 If any of you, not all of you, have restarted or have made some progress on homework assignment which was officially due to the last class, but because of the hurricane, you should be finishing some time before you leave today. If you're having trouble getting into web assignments, you're not going to know, so we can work around it. There's a few people that have issues where they try to log in and instead of being asked for their Blackboard Password, they're asked for the WebAssign Password. That's according to the WebAssign, that's due to the fact that you don't pick WebAssign's cookies. So you have to turn the cookies on when you're done. I'll probably do that. Or you can just sign in through WebAssign. Any questions? Is there a way to see if you directly submit your homework? So on your homework, if there's a little green check next to the question? No, I mean like in total, like I press submit homework. It submits as you go. So when you see a red X, and that means you submitted it but it's wrong, if you see a little green check, you can have those points. So it's not like you do your homework, you work on it, whatever. Okay, turn it in. It gets turned in as you go. So when I said that some people are working on it but have started it and some people have finished, that's because I can see that, you know, there's some people who have, I don't know, a point to ask. That means they got a question right and no question wrong and that's all they've done. So they may very well have 19 points by the end of the day. Okay, other questions? Yeah. No, the paper homework is due next week and there's no paper homework due this week. No, it says the 14th to the 16th. So I put the homework on the week that corresponds to the material that's on the homework. So it's not on the week that it's due, it's on the week that it's assigned. So, for example, homework 2 on the other side is up and it's on this week because it's due next week. So, yeah. And also there was some confusion, I know I announced this before, but sometimes people don't listen or it doesn't sink in, that if you do your homework problems more than 48 hours before they're due, you get extra credit. So it's quite possible for somebody to get a point and a half for a problem because a problem is worth one point. So there's a lot of people, first homework assignment is worth 20 points, but there's a lot of people who go more than 20 points because they did work. Okay? Other issues, questions, anything? All right. Then let's stop. No, that wasn't good. Okay, so let's do some math now. Okay, so we were talking about, we talked about what the integral is, how it represents an area when we were focusing on definite integrals, which is something like this, which corresponds to this area. We also used in the process a similar notation without putting little numbers here. So this is a number. Putting little numbers here means a function. So this could be some function, a, b, y would need a constant there. So this is a function, even though the notation is almost to say this is a function, and which we saw last time by the fundamental theorem, that derivative, so this is an anti-derivative. So sometimes this is called an anti-derivative. And there's lots of them. So for example, this should be nothing new to anybody. So for example, if I write the integral from 1 to 2 of x squared dx, then this is 1 third x cubed evaluated from 1 to 2, which is 8 thirds minus 1 third, which is 7 thirds. If I write this, then this is 1 third x cubed, and there are infinitely many of these. So I can add any constant here I like. So we usually write plus c. This is the statement that the derivative of this is this, no matter what the c is. Again, better not be anything new. Just remind me of what's going on here. A lot of students are sloppy, and they leave off this dx. The dx is very important because it tells you what you're integrating with respect to. But it also is something that's useful in calculating the integral, which is a lot of what I'm going to talk about today. So let me write something that looks almost the same. Yes, well, actually we're going to just write that. So I'm telling you what this is. You can't do it. I can do it. Yeah. Okay, so let's take a poll. There's an answer. B, you can't do it. Can you go back to that x squared z? C, x squared z. Anyone else? Any other suggestions? I'll ask a quicker question now. It's back. There are five answers here. Channel 41, as usual. What percentage of people think this? 8% think this. Big number. 28% think this. And 41% think this. Okay, this one's wrong. So can someone tell me what's wrong with this one? I need a constant. So the right answer is e because the right answer tells me that the thing that's changing is a d. Not an x. So x is just a number. 43. 