 In this video, let's discuss the solution to question 15 from the practice midterm exam for characters two, math 12-20. We're asked to show that the integral from zero to one, e to the x over x squared dx diverges. Well, you could try to do this directly, we really could, by doing some type of like integration by parts. We could do u equals e to the x. We could then do du is then e to the x dx. dv is gonna equal one over x squared dx. And then u is gonna equal, we're gonna end up with a negative one over x. Like so, then you could try to rinse and repeat and maybe eventually get something out of that. That might work, right? I mean, there's gonna be more to it than that. That might work, but turns out this is a much easier way. Cause it talks about using a major theorem here. We're gonna use the comparison test. And so what we're gonna say is the following. If you look at the graph y equals e to the x, notice that the standard graph is gonna look something like the following, e to the x. I have this increasing function, exponential growth, right? And so we wanna go from zero to one. This function is increasing on this bounds. And so what we can see that is on this interval, we're gonna see that e to the x is greater than or equal to one, great. Why that is useful is then if you divide everything by x squared, you're gonna get e to the x divided by x squared, which is a positive expression, is greater than or equal to one over x squared. And so this is a comparison between two functions. We have a comparison here. We have one function is greater than the other function. And so therefore the integral from zero to one of e to the x over x squared dx. This will be greater than or equal to the integral from zero to one of one over x squared dx. And as we saw already, as we were trying to do integration by parts, this is a much easier integral. You end up with negative one over x as you evaluate from zero to one. Flipping the direction, you get one over x as you evaluate from one to zero. And then as you approach zero here, you're of course approaching zero from the right. You're gonna end up with the limit as x approaches zero from the right of one over x. When you plug in one, there's no issue there. You get minus one. But the first limit is gonna turn out to be infinity. You get infinity minus one, which itself can be infinity. This is indication that this series diverge, or this integral diverges here. So we kind of comment. So note that the integral from zero to one of one over x squared dx, it diverges. And so therefore we can conclude, thus the integral from zero to one e to the x over x squared dx. It likewise diverges, and likewise diverges by the comparison test, comparison test. And that's all that one has to do to make this argument here. So this is the major theorem that we were using the comparison test. We actually have to have a comparison. That means an inequality of some kind. You then show that the other integral diverges or converges and makes the comparison there. But I should also mention that the comparison test has limitations. If you have two functions, let's say that f of x is less than or equal to g of x, like so. And so then that would tell us that comparison right there then tells us that the integral of f of x dx will be less than or equal to the integral of g of x dx. Now, if the, whoops, in this situation, if the lower function diverges, that basically means that this thing is equal to infinity. Well, if someone's greater than or equal to infinity, that means that likewise it's infinity and therefore we can infer that it likewise diverges. That's what we saw in this example. Divergence applies there. Well, on the other hand though, if the bigger one converges, if the bigger one converges, that actually means that the bigger integral is going to be strictly less than infinity. Well, if the bigger one's strictly less than infinity, the smaller one has to strictly be less than infinity as well. And that would then tell us that the smaller one converges, like so. So the larger one being convergent implies that the smaller one's convergent. And that's as well as the comparison test can do. So the comparison test gives us these two implications. If the smaller one diverges, the bigger one diverges. If the bigger one converges, the smaller one converges as well. What we cannot say is that if the smaller one converges, we can say nothing about the bigger one because if the smaller one converges, that means it's less than equal to infinity. But what's greater than a finite number? Well, a finite number could be or infinity, right? There's no inference there. And likewise, if the bigger one diverges, that means nothing for the smaller one because of the bigger one's infinity, we could be less than infinity or equal to infinity. And so there's no statement there. And so on question number 15, you wanna be prepared to say something maybe about the comparison tests, like in this example here.