Evaluating improper integrals
Divergent improper integral
Right here I have graphed part of the graph of y is equal to 1/x. And what I'm curious about is the area under this curve and above the x-axis between x equals 1 and infinity. So I want to figure out this area right over here. So let's try to do it. So we could set this up as an improper integral going from 1 to infinity of 1/x dx. Well once again-- actually, let me do that same yellow color. I like that more-- we can view this as the limit as n approaches infinity of the integral from 1 to n of 1/x dx, which we can write as the limit as n approaches infinity of the antiderivative of 1/x, which is the natural log of the absolute value of x. So this is going to be the natural log of the absolute value of x. And the absolute value of x won't really matter so much. We could just say x because we're dealing with positive values of x. But I'll just write it down as the natural log of the absolute value of x between x is 1 and x is n. And so this is going to be equal to the limit as n approaches infinity of-- you evaluate this at n, so you're going to get the natural log-- I could write the absolute value of n, but we know that n is going to be positive. So we could just write the natural log of n minus the natural log of the absolute value of 1, or the natural log of 1. Natural log of 1 is just a 0. e to the 0-th power is equal to 1. So this boils down to the limit as n approaches infinity of the natural log of n. Now this is interesting. Natural log function just keeps getting larger and larger and larger. The natural log function keeps growing and growing and growing like this, albeit at a slower and slower pace, but it keeps growing. The limit as n approaches infinity of the natural log of n is just equal to infinity. So here we do not have a finite area. This is an infinite. This is an infinite area. It's interesting. When this function decreased faster-- when it was 1 over x squared-- we had a finite area. Now we have an infinite area. And so we would say that this integral right over here, this improper integral, is divergent. And we're done.
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