43dz. The answer is 43z plus a constant. This one is morally right but wrong due to a typo. Probably the 41% of you, a lot of you, didn't realize the plus c, but you said, ah, it's not that crap. So if you put that answer in, you might see what the sort of most freak is. But that's okay. So it's important to remember not only as a clue to you what you are integrating, but also, as we'll see in a minute, something else can help you keep track of what's going on. So this rule, so if you have, we don't write something, it's just meaning us. We write a notation here. Every little symbol in here tells us something. The numbers here tell us what our balance of integration are. They're being stretched out and that tells us what we're doing in a new rule. This tells us our function. This tells us our variable. We put a d in front of it so that we know that it's the differential of action. It's a rate of change. A dx is not an x, but they're intimately related. So all of these symbols here play a role. So one of the problems that a lot of students have, where a lot of people have in math, is this is not true in writing prose, in English, and really in any language. Even in a very efficient, compact writing language like Chinese, there's redundancy there. There's no, or very little redundancy here. Every little mark here on the page means something. We don't just draw them to make it look pretty. Every piece of that symbol means something. And so when you're reading math, what a lot of students do, they're reading math, they read all the words, and then they say, oh, there's some symbol about something, let's give it. You can't do that. In fact, what you should do is, there's a bunch of words, let me skip them. Let me look at the symbols. Oh, that's really where all the content is. There are words there to help you understand the symbols. But both parts are important. And there's a lot of redundancy in the words. So probably you could understand this whole lecture. So if you go home and watch the video, and turn the sound off, you can probably still follow this lecture just fine. If you turn the video off and just listen to the sound, you're going to be confused. By the way, I'm sure you all know that I set up those stupid cameras in the back every day and every time. These things I put on the classbook page so that you can look at them later. I don't know if you find them useful, but I do. Okay. So now I lost track of what I'm doing. Okay. So we have all of these little things meaning something. Now, one of the first goals of this class is to look closely at integrals. You know what integrals are from your previous class. Well, most of this already. And now we're going to learn how to do a bunch of integrals. This is the last lecture in our review. One of the main techniques of doing integrals. So we already see that if we have, so we know from before, yeah, well, I wrote it there. But if we know the antiderivative, we know the integral, whether it's a symbolic or a definite one. And so if we have something like the integral of the square root of x, yeah, exactly, let me just... So how many of you know how to do this? Okay, so I'm not going to ask it. So to do this, we think that this is extra one-half. And now we just remember the power rule for derivatives and we turn it backwards. So we increase the power by one. So one-half plus one is three-half. And then we have to divide by this new power to adjust for the fact that we take the derivative and we want this to cancel that. And then we add a constant. So that's very easy. If we have something slightly different, so this is really morally the same question, but it's written in a different way that makes it look harder, use kind of a trick to change our point of view to make this easier. And the trick that we use is we make a substitution. The substitution that I'm showing is what I'm talking about today. Substitution, what we do, we write this not in terms of this x, but in terms of a new variable that makes our life more convenient. And someone tell me what that new variable is. You got to do it. Well, it's u, but tell me what u would be. You can call it w, you can call it j. Three x plus four. So I'm going to use w because you all said u. I don't care what letter it is. We let w be three x plus four because our life would be easy if this looked like that. So we're going to let w be three x plus four and then we just have now the integral of the square root of w to w, but I did it wrong. But in fact, when we leave the d w off, this is wrong. Why is this wrong? Yeah? I'm close to having one-third. I'm close to having one-third because when I change x a little bit, so if I change x from one to 1.001, w is going to change by three times that. w, if I increase x by 0.001, w will increase to 0.003. So that's the statement that three times the derivative, so w, sorry, brain freeze, one-third. d w is one-third, is three times however much x changes. We take the derivative of these things. Only I write it in differential notation. So that says when I wiggle x a little bit, w wiggles by one-third of that amount. So this is the same statement that you can solve this with your life. And so here this is wrong because I left off the dx. The dx here is a real part of the integral. It actually represents something. This part transforms to that, but this part will transform to this. So the dx becomes the d w over three and the 3x plus 4 becomes the number. And now this is easy. This is just the same integral I did before. So this is just one-third of two-thirds w to the three-halves. The cons can go inside or outside because it's an arbitrary number so multiply an arbitrary number by a third. You get another number. So