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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,
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,
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
Now we got rid of the integral and we have a limit to find:
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
Now we got rid of the integral and we have a limit to find:
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(4 votes)
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.(31 votes)
How can i know if the improper integral is divergent?(3 votes)
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.(19 votes)
- The solution of indefinite integral logsinx is ?(3 votes)
That function can't be expressed in terms of elementary functions.(8 votes)
- 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?(3 votes)
This integral has a problem whenx=1, so substituteafor1and find the limit of the integral asa-->1
If you usedu=(x-1)you should have gotten3/2(x-1)^(2/3) + C
Now evaluate the integral from0 to ato 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 asa-->1to get3/2(1-1)^(2/3) - 3/2 = 0 - 3/2 = -3/2(7 votes)
- For Practice Set 2, the limit should be as a approaches 0 from the right.(4 votes)
- 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?(4 votes)
You are correct. All Odd and Even Function rules apply(0 votes)
at practice set 2, why isn't it
lim a->0^+, instead, it's shown as
lim a->0
is it a typo?(2 votes)
You're right. It should be a one sided limit from the positive side(4 votes)
- 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?(1 vote)
- 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.(5 votes)
- 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)?(2 votes)
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.(2 votes)
- sir, please explain about the unbounded function?(2 votes)
