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### Course: Calculus, all content (2017 edition) > Unit 4

Lesson 18: 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?

- At03:09: Why do you substitute 5tan*delta for x?(65 votes)
- As Jacob said, it's called trig substitution.

For this specific case, tangent was used.

Here's a link to a video explaining trig substitution using tangent:

http://www.khanacademy.org/math/calculus/integral-calculus/trig_substitution/v/trig-substitution-with-tangent(65 votes)

- 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?(20 votes)
- 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.(12 votes)

- at9:51, how do you conclude that arctan of negative infinity is -pi/2?(14 votes)
- 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.(10 votes)

- Just to make clear... an improper integral is basically just a definite integral with one or both boundries being -infinite or +infinite?(7 votes)
- 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.(19 votes)

- Is trig substitution the only method of integration we could use here?(4 votes)
- 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!(16 votes)

- Where did the tangents and arctangents come into play? Everything was making sense up until that point(8 votes)
- 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! :)(0 votes)

- 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?(3 votes)- 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).(6 votes)

- Couldn't the improper integral be written as the lim as n--> infinity of the integral from -n to n?(5 votes)
- 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..(2 votes)

- At03:57: Why is dx equal to 5 sec^2*delta*d*delta?(4 votes)
- Since x=5tan(x), its derivative - which is dx - will be 5sec^2(x). The derivative of tan(x)=sec^2(x).(3 votes)

- It really blows my mind that a region with infinite bounds can have a finite area! How does that work? Does it just approach that value as its area? Sorry if my question doesn't make sense, I'm just trying to wrap my head around this.(2 votes)
- Yes. Improper integrals are always strictly defined in terms of a limit, even if the notation used does not indicate this.(5 votes)