that's why. So we can write this as two-ninths. We can go back here to terms of text because I may not tell you. Now again, this should be review-free. That's right everyone. Is anyone confused by this at all? I have a few questions. What I want to emphasize is that we need to pay attention not only to how the stuff inside changes but how the differential changes. I guess a little more formally what's really going on here is we're just using, we're just writing the chain rule in integral form. So if we know that from calculus from the first semester so we know that the chain rule says if I have one function inside another function and I take the derivative then this is the derivative of the outside function and the inside function of the inside function. If we want to write this in differential notation this would also say that's the chain rule and now let's just integrate both sides of this thing. So if I integrate with respect to x then it's still true. My g is well this example is w which means u which means j I don't care. So this is saying I have a function with something written inside it so f in this example is the square root and g is 3 x plus 4 and here this g not of x dx this is a 3 dx which is the same thing as the last one. Yeah, that's right. So that means that I have to write it in terms of w I write it in terms of e w. Right? Yes? No? Why? Okay. So this is just rewriting the substitution rule in a slightly more formal thing. Let's do another example. I'm having only one and I forgot it. Let's see. So suppose that I have the integral so this should be easy. One fifth e to the 5x and here what I'm doing behind the scenes I don't, I just I hit it right down and you don't see that immediately and what I'm doing behind the scenes is I'm writing u to the 5x so du is 5 dx which means that dx is one fifth which is du over 5 and so when I change this to that then I have the integral of one fifth e to the du and I just write it down because that's what I know. So if you can't just look at this and write this down please put this intermediate stuff down until you get good at it. But most of you are maybe to that point where you don't have to write it down but maybe not. Okay. Suppose I have something like the integral of sine of x cosine of x squared what would be a good substitution to make? Maybe, yeah. Okay, so let's make this be a what if I click or what? I should look at stuff to equal sine of x a cosine of x plus cosine of x squared of x to do for another part. This substitution is one of these things so I'm going to stop this because there's people still madly punching their butts inside. A small amount of people about nine percent think this, they're wrong. You can do this but it would be r. In fact, you can do sort of all of them except, well, no one chose this one. Okay, fine. So I'm really going to go down to these two. Okay, so let's see what happens. So here I have the integral and if I make the substitution u equals the cos let's do the one plus cos so the most popular choice but it's close is 41 to 49 percent so those are close so if I make the substitution u equals one plus cos squared then du is going to be the derivative of this which will be equal to that will die this will give me a two cosine x times the sine of x dx that is. And so you know this is sort of the most complicated bit here so this becomes well I have the sine x dx u here and I have the sine x dx then I have this sort of I don't know quite what to do with this cos x business. I have sort of a cos x left over but I don't know what to do with. Now I can work hard to find and figure it out but this is so so I sort of have minus two cosine x du and I can't get rid of this so maybe I can make this go somewhere with a lot of sweat but I don't want to work that hard so let's try the other choice so I'm going to let w be cosine and when w is cosine dw is minus the sine dx and that's good because I have sitting around here a sine x dx so this is really dw except I just have to take it so when I do this I see dw sitting there looking at my face so that means that this becomes the thing on top becomes dw but it's negative and the thing on the bottom becomes one plus w squared now this is the magic rule that probably you know how to do you should know how to do if you were sleeping when they did inverting functions well they even forgot but this is dr tan so this is dr tan w and now w was cosine really answer b is a better answer but it's not obvious so one of the things about substitution is it's not completely obvious what the right substitution is it's more obvious when you flame a little odd and get some background it's not completely obvious so it requires a little effort and maybe sometimes a little bit of a wrong turn but it's okay to make a wrong turn as long as you eventually get right yeah so you're saying it's one plus three times this square here this is what I want to integrate so here I would make a substitution the view is but I'll see I want this to look like one plus u squared so I wouldn't make the substitution u equals three w I want it to be square root of three w because now this looks like one plus u squared so that du is root three dw so I pick up a factor of one over root three dw is one over