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## AP®︎ Calculus BC (2017 edition)

### Unit 9: Lesson 6

Improper integrals

# Improper integral with two infinite bounds

A worked example of a challenging improper integral that involves two infinite bounds and an inverse trig substitution. Created by Sal Khan.

## Want to join the conversation?

• At : Why do you substitute 5tan*delta for x?
• Could we also have said we compute the improper integral with one infinite bound and multiply the result by 2 if the graph is symmetric to the x - axis?
• In general, yes, as long as you are certain that your function is even, i.e. symmetric about the y axis. Proving a function is even is not the goal of this video, so Sal ignored that useful tidbit you noticed.
In Sal's example, the function turns out to be even, and as the calculations in the last minute of the video show, computing one of the integrals and doubling it gives the correct answer.
• at , how do you conclude that arctan of negative infinity is -pi/2?
• Remember that the tan or arctan of an angle on the unit circle represents its slope. A slope of negative infinity is a vertical line drawn downwards from the origin ( if using the center of the unit circle as the reference point). This angle is 270 degrees which is expressed as 3pi/2 or -pi/2 depending on which way you approach it. In this case, we approached it from the negative direction.
• Just to make clear... an improper integral is basically just a definite integral with one or both boundries being -infinite or +infinite?
• Close. An improper integral can also have a singularity within the bounds. For example:
∫ 1/x from 0 to 1
As x → 0, 1/x → ∞
So an improper integral is either bounded by ±infinity, or approaches infinity at one or both of the bounds.
• Is trig substitution the only method of integration we could use here?
• No, you could also have used a u-substitution. Divide the numerator and denominator by 25, and you can write the integrand as 10/(1 + x^2 /25) dx, or 10/(1 + (x/5)^2) dx. If we substitute u = x/5, so that du = (1/5)dx, we see that the integrand becomes 50/(1 + u^2) du. Integrating this function is easy--its antiderivative is 50arctan(u) + C or 50arctan(x/5) + C, which is what Sal obtained on the video via his trig substitution. Hope that helps!
• Where did the tangents and arctangents come into play? Everything was making sense up until that point
• well, to understand the role of tangents and arctangents you need to have some basic
knowledge about integration and calculus...if you do then observe sal using trig
substitution
to solve the integration of 250/25 + x^2. you can even use a direct formula ;
integration 1/a^2 + x^2 where a=5 & x=x
or you can use u substitution . Hope that helps! :)
• The integral in this video demonstrates an area under the curve of 50pi. But the very next video "Divergent Improper Integral" shows an area of infinity under the curve of 1/x. The curve on this page (250/(25+x^2)) looks like it should be at least twice as large as that under the curve of 1/x. It goes to infinity in two different directions, whereas 1/x only goes to infinity in one direction.
How can the curve on this page yield an area of only 50pi, where as the curve of 1/x yields an area of infinity?
• 1/x^2 becomes smaller faster than 1/x, and all the other numbers (multiplying by 250 and adding 25) are just "applied" to a finite number, thus the results can be only finite.

You can compare that to the discrete case. The sum from 1 to infinity of 1/x is infinity (the sum is divergent), but the sum of 1/x^2 is finite (=pi^2/6).
• Couldn't the improper integral be written as the lim as n--> infinity of the integral from -n to n?
• To my knowledge yes it could be, I think you don't even need a limit for this problem because you don't get an indeterminate form when inputting the boundaries into the function..