 Okay, yes. Well, I have no control over that. So it's before the physics exam. We're going to do two review sessions. I'll do one Sunday afternoon. And Professor Bonifat will do one Tuesday evening. We're sort of constrained by many who can get rooms large enough to hold more than 100 people. So the review session that we're going to do on the Tuesday before, I think it's at, what is the class slot that ends at 8 something? 650. 650, right? Yeah. So this is at 650. And I still need to get in time for this one, but Sunday afternoon. Yeah. Yes, there are, well, no, there are classes. There are no other exams. If you have a class at that time, you need to make arrangements not to go. So this time was announced when you, this time, and the time of the other one, was scheduled last January. So even before I knew I was teaching this class, we had no control over that. The registrar can't say no. Yeah. Can you say no? Oh, the review. I could review the Sunday, I could do the Sunday one. I'm not so sure about the Tuesday one. I will, I will try to review, I have to review. Video for the Sunday one. It probably won't be up until Tuesday or so. Maybe Monday. Did people watch the videos? Yeah. Okay. Once in a while, it's like, geez, it's him again. All right, we're done. Okay. So for those of you that are using laptops or phones or clickers, I know that networking in this room is a little problematic. They broke WolfiNet Secure, and then they tried to fix it. I'm sure you've got messages about it. It kind of sort of works if you enforce it. Sometimes it works, sometimes it doesn't. You could also try to WolfiNet Guest. I don't think WolfiNet opening system is there. Supposed to. So, okay. So let's go on from where we were. So what we were talking about was this idea of these proper integrals, which is if we have either an unbounded domain. So, for example, so we can define, so I said this last time, the integral from anything to infinity of some function to just be the limit as the upper bound goes to infinity. This is just saying if I want to integrate some function that stretches off that way, I just cut it off further and further out, and I send the cutoff off to infinity. And so if this limit exists, and then we say it converges. And if not, we say it diverges. Okay, so I did an example of that last time at the end of the class. I integrated e to the minus x from zero to infinity. The answer is one. Everybody okay with this idea? It's relatively straightforward. So, in fact, I'm going to make this even clearer for the question. 0 to infinity of 1 over 1 plus s squared is... You've got it right. Well, half of you. So, I think this is the right answer. You know, 17% think this is the right answer. Whatever this number is, this left over, 50% think this is the right answer. And this must be something small. So, this is supposed to be relatively easy because what we do is we just... We can do this integral. I hope you all know this integral. This is the arc tangent. This is the arc tangent, the value of x, evaluated from zero to m, which is the arc tangent of m. Take the limit as n gets 8 minus the arc tangent of zero. So, the inverse tangent of zero is zero. The inverse tangent... You can think of the graph of the inverse tangent with the limit here on pi over 2. Because the tangent of pi over 2 is infinite. So, this is pi over 2 minus zero. So, the answer is pi over 2. So, this is... I thought fairly easy, but I guess not so much for you. It took a long time. I don't know why it took a long time. So, I have to do the right answer. I thought it was divergent. Okay. Let's try another one. So, let's try the integral for 1 to infinity of 1 over x dx. And so, I'm going to ask you again. So, a, this is divergent. b, this is 1. d, this is log 2. d, very few people said this one. Less than 5 percent between these two. So, how do we do this? We do this just the same way. What's the integral of 1 over... Alright, the log. So, it's the log minus the log of 1. So, the log of 1 is still zero. This is zero. And now we need to know what's the limit as m goes to infinity of the log. So, what's the graph of the log look like? Like that. It doesn't have a limit. It grows forever. The log is just what exponent you need to raise e to to get some big number. So, you take your big number and you raise e to a big exponent. So, this is infinity minus zero. So, this campus does not exist. So, this time... 47 percent of you thought... No, sorry. 41 percent of you thought it diverged. And 47 percent of you thought it converged. So, these two aren't the two of my distractions. I think I handled these right now. Some people came in at the last moment and got it wrong. Before it was like 85 percent right answer. And now it went down to 60 because of the log that got it wrong. So, anyway, let's just do this problem. So, the integral of 1 over x squared is minus 1 over x, right? Because this is x to the minus 2. We add 1 to the power. We get x to the minus 1. And we have to divide it by the power. So, this is... this guy evaluated from 1 to n. We make the limit to infinity. So, this is the same as minus 1 over n minus a minus 1 limit. It goes to infinity. 