root three so this is one over root three r10 root three so it's a one third square root yeah this won't work it's not r so really b is not wrong sorry c is not wrong c is not wrong it's just so I just gave up I just gave up you could maybe make this work but if I ask you to drive to court jefferson or to walk to court jefferson it's not wrong to go to hunting to the furthest it's just not really a good idea perfect you could maybe make this work so in some sense you're not completely correct answer to this it's just that the best substitution is certainly the perfect one so substitution I want to emphasize here that by analogy I said this before I guess I'll use this one again differential form or dx or whatever you're calling it which everyone always seems to forget not everyone, but many people seem to forget plays two roles this dx it tells us the variable like a unit it's telling us when we make the substitution it's telling us how to transform the substitution it acts like by analogy this is like this one this one is like the dx is what I'm saying inches or meters it's a unit when we transform it to some other units we write this function in terms of some other variable u we have to say what the relationship is between the original quantity and that's all the chain rule is saying too the chain rule is saying if we measure if our g is in feet and we change it to inches we can multiply by 12 now sometimes in math not so much in science the difference in the units are not linear sometimes in science so for example when we're measuring an earthquake the energy is related to the value on the Richter scale logically so the chain of units there is nonlinear but typically in science and in life in general most of the units are related by a linear relationship not so here in fact the linear ones you can do in your head the nonlinear ones you are right at so if you have a question can you raise your hand because I was talking so I had no idea okay sorry I had no idea and you forgot your question and I forgot that you were so real um okay so uh so this one is a nonlinear change of variables I mean in some sense all of the substitution problems are the same once you get the hang of it you're done so and supposedly you know that already so I think I've not actually well okay so the idea in substitution so you won't know of course whether it's a substitution problem or not but when you're doing a substitution you want to look for let me call it u so you want to let it be something and the thing that you want to let it be is something that number one simplifies the integral the derivative is laying around in some sense what I mean by that so for example in this guy letting u be one plus cosine simplifies this integral but the derivative is not here nor is it easily made to be here letting u be cosine simplifies the integral to be a harder integral but its derivative is right here um so you know let me do one more example did that stew next week um so okay so we have this thing laying around uh there's not always a unique substitution right if I have and I'm purposefully not using nx I can't use u versus my substitution so I shouldn't use u I'll use x or I can use whatever I'll use x what's the obvious substitution thing yeah so if I let u be x be u squared plus one then dx is 2u du and I have a u du laying around and so then this just becomes uh the square root and there's the one half dx and then u du and so this we already did this this is one half of x2 so this is x in the one half so this is two out of three halves so this is two to cancel one third of u squared plus one three halves right now this example was a stupid one because I remembered the wrong example um this was an easy one I should have done a harder one so in fact most integrals are attackable by this technique of substitution what if I have something oh I guess I need to say something about the balance sorry so I forgot that suppose I have a definite integral for example the integral from suppose I have some integral like this then I can do this in two completely equivalent ways and I prefer the first one but many people do the second one so let me do it both ways I suppose there's an obvious substitution here so we want to tell me what it is right so if I let u be the natural of 2x then u except for a factor of 2 is sitting around because I remember that the derivative of the log is 1 over x so here if I let my substitution be the natural of 2x and the differential is 1 over 2x times the derivative of 2x which is 2 times dx 1 over x dx which is sitting right here so this just becomes the integral of u to u and I did this last time and some people objected this says x is 1 and this says x is 5 these numbers if I write 1 and 5 here this is wrong because this is saying u is 1 which is wrong so 5 so let me do this way I don't like first the way I don't like I can write for myself x is 1 and x is 5 then I don't forget now it's not wrong it's just unconventional because I have to remember what x is but ok so if I do this then this is 1 and x equals 5 which gives me well I have to remember what x was x is the log of 2x sorry u is the log of 2x thing so this is 1 half 2x square I don't have to write the x anymore because it's the same and so