetric interval $[- a, a]$ , symmetry simplifies the integral.

The reason is geometric: for an even function the two halves have the same area; for an odd function the right-hand half is the negative of the left-hand one, and they cancel out.

Mean value of a function

The constant height that gives the same area as $f$ over the interval is called the mean value.

For example, the mean value of $x^{2}$ on $[0, 3]$ is $\frac{1}{3} \int_{0}^{3} x^{2} d x = \frac{1}{3} \cdot 9 = 3$ .

Worked examples

Example 1 Basic

Three integrals with Barrow

Calculate the exact value of:

a) $\int_{1}^{3} (2 x + \frac{1}{x}) d x$ .

b) $\int_{0}^{π /4} cos 2 x d x$ .

c) $\int_{0}^{2} e^{x /2} d x$ .

Solution

a) An antiderivative is $x^{2} + ln x$ . Then $[x^{2} + ln x]_{1}^{3} = (9 + ln 3) - (1 + 0) = 8 + ln 3.$

b) An antiderivative is $\frac{1}{2} sin 2 x$ . Then $[\frac{1}{2} sin 2 x]_{0}^{π /4} = \frac{1}{2} sin \frac{π}{2} - \frac{1}{2} sin 0 = \frac{1}{2} .$

c) An antiderivative is $2 e^{x /2}$ . Then $[2 e^{x /2}]_{0}^{2} = 2 e - 2 = 2 (e - 1) .$

Answer: a) $8 + ln 3$ ; b) $\frac{1}{2}$ ; c) $2 e - 2$ .

Example 2 Medium

Properties with given data

You know that $\int_{0}^{3} f (x) d x = 5$ , $\int_{3}^{7} f (x) d x = - 2$ and $\int_{0}^{3} g (x) d x = 4$ . Calculate:

a) $\int_{0}^{7} f (x) d x$ .

b) $\int_{7}^{3} f (x) d x$ .

c) $\int_{0}^{3} [2 f (x) - 3 g (x)] d x$ .

d) $\int_{0}^{1} f (3 x) d x$ .

Solution

a) By the interval property: $\int_{0}^{7} f = \int_{0}^{3} f + \int_{3}^{7} f = 5 + (- 2) = 3$ .

b) Reversing the limits changes the sign: $\int_{7}^{3} f = - \int_{3}^{7} f = 2$ .

c) By linearity: $2 \cdot 5 - 3 \cdot 4 = 10 - 12 = - 2$ .

d) With $u = 3 x$ , $d u = 3 d x$ and the limits change from $0$ and $1$ to $0$ and $3$ : $\int_{0}^{1} f (3 x) d x = \frac{1}{3} \int_{0}^{3} f (u) d u = \frac{5}{3} .$

Answer: a) $3$ ; b) $2$ ; c) $- 2$ ; d) $\frac{5}{3}$ .

Example 3 Hard

A function defined by an integral

Let $G (x) = \int_{1}^{x} (t^{2} - 4 t + 3) d t$ .

a) Write $G^{'} (x)$ .

b) Find the $x$ -coordinates of the local maximum and the local minimum of $G$ .

c) Calculate $G (3)$ .

d) Let $H (x) = G (x^{2})$ . Calculate $H^{'} (2)$ .

Solution

a) By the fundamental theorem, $G^{'} (x) = x^{2} - 4 x + 3 = (x - 1) (x - 3)$ .

b) $G^{'} (x) = 0$ for $x = 1$ and $x = 3$ . The sign of $G^{'}$ is $+$ for $x < 1$ , $-$ for $1 < x < 3$ and $+$ for $x > 3$ . Therefore $G$ changes from increasing to decreasing at $x = 1$ (local maximum) and from decreasing to increasing at $x = 3$ (local minimum).

c) Using Barrow's rule: $G (3) = [\frac{t ^{3}}{3} - 2 t^{2} + 3 t]_{1}^{3} = (9 - 18 + 9) - (\frac{1}{3} - 2 + 3) = 0 - \frac{4}{3} = - \frac{4}{3} .$

d) By the chain rule, $H^{'} (x) = G^{'} (x^{2}) \cdot 2 x = (x^{2} - 1) (x^{2} - 3) \cdot 2 x$ . At $x = 2$ : $H^{'} (2) = 3 \cdot 1 \cdot 4 = 12.$

Answer: a) $x^{2} - 4 x + 3$ ; b) maximum at $x = 1$ , minimum at $x = 3$ ; c) $- \frac{4}{3}$ ; d) $12$ .

In the exam: in b) justify the type of point with the sign of $G^{'}$ on each side (or with $G^{''}$ ). And in d) do not work out $H$ in full: the chain rule is enough.

Common mistakes

Summary

$G (x) = \int_{a}^{x} f (t) d t$ satisfies $G^{'} (x) = f (x)$ (fundamental theorem). With upper limit $g (x)$ : $G^{'} (x) = f (g (x)) g^{'} (x)$ .
Barrow: $\int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} = F (b) - F (a)$ , with $F$ any antiderivative.
Properties: linearity, additivity over intervals, $\int_{a}^{a} = 0$ , $\int_{b}^{a} = - \int_{a}^{b}$ and monotonicity.
Even function: $\int_{- a}^{a} = 2 \int_{0}^{a}$ . Odd function: $\int_{- a}^{a} = 0$ .
Mean value of $f$ on $[a, b]$ : $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

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