1 over n goes to zero. Minus minus 1 is plus 1. This is the right answer. And 63 percent of you thought it was the right answer. And the second prize goes to this. So, it's good that you know it converged, but it would be better if you know it. Okay. So, these things, I think, are relatively straightforward. It's just a matter of doing an integral as before. But one of the limits is just let go to infinity. It's a little... So, it's a little... So, now if we have something like... let's just do this. If we have something like this where we're going on both sides, so this is saying the graph of 1 over 1 plus x squared looks kind of like V bell curve, but it sort of looks like that. And here we're saying both sides. We can do this in two ways. One way we can notice that this is an even function. And so, this is the same thing as twice the integral from zero to infinity to 1 over 1 plus squared. We can do that because it's even. That is, the area over here, the same as the area over here. But even if it weren't an even function, we could still do this by splitting it up, by splitting it up, saying the integral from minus infinity to hit any place, about zero, plus the integral from zero to infinity. And now we can do this in the same way we take the limit as, let's say, L goes to minus infinity on this side and N goes to plus infinity on this side. So this is the r tan of zero minus the r tan of L. That's the first integral, plus M goes to infinity of the r tan of M minus infinity to zero. So, anyway, this is high over 2 plus 5 over 2. Okay, so I'm saying I can do it in two ways. One way, I can notice that the function is even. Half of the integral started zero, go to infinity. I just did it, so right here, I got 5 over 2, so it's twice what I got before. If this function, even if this function is not even, so I can't use this trick, I can still always split it in half. I split it at zero, I do one side, and I do the other side. Yeah, so even means symmetrical with respect to the y-axis. Odd means symmetrical with respect to zero or it's even but flipped on the other side. There are certainly plenty of functions that are either even nor odd, such as even the x, log, most polynomial, but not all things like that. Most functions are neither even nor odd. Just like most numbers are neither even nor odd because you're a fraction. If it's odd, you can play the same trick and just change the sum. If the function is odd, then I can still split it, but I only have to do one half of the integral because I know the other one, but it doesn't make it. And it would be zero. Okay, so again, there's not a whole lot to these other than wrapping your head around the idea that something with an infinite domain can have a finite area, but we see we have several to do and several to don't. Now, another thing that can go on is we might have an integral like it has a little bit of an issue at zero. It's okay at one, but it has a problem at zero. But it's the same trick. The graph of one over the square root of x looks like this. Now to one, the area that we're looking at here could be infinite because it blows up here. So what we can do the exact same trick, we just take this to be the limit as, I don't know, a goes to zero from above of a to one. Let me write it this way. This is the x to the minus one half, dx. And then we can just do this as before. The integral of x to the minus one half is x to the plus one half, but I have to divide by minus by two. Like, divide by a half. So this is limit a goes to zero from above of x to the plus one half but then I divide by the value of two from a to one. Everyone okay with this? So then this is, I take the limit of two times the square root of one minus two times the square root of a as a goes to zero, so this is two goes to zero and we say two is what the answer is. Once in a while things can be hiding. Keep this saying, suppose I have something like the integral from zero to two of one over one minus x. Suppose I have something like that. Now this looks like it's okay. So here's the wrong answer. You just do the integral. The integral of one over x minus one makes substitution u equals x minus one to the x is what we use. So this is the log. So this is the log of three minus the log of the, that's the number. But this is wrong. This is wrong. So this is just the number and this is wrong because when x is one we have trouble. Since the graph here is from one half to three so one over a half is two and it blows up at one because x minus one. The graph looks like so one. Yes, I've got this. So the graph looks like this. It has a problem here at x equals one. So we can't do it this way. Yeah. No matter if it's still wrong. Well it's especially wrong now we can give the log a negative half if it's even messier. So anyway it's all wrong because this is completely wrong. But the point that I'm trying to get at is that we can't just jump over. And in fact notice that the integral for one to three one over x minus one dx so