 Let us look at one more topic about integration techniques in chapter seven from James Stewart's calculus textbook. This is commonly referred to as the comparison test. Suppose we have two continuous functions, F and G, with the property that F is always bigger than G. I mean, they could be equal, but F is always greater than equal to G. They're both positive functions on the interval X is greater than equal to A. So we have some comparison about the functions. F is bigger than the, F is bigger than G. So if we were to graph this thing, we're saying something like the following, F is this function that's blue right here. And let me, let me reschedule. We're going to have F as something like this. This is F and then G might do something like the following, but G always sits below F. And so, and then, and this will be true as X is greater than or equal to, so like we see in this illustration. So what does this say about improper integrals? If we take the integral from A to infinity of F of X, if this thing is convergent, then that means the integral from A to infinity of G of X is likewise convergent. So what does convergent mean in this setting? We're saying if the area below F is finite, then the area below G must also be finite. If a number is less than a finite number, then it also has to be finite. Oh, you might think there's a loophole there. The only exception of that would be negative infinity, but that's why this is so important. If the numbers are positive, the smallest it could be was zero. And so negative infinity is not an escape hatch we can use here. So if the bigger function integrals converges, then the smaller functions integral must also converge. But in contrast, if the integral from A to infinity of G of X DX is divergent, then the integral from A to infinity of F of X must also diverge. Now, because these functions are both positive, the only way they could diverge is if they go off towards infinity. So if the area under the smaller one is infinite, then the area under the bigger one must also be infinity because there's no real number that is greater than infinity. So let me show you how you can use this. So let's consider the integral from zero to infinity of E to the negative X squared. Well, E to the negative X squared, that's a hard integral to do. It doesn't have an elementary anti-derivative. But some things we can do are the following. It's like, okay, notice that if I go from zero to infinity of E to the negative X squared DX, what I can do is I can break this up from zero to one E to the negative X squared DX plus integral from one to infinity, right? So that might seem like a weird thing to do at first, but it's perfectly legitimate. You can break up the interval however you want. Notice that zero to infinity, I'm sorry, zero to one is the first one we did. And I can guarantee without calculating this that this is gonna be convergent. How do I know that? Well, zero to one is a finite interval. So there's no issues about going to infinity. And E to the negative X squared is continuous on that interval. So there's no discontinuities. So this will be a proper integral. Proper integrals are always convergent. So if I'm worried about convergence, I don't have to worry about that first part. So I'm kicking the can down the road a little bit here and why is that? Well, that's cause I wanna make a comparison. Although I don't know how to integrate, I don't know how to find an anti-derivative for E to the negative X squared. At least I don't know an elementary anti-derivative. What I can do is the following comparison. So on this interval one to infinity, right? We're assuming that X, that X is greater than or equal to one. And so if X is greater than or equal to one, this would imply that X is less than X squared. That's why we actually had to kick the can down to one. X is gonna be less than X squared. And therefore, if you times both sides by negative one, negative X is greater than or equal to negative X squared. Again, this is on the assumption that X is greater than one. And then if you take the exponential, we're gonna get E to the negative X is greater than or equal to E to the negative X squared. And so what this tells us by comparison, comparing these two things, the integral from one to infinity of E to the negative X DX, this is gonna be greater than or equal to the integral from one to infinity of E to the negative X squared DX. So that's the comparison we can make there. And so the one on the left, I believe we can calculate this one. The antiderivative is much, much easier. The antiderivative E to the negative X is gonna be negative E to the negative X. Evaluate that from one to infinity. Plugging in, I think we can reverse the order, right? That's gonna be E to the negative X as you go from infinity to one. So we get E to the negative one minus E to the negative infinity. Or if you prefer, this is one over E minus one over infinity. Dividing by infinity will just be the same thing as going towards zero. And so we get that this integral is one over E, which is in particular a finite value. What this tells us is that this integral right here, the integral from one to infinity of E to negative X DX, this is a convergent integral. And so by the comparison test, if you are smaller than a convergent integral, then you yourself are convergent. You're convergent here. And so then if I come back up to the original one, we broke up our integral into two pieces. The first one was convergent because it was a proper integral. The second one is likewise convergent by the comparison test, the comparison test. So by the comparison test, the second one was convergent. So when you glue the two things back together, we see that this integral is in fact a convergent integral. It's really nice there. Now be aware that we know that this integral is convergent, but we don't know what the value is. We do not know what this thing adds up to be, at least not without a separate argument. Now we did compute that this integral will add up to be, this improper integral will be one over E. That doesn't tell us what this is. This just tells us that this one is less than one over E, right? So we have an upper bound for this thing, but we don't know what it is. We've determined convergence, we haven't actually computed it, but we do know that it's a finite number. As a second example of this, let's show that the integral from one to infinity of the function one plus E to the negative X over X DX is a divergent integral, all right? We're gonna show divergence. Now, before we do that one, let me go back to this one right here. We showed that this one was convergent, so the smaller one was convergent. One has to be cautious. If we had shown that this one was actually divergent, that actually says nothing about this one over here. We have to be cautious because if this thing was divergent, the blue one, that means its area turned out to be infinity. Well, what about the one on the right? It could be infinity too, which makes it divergent, but it could also be finite because finite's less than infinity. So be aware that the comparison test does not say anything about if the bigger one is divergent, I can't say anything about the smaller one. Likewise, if the smaller one is convergent, that doesn't say anything about the bigger one because convergent just means it's finite and the bigger one could be finite or it could be infinite. There are only two directions that the comparison test makes. Don't make the error of assuming, you know, applying the comparison test into situation it doesn't apply. So coming back to this example right here, let's show that it's divergent. And to do that, we're gonna make the following observation. E to the negative X is greater than equal to zero, and this is true for all X. In particular, this is true when X is greater than equal to one, right? E to the negative X is always positive. So if I add one to it, E to the negative X will be greater than equal to one. This is true for all X. And if X is positive, this then means that one plus E to the negative X over X will be greater than equal to one over X. I divided both sides of the inequality by X, which if that was a negative number it would switch the direction, but since we're assuming that X is greater than one, that's a positive value, no directions get switched around right here. And so this is the comparison we wanna do. One plus E to the negative X over X is greater than equal to one. So this tells us that the integral from one to infinity of one plus E to the negative X over X DX, this will be greater than or equal to the integral from one to infinity of DX over X. And we actually saw this one on a previous slide, but if we were to do it again, we're gonna get the natural log of X as you go from one to infinity. This is gonna give you the natural log of infinity minus the natural log of one, which ends up being infinity itself. So what this tells us is that the smaller integral, this one right here, it turned out to be divergent. And if it's divergent, that means it was infinite here. That then implies that the larger one is likewise divergent. And therefore, this integral is divergent by, again, the comparison test. And that brings us to the end of lecture 19. It also brings us to the end of chapter seven in James Stewart's calculus textbook. So that's a good place to sign off for today. In chapter eight, we're gonna return to some applications of anti-derivatives and integrals that we didn't see in chapter six of Stewart's textbook. So stay tuned for that. Some of those applications will involve numerical approximations. Some of them will involve actually improper integrals, believe it or not. And so these applications we had to postpone until now because now that we've developed techniques of integration, like we did in chapter seven, we're now ready to approach story problems which we were not able to do beforehand. So check out the link that you can see right now to see those videos and I'll hopefully see you then. Bye everyone.