 In this video, we provide the solution to question number 12 for the practice exam number two for math 12-20, in which case we have to show that the integral from zero to one of e to the x over x squared dx diverges. Notice that this is an improper integral, excuse me, because as x goes to zero, you're gonna have to divide by zero right here. There's a vertical acetote on that graph. So let's show that it's divergent. And it says to name the major theorem that we're gonna use. We're gonna probably have to use the comparison test. Now to make the comparison test work for a divergent integral here, what we have to do is we have to show that a smaller integral is divergent than this one right here. Because if this thing is divergent, it's going off towards infinity, and therefore we wanna find something smaller, something simpler. And so we have to find something to compare two. So we might write something like the following. Note that on the interval and the interval in play here is gonna be the interval zero to one. This is the interval for which we're integrating across. On the interval zero to one, we have that e to the x is greater than or equal to one. Notice that when x equals zero, e to the zero is equal to one, e to the x is an increasing function. So as it ranges from zero to one, e to the x is only getting bigger. And so e to the x is always greater than or equal to one. Thus, if you take this inequality and if you divide everything by the positive quantity x squared, you're gonna get that e to the x divided by x squared is greater than or equal to one over x squared. And so this right here is the critical observation because this is the function we were given. It's anti-derivative is too hard to approach directly. But by comparison, this is a smaller function. And so I wanna consider the limit or the integral of the one over x squared here. So then the next thing we say something's like now, observe the following. If I take the integral from zero to one of one over x squared dx in this situation, I can actually calculate an anti-derivative. The anti-derivative of one over x squared will be negative one over x. You evaluate this from zero to one. Now, because of this negative sign here, I can actually swap the bounds to make it positive one over x. You evaluate it from one to zero in that situation for which if you were to plug in zero, you divide by zero, that's really a limit that we're trying to compute here. You're taking the limit of one over x as x approaches zero from, well, I mean, we have to be very careful on this one. We wanna approach it from the positive side there. And then we're gonna subtract from this. Well, what happens as x approaches one there but that one's just one like so. Now this limit right here is the one we're interested in as x approaches zero one over x is gonna approach infinity. So you end up with this infinity minus one, which I should mention is just the same thing as infinity. And so therefore the integral from zero to one of one over x squared dx diverges by the calculation we just said. So that's what we're trying to do here since this integral turned out to be infinity. And since the integral we care about is larger than the integral we just computed, we can then say by the comparison test, make sure you actually state the comparison test here and don't just say it, like actually put it in the statement where you use it by the comparison test, the integral from zero to one of e to the x over x squared dx likewise diverges. And so now we have proven that this integral diverges using the comparison test. So things that you need to see here is first of all to use the comparison test you need a comparison. You need to compare your function to something smaller if you want it to be divergent or you compare it to something bigger if you want to be convergent. You then, so after you have a comparison there and that comparison should be valid on the interval that you're considering here, once you then have the comparison then you compute the improper integral for the one you're comparing to and argue that it's either convergent or divergent. And then once you know that it's divergent or convergent then you invoke the comparison test to show that yours is also likewise divergent depending on which direction you're going there. So those are the three ingredients you need. You need a comparison, you need to evaluate the integral you're comparing to and then you invoke the comparison test to get that the integral you care about is either convergent or divergent.