this must equal if it exists at all the integral from one half to one x minus one plus the integral from one to three and here this diverges because so this guy we can do this is the limit as I don't know a equals to one from below of the natural log x minus one evaluated from one half to k plus the limit as b goes to one half of the natural log to minus one evaluated from and so this is the natural log one half minus one is one half when you take the absolute value minus the natural log to the limit a equals to one minus the natural log x minus one and this blows up and so does this and so this diverges be a little bit careful you have to make sure that when you're integrating your function is defined on the whole domain and as it blows up somewhere within the domain you have to split it there and do it improperly as you approach that now sometimes so for example suppose that you're going to do something numerically it's probably a good idea to see if the integral makes sense or not even if you can't do the integral in this way so one of the questions that you might ask is does the integral exist or not so if I have something like the integral from zero to infinity of e to the minus x squared dx you might want to know does this converge or does it not converge before we try and do it numerically this we cannot do by any means except numerically we already know this this is the bell curve but we know that this makes sense how do we know that this makes sense even though we can't calculate because I told you so what's the biggest this could possibly be why? one is good why one because it's not seven and I just keep writing one on the board nobody said one two people said one just because you know oh there you are well this is not quite the bell curve you have to divide by pi over two so the real one for the statistic bell curve it's over two here and then there's either pi over two so how do we so people know this is true but why? okay that's true but if I add up a lot of things less than one they can add up to b so I can add up for example one is no bigger than one plus a half plus a fourth plus an eighth and so on 52 or I can add one plus a half plus a third plus a fourth so just because everything under here is less than one doesn't mean this is less than one but you have the right idea even though maybe you don't know that we have it so do you need a hint? someone else have an idea yet? but it's a number how can I take the derivative? take the derivative open what the critical point is I can tell you right now the critical point is at zero the graph of this function that's inside here looks like this okay so this is not completely obvious so it's actually bigger than one right? maybe it's equal to x is one and we already know what the integral of this is integral from one to infinity of e to the minus x it's one we already did that so maybe one is not quite big enough maybe I need two so the idea here is we know that this graph e to the minus x squared eventually slips underneath e to the minus x they cross at one and this guy my scale is going off the shaded area that we want sits underneath the area of e to the minus x so that means that whatever this area is it has to be less than this piece this problem a little bit so this has to be less than or maybe equal to not equal to the integral from one to infinity of e to the minus x dx and we did this yesterday not yesterday last week this is the limit as m goes to infinity of minus e to the minus x by one m which is the one minus one of m limit one so because the function here is always smaller than the function here the integral here has to be smaller than the integral here even if you can't calculate and write it down like that so the function here is always smaller than the function here so the integral here has to be smaller than the integral here right if I have a room that's higher than a smaller room I can fill up the taller room with more stuff than the smaller room so if I were to fill this room half way full I would have less stuff than if I filled this room all the way full so we know that if I fill this room half way full of those little balls that you play around in you fill this with those little red yellow and green balls the volume of the balls is less than the entire volume of this room so we can say that a little more formally we call this a comparison theorem so if I have one function that's bigger than another function and the big function I don't care where it starts converges so it's the little function and converges to something less bad if we have an upper bound that we know with some number we're trying to grow the infinity we can't get bigger than the top function I can't put that up there is another part to this which says that if the little function doesn't converge so so again I still have an X bigger than G of X the integral diverges and so does the we've got X is going to the moon and you're sitting on top of G of X guess where you're going this guy is bigger than that one this guy goes to infinity area and you're at that and that's the infinity