 In this video, we'll discuss why the integral test makes intuitive sense. While this is not a formal proof, it provides some justification for why the test works in determining whether a series converges or diverges. As you read, the integral test essentially tells us that an infinite series and the improper integral associated with that series, either both converge or both diverge. Let's consider an example. Suppose we look at the infinite series, the sum from k equals 1 to infinity of k squared. And it's associated integral, the integral from 1 to infinity of x squared dx. If we evaluate the improper integral, we find that this is equal to the limit as b approaches infinity of 1 third x cubed, evaluated from 1 to b, is equal to the limit as b approaches infinity of 1 third b cubed minus 1 third. Now, this limit as b approaches infinity of 1 third b cubed is actually infinity. So this integral diverges. Consider the graphical relationship between this integral and the series, the sum from k equals 1 to infinity of k squared. Well, let's build using a right-hand sum. We'll build our rectangles such that delta x is 1, so the width of each of these rectangles is 1, and the height is determined by the function y equals x squared. So for example, the area of this first rectangle here is 1 times its height, since this x value is 2, this height is 2 squared. So the area of this rectangle is 2 squared. The area of this rectangle is 1 times 3 squared, and this rectangle is 1 times 4 squared, and so on. What we notice is that this can be written as the sum from k equals 2 to infinity of k squared. Now, it doesn't include the very first term of the series we were investigating, but what we find is that the sum of these rectangles is going to be greater than the area under this curve, y equals x squared from 1 to infinity. What we found in our previous slide is that the integral associated, or the area under the curve from 1 to infinity, diverges. Now, this sum of the areas is greater than that, so this sum must also diverge. So if this sum also diverges, then certainly if we add one more term, which would be the 1 squared, then the entire series diverges. Let's consider a second example. The series, the sum from 1 to infinity of 1 over k squared, and its associated integral, the integral from 1 to infinity of 1 over x squared dx. Now, if we evaluate this integral, we find that it's equal to the limit as b approaches infinity of negative 1 over x evaluated from 1 to b using the fundamental theorem of calculus. We find that this is equal to the limit as b approaches infinity of negative 1 over b minus a negative 1 over 1. Now, the limit as b approaches infinity of this term is 0. So this integral is equal to 1, which means that it converges. Now, consider the graphical relationship between this integral and the series k equals 1 to infinity of 1 over k squared. So here's a graph of 1 over x squared, and we're interested in the area under this curve, or that's what we evaluated in the previous slide, the area under the curve from 1 to infinity. Now, as before, if we construct rectangles using a right hand sum with a delta x of 1 as before, we find that the sum of the areas of these rectangles is going to be less than the area under the curve of 1 over x squared from 1 to infinity. For example, the area of that rectangle is 1 times 1 over 2 squared, because the height is determined by dysfunction value. The area of that rectangle is 1 times 1 over 3 squared. Similarly, the area under the area of this rectangle is 1 times 1 over 4 squared, and so on. We recognize that to be the series from k equals 2 to infinity of 1 over k squared. Now, our series of interest began where k was 1. We can write this as 1 over 1 squared plus this series, which is the sum of the areas of these rectangles. Now, in the previous slide, we proved that the integral from 1 to infinity of this function, 1 over x squared, converges. And what we're seeing here graphically is that the sum of these rectangles of the area of the rectangles is less than the area under the curve of 1 over x squared. Since the integral converges, then so must the series. And if this converges, then so must 1 plus it. This should give us a sense of why the integral test works. The test is quite powerful as it gives us a definitive answer about whether a series converges or diverges when we use it. The challenge is knowing when the test is appropriate to use. If we can integrate the associated function f of x, then the test is a good candidate to try. So in summary, the integral test tells us that if the series a sub k has positive terms and a sub k is equal to f of k, where f is a continuous positive decreasing function, where k is greater than or equal to some real number c, if the integral from c to infinity of f of x dx converges, then so does the associated series. And if the integral diverges, then so does the associated series.