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.
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.