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.

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`

The solution of indefinite integral logsinx is ?

That function can't be expressed in terms of elementary functions.

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

For improper integrals that don't include infinity is it necessary to rewrite them as limits? So far all the examples I have done for those I evaluated them as if they were normal definite integrals and got the same answer.

You will still get the same answer, but it’s not as formal to do it that way. If you are taking the AP exam, make sure you write them with limits because that’s technically how they are supposed to be, and you will be graded for it.

I do not understand why the integral of 1/sqtx from 0 to 1 is an unbounded function. Is it because it was a restriction in its domain (x cannot be equal to 0 even though 0 is in the domain of sqtx)?

The problem is that 0 is not in the domain of the function we're integrating 1/√x, and consequently the fundamental theorem of calculus does not apply. This is why we have to be a bit more ingenious and use the improper integral: Lim a->0 ∫ a, 1 (1/√x) dx to convince ourselves that the integral converges.

This is a side question. If f(x)=sin (pi*x), then my graph is a wave with a range of -1 to 1. If instead f(x) = sin x, does the graph then have a range of -2pi to 2pi?

No, the range of f(x) = sin(k*x) is ALWAYS from y = -1 to y = 1, for any value of k (except the silly value of k = 0). What the k does is change the *period* of the wave. f(x) = sin(x) has a period (or wavelength) of 2pi, while f(x) = sin(k*x) has a period of 2pi/k. Thus, f(x)=sin (pi*x) has a period of 2. In order to change the range, you would put a coefficient in front. f(x) = A*sin(k*x) has a range from y = -A to y = A.

Main content

Integral Calculus

Course: Integral Calculus > Unit 1

Lesson 19: Improper integrals

Improper integrals review

Google Classroom

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, integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x is an improper integral. It can be viewed as the limit limit, start subscript, b, \to, infinity, end subscript, integral, start subscript, 1, end subscript, start superscript, b, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x.

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, integral, start subscript, 0, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x is an improper integral. It can be viewed as the limit limit, start subscript, a, \to, 0, start superscript, plus, end superscript, end subscript, integral, start subscript, a, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x.

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.

Want to learn more about improper integrals? Check out this video.

Practice set 1: Evaluating improper integrals with unbounded endpoints

Let's evaluate, for example, the improper integral integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x. As mentioned above, it's useful to view this integral as the limit limit, start subscript, b, \to, infinity, end subscript, integral, start subscript, 1, end subscript, start superscript, b, end superscript, start fraction, 1, divided by, x, squared, end fraction, d, x. We can use the fundamental theorem of calculus to find an expression for the integral:

\begin{aligned} \displaystyle\int_1^b\dfrac{1}{x^2}\,dx&=\displaystyle\int_1^b x^{-2}\,dx \\\\ &=\left[\dfrac{x^{-1}}{-1}\right]_1^b \\\\ &=\left[-\dfrac{1}{x}\right]_1^b \\\\ &=-\dfrac{1}{b}-\left(-\dfrac{1}{1}\right) \\\\ &=1-\dfrac{1}{b} \end{aligned}

Now we got rid of the integral and we have a limit to find:

\begin{aligned} \displaystyle\lim_{b\to\infty}\int_1^b\dfrac{1}{x^2}\,dx&=\displaystyle\lim_{b\to\infty}\left(1-\dfrac{1}{b}\right) \\\\ &=1-0 \\\\ &=1 \end{aligned}

Problem 1.1

Current

integral, start subscript, 1, end subscript, start superscript, infinity, end superscript, start fraction, 1, divided by, x, cubed, end fraction, d, x, equals, question mark

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 integral, start subscript, 0, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x. As mentioned above, it's useful to view this integral as the limit limit, start subscript, a, \to, 0, end subscript, integral, start subscript, a, end subscript, start superscript, 1, end superscript, start fraction, 1, divided by, square root of, x, end square root, end fraction, d, x. Again, we use the fundamental theorem of calculus to find an expression for the integral:

\begin{aligned} \displaystyle\int_a^1\dfrac{1}{\sqrt x}\,dx&=\displaystyle\int_a^1 x^{^{\large -\frac{1}{2}}}\,dx \\\\ &=\left[\dfrac{x^{^{\large\frac{1}{2}}}}{\frac{1}{2}}\right]_a^1 \\\\ &=\Bigl[2\sqrt x\Bigr]_a^1 \\\\ &=2\sqrt 1-2\sqrt a \\\\ &=2-2\sqrt a \end{aligned}

Now we got rid of the integral and we have a limit to find:

\begin{aligned} \displaystyle\lim_{a\to 0}\int_a^1\dfrac{1}{\sqrt x}\,dx&=\displaystyle\lim_{a\to 0}(2-2\sqrt a) \\\\ &=2-2\cdot 0 \\\\ &=2 \end{aligned}

Problem 2.1

Current

integral, start subscript, 0, end subscript, start superscript, 8, end superscript, start fraction, 1, divided by, cube root of, x, end cube root, end fraction, d, x, equals, question mark

