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

Improper integrals

3 worked examples · By Álvaro López

Key idea

An improper integral is one in which the interval has no end (it goes on to $\infty$ ) or the function blows up at one end. It seems the area should be infinite, but sometimes it is a finite number. They are solved with a limit applied to the usual definite integral.

Infinite intervals

For an integral such as $\int_{1}^{\infty} \frac{1}{x ^{2}} d x$ you cannot apply Barrow's rule directly, because there is no “upper limit $\infty$ ” to evaluate. You replace it with a letter and let that letter tend to infinity.

When both limits are infinite, split the integral at any point $c$ (usually $c = 0$ ), and each part must converge: $\int_{- \infty}^{\infty} f (x) d x = \int_{- \infty}^{c} f (x) d x + \int_{c}^{\infty} f (x) d x .$ If one of the two diverges, the whole integral diverges, even if the two “cancel each other out”.

Graphs of y=1/x and y=1/x^2 for x greater than or equal to 1; the area under 1/x^2 is finite, although the region continues towards infinity

A useful test: the powers $\frac{1}{x ^{p}}$

The graph of $\frac{1}{x ^{2}}$ falls faster than that of $\frac{1}{x}$ and encloses a finite area. That of $e^{- x}$ falls even faster: decreasing exponentials always give convergent integrals.

A curious consequence: if $y = \frac{1}{x}$ ( $x \geq 1$ ) is rotated about the $x$ -axis, the volume $π \int_{1}^{\infty} \frac{1}{x ^{2}} d x = π$ is finite, even though the area of the region that generates it is infinite.

Unbounded functions (extension)

Integrals of a function that is not bounded at one end, such as $\int_{0}^{1} \frac{1}{x} d x$ , are also improper. They are solved in the same way, with a limit towards the problematic end: $\int_{0}^{1} \frac{1}{x} d x = t \to 0^{+} lim \int_{t}^{1} \frac{1}{x} d x = t \to 0^{+} lim [2 x]_{t}^{1} = t \to 0^{+} lim (2 - 2 t) = 2.$

Worked examples

Example 1 Basic

Three integrals with an infinite limit

Calculate, if they exist:

a) $\int_{1}^{\infty} \frac{1}{x ^{3}} d x$ .

b) $\int_{0}^{\infty} e^{- 2 x} d x$ .

c) $\int_{1}^{\infty} \frac{1}{x} d x$ .

Solution

a) $\int_{1}^{t} x^{- 3} d x = [- \frac{1}{2 x ^{2}}]_{1}^{t} = \frac{1}{2} - \frac{1}{2 t ^{2}} t \to \infty \frac{1}{2}$ . It converges to $\frac{1}{2}$ .

b) $\int_{0}^{t} e^{- 2 x} d x = [- \frac{1}{2} e^{- 2 x}]_{0}^{t} = \frac{1}{2} - \frac{1}{2} e^{- 2 t} t \to \infty \frac{1}{2}$ . It converges to $\frac{1}{2}$ .

c) $\int_{1}^{t} \frac{1}{x} d x = ln t - ln 1 = ln t t \to \infty \infty$ . Diverges.

Answer: a) $\frac{1}{2}$ ; b) $\frac{1}{2}$ ; c) diverges ( $+ \infty$ ).

Example 2 Medium

Other infinite intervals

Calculate:

a) $\int_{- \infty}^{0} e^{3 x} d x$ .

b) $\int_{- \infty}^{\infty} \frac{1}{x ^{2} + 4} d x$ .

c) $\int_{0}^{\infty} x e^{- x} d x$ .

Solution

a) $\int_{t}^{0} e^{3 x} d x = [\frac{1}{3} e^{3 x}]_{t}^{0} = \frac{1}{3} - \frac{1}{3} e^{3 t} t \to - \infty \frac{1}{3}$ .

b) A primitive is $\frac{1}{2} arctan \frac{x}{2}$ . As the function is even, one half is enough: $\int_{0}^{\infty} \frac{1}{x ^{2} + 4} d x = t \to \infty lim \frac{1}{2} arctan \frac{t}{2} = \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4}$ , and the complete integral is twice that: $\int_{- \infty}^{\infty} \frac{1}{x ^{2} + 4} d x = \frac{π}{2} .$

c) By parts with $u = x$ and $d v = e^{- x} d x$ : a primitive is $- (x + 1) e^{- x}$ . Then $\int_{0}^{t} x e^{- x} d x = [- (x + 1) e^{- x}]_{0}^{t} = 1 - (t + 1) e^{- t} t \to \infty 1,$ because $t e^{- t} \to 0$ .

Answer: a) $\frac{1}{3}$ ; b) $\frac{π}{2}$ ; c) $1$ .

Example 3 Hard

Finite and infinite areas and volumes

a) Find the area of the region under $y = x^{- 3/2}$ for $x \geq 1$ .

b) Find the area of the region under $y = \frac{1}{x}$ for $x \geq 1$ .

c) Calculate the volume of the solid obtained by rotating the region in b) about the $x$ -axis.

d) Find the value of $k > 0$ for which $\int_{0}^{\infty} e^{- k x} d x = 5$ .

Solution

a) $\int_{1}^{t} x^{- 3/2} d x = [- 2 x^{- 1/2}]_{1}^{t} = 2 - \frac{2}{t} t \to \infty 2$ . The area is $2$ .

b) $\int_{1}^{t} \frac{1}{x} d x = ln t \to \infty$ . The area is infinite (the integral diverges).

c) $V = π \int_{1}^{\infty} \frac{1}{x ^{2}} d x = π t \to \infty lim [- \frac{1}{x}]_{1}^{t} = π t \to \infty lim (1 - \frac{1}{t}) = π$ . The volume is finite.

d) $\int_{0}^{t} e^{- k x} d x = [- \frac{1}{k} e^{- k x}]_{0}^{t} = \frac{1}{k} - \frac{1}{k} e^{- k t} t \to \infty \frac{1}{k}$ (with $k > 0$ ). Then $\frac{1}{k} = 5$ and $k = \frac{1}{5}$ .

Answer: a) $2$ ; b) infinite (diverges); c) $π$ ; d) $k = \frac{1}{5}$ .

In the exam: always write the limit $lim_{t \to \infty}$ and justify why the leftover term tends to $0$ (for example $e^{- k t} \to 0$ if $k > 0$ ). Without the limit written you lose the marks for the method, even if the result is correct.

Common mistakes

Summary

$\int_{a}^{\infty} f = t \to \infty lim \int_{a}^{t} f$ and $\int_{- \infty}^{a} f = t \to - \infty lim \int_{t}^{a} f$ ; with both infinite, split at a point and each piece must converge.
Convergent: the limit is a real number; divergent: it is $\pm \infty$ or does not exist.
$\int_{1}^{\infty} \frac{1}{x ^{p}} d x$ converges if and only if $p > 1$ (and then equals $\frac{1}{p - 1}$ ).
An infinite area can give a finite volume of revolution.
A function that is unbounded at one end also gives an improper integral: take a limit towards that end.

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

Improper integrals

Infinite intervals

A useful test: the powers xp1​

Unbounded functions (extension)

Worked examples

Common mistakes

Summary

Want to work on it together?

A useful test: the powers $\frac{1}{x ^{p}}$