Main content

# Introduction to improper integrals

AP.CALC:

LIM‑6 (EU)

, LIM‑6.A (LO)

, LIM‑6.A.1 (EK)

, LIM‑6.A.2 (EK)

Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. As crazy as it may sound, we can actually calculate some improper integrals using some clever methods that involve limits. Created by Sal Khan.

## Video transcript

What I want to figure
out in this video is the area under the curve
y equals 1 over x squared, with x equals 1 as
our lower boundary and have no upper
boundary, just keep on going forever and forever. It really is essentially
as x approaches infinity. So I want to figure out
what this entire area is. And one way that
we can denote that is with an improper
definite integral, or an improper integral. And we would denote it as
1 is our lower boundary, but we're just going to
keep on going forever as our upper boundary. So our upper
boundary is infinity. And we're taking the integral
of 1 over x squared dx. And so let me be very clear. This right over here is
an improper integral. Now how do we actually
deal with this? Well, by definition
this is the same thing as the limit as n
approaches infinity of the integral from 1 to
n of 1 over x squared dx. And this is nice, because we
know how to evaluate this. This is just a definite integral
where the upper boundary is n. And then we know
how to take limits. We can figure out what the limit
is as n approaches infinity. So let's figure out if we can
actually evaluate this thing. So the second fundamental
theorem of calculus, or the second part of
the fundamental theorem of calculus, tells us that
this piece right over here-- just let me write
the limit part. So this part I'll just rewrite. The limit as n
approaches infinity of-- and we're going to use the
second fundamental theorem of calculus. We're going to evaluate
the antiderivative of 1 over x squared or x
to the negative 2. So the antiderivative
of x to the negative 2 is negative x to the negative 1. So negative x to the negative
1 or negative 1 over x. So negative 1/x is
the antiderivative. And we're going to evaluate
at n and evaluate it at 1. So this is going to be equal
to the limit as n approaches infinity. Let's see, if we evaluate this
thing at n, we get negative 1 over n. And from that we're
going to subtract this thing evaluated at 1. So it's negative 1 over
1, or it's negative 1. So this right over
here is negative 1. And so we're going to find the
limit as n approaches infinity of this business. This stuff right here is
just the stuff right here. I haven't found the limit yet. So this is going to be equal
to the limit as n approaches infinity of-- let's see,
this is positive 1-- and we can even write that minus
1 over n-- of 1 minus 1 over n. And lucky for us, this
limit actually exists. Limit as n approaches infinity,
this term right over here is going to get closer and
closer and closer to 0. 1 over infinity you can
essentially view as 0. So this right over
here is going to be equal to 1, which
is pretty neat. We have this area that
has no right boundary. It just keeps on going forever. But we still have a
finite area, and the area is actually exactly equal to 1. So in this case we had
an improper integral. And because we were actually
able to evaluate it and come up with the number that this
limit actually existed, we say that this improper
integral right over here is convergent. If for whatever reason
this was unbounded and we couldn't come up with
some type of a finite number here, if the area
was infinite, we would say that it is divergent. So right over here we figured
out a kind of neat thing. This area is exactly 1.