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

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Miyoba G. Mwiinga
Posted 7 years ago. Direct link to Miyoba G. Mwiinga's post “how to evaluate an improp...”
how to evaluate an improper integral whose limits have been given
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(4 votes)
Answer
- Cassie
  Posted 7 years ago. Direct link to Cassie's post “I assume you're asking ho...”
  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.
  Comment on Cassie's post “I assume you're asking ho...”
  (27 votes)
camille benoit
Posted 7 years ago. Direct link to camille benoit's post “How can i know if the imp...”
How can i know if the improper integral is divergent?
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(3 votes)
Answer
- Lydia Wood
  Posted 6 years ago. Direct link to Lydia Wood's post “If the limit doesn't exis...”
  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.
  Comment on Lydia Wood's post “If the limit doesn't exis...”
  (14 votes)
D. Ashley Nelson
Posted 7 years ago. Direct link to D. Ashley Nelson's post “My calculations for the r...”
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?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Stefen
  Posted 7 years ago. Direct link to Stefen's post “This integral has a probl...”
  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
  Comment on Stefen's post “This integral has a probl...”
  (7 votes)
umeshbhattarai2014
Posted 5 years ago. Direct link to umeshbhattarai2014's post “The solution of indefinit...”
The solution of indefinite integral logsinx is ?
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(3 votes)
Answer
- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “That function can't be ex...”
  That function can't be expressed in terms of elementary functions.
  Button navigates to signup page
  (7 votes)
Nathan Mackouse
Posted 6 years ago. Direct link to Nathan Mackouse's post “Is the integral from -1 t...”
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?
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(4 votes)
Answer
- Wayne Clemensen
  Posted 6 years ago. Direct link to Wayne Clemensen's post “You are correct. All Odd ...”
  You are correct. All Odd and Even Function rules apply
  Button navigates to signup page
  (0 votes)
steve.grosteffon
Posted 4 months ago. Direct link to steve.grosteffon's post “For Practice Set 2, the l...”
For Practice Set 2, the limit should be as a approaches 0 from the right.
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(3 votes)
Answer
D. Ashley Nelson
Posted 7 years ago. Direct link to D. Ashley Nelson's post “This is a side question. ...”
This is a side question. If f(x)=sin (pi*x), then my graph is a wave with a range of -1 to 1. If instead f(x) = sin x, does the graph then have a range of -2pi to 2pi?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- robshowsides
  Posted 7 years ago. Direct link to robshowsides's post “No, the range of f(x) = s...”
  No, the range of f(x) = sin(k*x) is ALWAYS from y = -1 to y = 1, for any value of k (except the silly value of k = 0). What the k does is change the period of the wave. f(x) = sin(x) has a period (or wavelength) of 2pi, while f(x) = sin(k*x) has a period of 2pi/k. Thus, f(x)=sin (pi*x) has a period of 2.
  
  In order to change the range, you would put a coefficient in front. f(x) = A*sin(k*x) has a range from y = -A to y = A.
  Comment on robshowsides's post “No, the range of f(x) = s...”
  (5 votes)
Liang
Posted 3 months ago. Direct link to Liang's post “at practice set 2, why is...”
at practice set 2, why isn't it
lim a->0^+, instead, it's shown as
lim a->0
is it a typo?
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(2 votes)
Answer
- Venkata
  Posted 3 months ago. Direct link to Venkata's post “You're right. It should b...”
  You're right. It should be a one sided limit from the positive side
  Comment on Venkata's post “You're right. It should b...”
  (3 votes)
samirchokshi11
Posted 2 months ago. Direct link to samirchokshi11's post “For improper integrals th...”
For improper integrals that don't include infinity is it necessary to rewrite them as limits? So far all the examples I have done for those I evaluated them as if they were normal definite integrals and got the same answer.
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(2 votes)
Answer
- Tanner P
  Posted 2 months ago. Direct link to Tanner P's post “You will still get the sa...”
  You will still get the same answer, but it’s not as formal to do it that way. If you are taking the AP exam, make sure you write them with limits because that’s technically how they are supposed to be, and you will be graded for it.
  Button navigates to signup page
  (3 votes)
BassDant21
Posted 5 years ago. Direct link to BassDant21's post “I do not understand why t...”
I do not understand why the integral of 1/sqtx from 0 to 1 is an unbounded function. Is it because it was a restriction in its domain (x cannot be equal to 0 even though 0 is in the domain of sqtx)?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Howard Bradley
  Posted 5 years ago. Direct link to Howard Bradley's post “The problem is that 0 is ...”
  The problem is that 0 is not in the domain of the function we're integrating 1/√x, and consequently the fundamental theorem of calculus does not apply. This is why we have to be a bit more ingenious and use the improper integral:
  Lim a->0 ∫ a, 1 (1/√x) dx to convince ourselves that the integral converges.
  Button navigates to signup page
  (2 votes)