## Video transcript

Right here we have the graph
of y is equal to 250 over 25 plus x squared. And what I'm curious
about in this video is the total area under this
curve and above the x-axis. So I'm talking about everything
that I'm shading in white here, including what we can't see,
as we keep moving to the right and we keep moving to the left. So I'm talking about from
x at negative infinity all the way to x at infinity. So first, how would we
actually denote this? Well, it would be an
improper integral. We would denote this area
as the indefinite integral from x is equal to
negative infinity to x is equal to
infinity of our function, 250 over 25 plus x squared, dx. Now, we've already
seen improper integrals where one of our
boundaries was infinity. But how do you do
it when you have one boundary at
positive infinity and one boundary at
negative infinity? You can't take a limit
to two different things. And so the way that we're
going to tackle this is to actually
break up this area into two different
improper integrals, one improper integral that describes
this area right over here in blue from negative
infinity to 0. So we'll say that this is equal
to the improper integral that goes from negative infinity
to 0 of 250 over 25 plus x squared dx, plus
the improper integral that goes from 0 to
positive infinity. So plus the improper,
or the definite, integral from 0 to positive
infinity of 250 over 25 plus x squared dx. And now we can start
to make sense of this. So what we have in
blue can be rewritten. This is equal to the limit as
n approaches negative infinity of the definite integral
from n to 0 of 250 over 25 plus x squared dx. Plus-- and I'm running
out of real estate here-- the limit as--
since I already used n, let me use m now-- the
limit as m approaches positive infinity of the
definite integral from 0 to m of 250 over 25
plus x squared dx. So now all we have
to do is evaluate these definite integrals. And to do that, we
just have to figure out an antiderivative of 250
over 25 plus x squared. So let's try to figure
out what that is. I'll do it over
here on the left. So we need to figure out the
antiderivative of 250 over 25 plus x squared. And it might already
jump out at you that trig substitution
might be a good thing to do. You see this pattern of
a squared plus x squared, where in this
case, a would be 5. So we can make the
substitution that x is equal to a tangent
theta, 5 tangent theta. And since we're going to have
to reverse substitute later on, we can also put in
the constraint-- well, we'd say x/5 is equal to tangent
of theta, which is completely consistent with this
first statement. And so if we wanted to
have theta expressed as a function of x, we
can put the constraint that theta is equal
to arctangent of x/5. So once again, this is
completely consistent with this over here. x
can be 5 tangent of theta, and theta can be equal
to arctangent of x/5. So now let's do
the substitution. Actually, before
that, we also have to figure out what
dx is equal to. So dx is equal to-- I'll do
it right over here-- well, the derivative of this
with respect to theta is 5 secant squared
theta d theta. So now we're ready to
substitute back in. So all of this business is going
to be equal to 250 times dx. Well, 250 times dx is 250 times
5 secant squared theta d theta. So that's 250 dx is
this business up here. All of that over
25 plus x squared. Well, x squared is going to be
25 tangent squared of theta. And now we can try to
simplify all of this business. This is equal to
250 times 5 secant squared theta over
25 times 1 plus tangent squared theta d theta. So 25 goes into 250 10 times. And then 1 plus tangent squared
theta is secant squared theta. You can prove it for yourself
if you write tan squared theta as sine squared theta
over cosine squared theta. Add that to 1, which
is the same thing as cosine squared theta
over cosine squared theta. And then you can use some
basic trig identities to realize that this is equal
to secant squared theta, which simplifies our
expression a good bit. You get secant squared theta
over secant squared theta, which is just 1. And so you're just left with
the 10 times the 5, 50 d theta. So this is equal
to 50-- I'll take the 50 outside the
integral-- 50 d theta, which is equal to 50 theta. We can write the
plus C, just to show that these are all of
the antiderivatives. But we only need the
most basic antiderivative to evaluate these
definite integrals. But we right now only
have it in terms of theta. Let's write this in terms of x. We set the constraint that theta
is equal to arctangent of x/5. So this is equal to 50
arctangent of x/5 plus C. These are all the
antiderivatives. We can pick C is
equal to 0 to find an antiderivative
of these things to evaluate the
definite integrals. So let's do that. So what we have in blue, we
can rewrite as the limit as n approaches negative infinity
of the antiderivative of this, or an antiderivative of
this, which we could say is 50 arctangent of x/5. And we're going to
evaluate it at 0 and n. And to this, we're going to
add the limit as m approaches positive infinity of just
the antiderivative of this, 50 arctangent-- not "actangent,"
arctangent-- arctangent of x/5 evaluated from 0 to m. And let me put a little
parentheses right around this x/5. So what's this going to be? This is going to be 50--
so let me write this. This is going to be the
limit as n approaches negative infinity of
50 arctangent of 0/5 / minus 50 arctangent of n/5. And then to that, we're going
to add-- let me give myself even a little bit more space--
the limit as m approaches infinity of 50 arctan of m/5
minus 50 arctan-- I think you see where all this is
going-- minus 50 arctan of 0/5. So now we can
evaluate these things. And to help us
evaluate them, let's think about a unit circle. So let's think
about a unit circle so we can really visualize
the arctangent function. So one way to think
about the tangent is the slope of the line
that is on the angle or that helps define the
terminal side of an angle. So for example, if we have an
angle that looks like this-- this is an angle between
the positive x-axis and this line right over here. If we have an angle just like
that, the tangent of that angle is going to be the
slope of this line. And so one way to
think about it is, OK, if I want the arctangent
of 0, it's essentially saying, OK, let's get an angle where
the terminal side has slope 0. Well, an angle where the
terminal side has slope 0 is an angle of 0. So arctangent of 0 is 0 radians. So it's 50 times 0. That's just going to be 0. And we could also
write that over here. This is just going to be 0. So what we're left
with is the limit as n approaches
negative infinity of negative 50 arctangent of n/5
plus the limit as m approaches positive infinity of
50 arctangent of m/5. So let's think about what
these are going to be. So it's the limit as n
approaches negative infinity. One way to view
that is the limit as the slope of
this terminal side approaches negative infinity. So that's getting more
and more negative, more and more negative. It's at negative
infinity, or it's approaching negative infinity
when our angle right over here is negative pi/2 radians. Or you could say the
limit of arctan of n/5 as n approaches negative
infinity, this part right over here, is
going to approach, as n approaches
negative infinity, it's going to be negative pi/2. And then we're just
going to multiply that times the negative 50. So this is going to be negative
50 times negative pi/2, which is going to be positive 25 pi. So this is going to be 25 pi. And similarly,
arctan of m/5 as m approaches infinity-- well,
as m approaches infinity, the slope of the terminal
side of the angle approaches infinity. So the slope is
getting higher, higher. It approaches infinity as
it gets closer and closer to vertical. So the arctan of
m/5 as m approaches infinity is going
to be positive pi/2. This is an angle
of positive pi/2. So what we see in
orange right over here is going to be
positive 50 times pi/2, which is going to be 25 pi. So the area that
we have in blue, going back to our original
problem, is 25 pi. The area that we have
in orange is 25 pi. And so if we wanted to
answer our original question, what's the total area
under this curve-- which is kind of a cool
question to try to answer-- we get it's 25 pi
plus 25 pi, which is 50 pi. And we're done.