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### Course: AP®︎/College Calculus BC>Unit 6

Lesson 15: Evaluating improper integrals

# Improper integrals review

Review your knowledge of improper integrals.

## What are improper integrals?

Improper integrals are definite integrals that cover an unbounded area.
One type of improper integrals are integrals where at least one of the endpoints is extended to infinity. For example, ${\int }_{1}^{\mathrm{\infty }}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx$ is an improper integral. It can be viewed as the limit $\underset{b\to \mathrm{\infty }}{lim}{\int }_{1}^{b}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx$.
Another type of improper integrals are integrals whose endpoints are finite, but the integrated function is unbounded at one (or two) of the endpoints. For example, ${\int }_{0}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx$ is an improper integral. It can be viewed as the limit $\underset{a\to {0}^{+}}{lim}{\int }_{a}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx$.
An unbounded area that isn't infinite?! Is that for real?! Well, yeah! Not all improper integrals have a finite value, but some of them definitely do. When the limit exists we say the integral is convergent, and when it doesn't we say it's divergent.

## Practice set 1: Evaluating improper integrals with unbounded endpoints

Let's evaluate, for example, the improper integral ${\int }_{1}^{\mathrm{\infty }}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx$. As mentioned above, it's useful to view this integral as the limit $\underset{b\to \mathrm{\infty }}{lim}{\int }_{1}^{b}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx$. We can use the fundamental theorem of calculus to find an expression for the integral:
$\begin{array}{rl}{\int }_{1}^{b}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx& ={\int }_{1}^{b}{x}^{-2}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & ={\left[\frac{{x}^{-1}}{-1}\right]}_{1}^{b}\\ \\ & ={\left[-\frac{1}{x}\right]}_{1}^{b}\\ \\ & =-\frac{1}{b}-\left(-\frac{1}{1}\right)\\ \\ & =1-\frac{1}{b}\end{array}$
Now we got rid of the integral and we have a limit to find:
$\begin{array}{rl}\underset{b\to \mathrm{\infty }}{lim}{\int }_{1}^{b}\frac{1}{{x}^{2}}\phantom{\rule{0.167em}{0ex}}dx& =\underset{b\to \mathrm{\infty }}{lim}\left(1-\frac{1}{b}\right)\\ \\ & =1-0\\ \\ & =1\end{array}$
Problem 1.1
${\int }_{1}^{\mathrm{\infty }}\frac{1}{{x}^{3}}\phantom{\rule{0.167em}{0ex}}dx=?$

Want to try more problems like this? Check out this exercise.

## Practice set 2: Evaluating improper integrals with unbounded function

Let's evaluate, for example, the improper integral ${\int }_{0}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx$. As mentioned above, it's useful to view this integral as the limit $\underset{a\to 0}{lim}{\int }_{a}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx$. Again, we use the fundamental theorem of calculus to find an expression for the integral:
$\begin{array}{rl}{\int }_{a}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx& ={\int }_{a}^{1}{x}^{{}^{-\frac{1}{2}}}\phantom{\rule{0.167em}{0ex}}dx\\ \\ & ={\left[\frac{{x}^{{}^{\frac{1}{2}}}}{\frac{1}{2}}\right]}_{a}^{1}\\ \\ & =\left[2\sqrt{x}{\right]}_{a}^{1}\\ \\ & =2\sqrt{1}-2\sqrt{a}\\ \\ & =2-2\sqrt{a}\end{array}$
Now we got rid of the integral and we have a limit to find:
$\begin{array}{rl}\underset{a\to 0}{lim}{\int }_{a}^{1}\frac{1}{\sqrt{x}}\phantom{\rule{0.167em}{0ex}}dx& =\underset{a\to 0}{lim}\left(2-2\sqrt{a}\right)\\ \\ & =2-2\cdot 0\\ \\ & =2\end{array}$
Problem 2.1
${\int }_{0}^{8}\frac{1}{\sqrt[3]{\phantom{A}x}}\phantom{\rule{0.167em}{0ex}}dx=?$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• how to evaluate an improper integral whose limits have been given
• I assume you're asking how it is an improper integral if it is being evaluated using defined numbers, rather than infinity?

To be a proper integral, the area being calculated must be an enclosed space (bounded on all sides) - you need to be able to draw an outline with no openings around the area. When you are integrating between two x-values, the right and left side are enclosed by vertical lines at those x-values. But if the curve itself has a vertical asymptote, then the top edge of the area is open (instead of an x-value of infinity, you have a y-value of infinity - infinite height instead of infinite width, if you will).

Luckily if you have a vertical boundary set at the vertical asymptote, the vertical asymptote will converge on the vertical line, so it's possible to calculate the area just like with converging horizontal asymptotes. For evaluation, you calculate it just like any other definite integral. If the x-value boundaries are not at the asymptote, split it into two integrals, one evaluated from the lower bound to the asymptote and the other from the asymptote to the upper bound.
• How can i know if the improper integral is divergent?
• If the limit doesn't exist! Say, you evaluate the limit and get infinity (+ or -) then the integral will be divergent. Otherwise the limit should exist and it will be convergent.
• The solution of indefinite integral logsinx is ?
• That function can't be expressed in terms of elementary functions.
• My calculations for the review problem 2.2 give an answer of 3/2. Sal's answer is -3/2. The question is what is the area over the interval [0,1] for the function y=1/((x-1)^1/3). Can you confirm which is the correct answer?
• This integral has a problem when x=1, so substitute a for 1 and find the limit of the integral as a-->1
If you used u=(x-1) you should have gotten 3/2(x-1)^(2/3) + C
Now evaluate the integral from 0 to a to get:
3/2(a-1)^(2/3) - 3/2(0-1)^(2/3) = 3/2(a-1)^(2/3) - 3/2(-1)^(2/3)
= 3/2(a-1)^(2/3) - 3/2(-1)^(2/3)
= 3/2(a-1)^(2/3) - 3/2(³√(-1)²)
= 3/2(a-1)^(2/3) - 3/2(³√1)
= 3/2(a-1)^(2/3) - 3/2(1)
= 3/2(a-1)^(2/3) - 3/2

Now take the limit as a-->1 to get 3/2(1-1)^(2/3) - 3/2 = 0 - 3/2 = -3/2
• For Practice Set 2, the limit should be as a approaches 0 from the right.
• Is the integral from -1 to 1 of of 1/x equal to 0? Is this (cancelling out equal but opposite sized infinite regions) allowed for integrating all unbounded odd functions?
• You are correct. All Odd and Even Function rules apply
• at practice set 2, why isn't it
lim a->0^+, instead, it's shown as
lim a->0
is it a typo?
• You're right. It should be a one sided limit from the positive side