this is 1 half log of 10 1 half log of 2 which you can simplify a little bit as the log of 10 over 2 which is the log of 5 which is the same as the log of square root of 5 so that's one way to do the problem it has a lot of extra garbage in it so what I think is a slightly more efficient and also mentally easier way to do the problem is just forget about x once you change the u it's really the same so here I have the integral from 1 to 5 log of 2x x to x make the same substitution and notice that when x equals 2 and when x equals 5 u is the log of 10 I don't have to write this down I can just think it from then the integral from the log of 2 to the log of 10 u 1 half to the log of 2 log of 10 which is just 1 half wait can I lose the square I sure did why didn't nobody stop me so this is all crap we'll see this in a minute log of 10 squared minus log of 2 square so now we'll write the integral to say this way is a lot more efficient in my mind and it's sort of what you do anyway if you're working with units suppose you're doing something in feet and you discover that it's easier to work in inches you don't change the inches for part of your calculation and then change back to feet the numbers are the same we're done so I have to change to a more convenient measurement system and then just change everything so when you're doing definite integral you can treat them as indefinite integral do it all out and then plug in again or you can just change over just convert your mind to the new system you get the same answer or you get an equivalent answer because you're doing the same thing but I'm tired of them and we ought to do both have an integral that looks like a good substitution I can make so some of you know and you're saying no some of you don't know and you're just looking blank but there's no good substitution I can make here well I can't find a good substitution so I have to be a little more clever I want to transform this into something else now when we came up with the substitution rule we had to look at the chain rule chain rule is an important part of the correction calculus there's another rule that you learn when you're learning derivatives that is extremely important use the product rule we didn't use the product rule but maybe the product rule would be our strength here I can integrate that piece I can integrate this piece I would like to write this as so this is not true you try that, you fail, you're done go away you're done, don't try that because it's just wrong but we can manipulate some kind of product rule so let's remember the product rule which is for derivatives and we want to somehow get a product rule for integrals the product rule for derivatives says that if I have one thing multiplied by another thing and I want to take its derivative well then I can write this as the derivative I know I'm getting to notation the derivative of the first thing this is not a line plus the first thing find the derivative of the second thing so maybe I can integrate this and get something useful so if I integrate everything inside and it's built through x e to the x this is two things that I want to integrate separately I could maybe think of one of these as being a derivative and the other one as being an integral that is when I integrate e to the x I get it back so I could have this as the derivative of something when I integrate x it gets more complicated when I take the derivative of x and I get the derivative of one so I could restructure this guy to say that well I want this it doesn't matter if I have something like this I mean since I have an equal sign I can subtract this and this is e this is the integral of the product I mean this is the product as it's e integral of the truth there's a more familiar form that is if we think of and unfortunately there's an unfortunate notation of collision if we call one of the functions u and the other one v this says that the integral let's let it have to be u instead the integral of a thing comes from the derivative of another thing is the same as the product of those two things minus the integral that's right so I can transform I can trade a product of something I can take the derivative of and something I can integrate into the things and integrate that guy would take the derivative now this is more you've never seen this before so this is called integration of my parts maybe you've seen before but I don't want it let me just do this x e to the x example and then we'll call it a so in this case if I left then du is just dx and if I call this part my dv because I can integrate it then when I integrate it d is just d dx so this is not a substitution and it's unfortunate that u is typically for both substitution and for the parts of the integration of the parts so that's the way it is and so that this formula over there tells me that this integral is now the product of this and this a different integral the product of dv of this and this but this integral is easy something that goes hard into something a little easier it takes a lot of practice but it needs to be taken out when you watch this line it takes a lot of practice in substitution this is not one of the homeworks that we're going to do on this day so if you don't understand yet don't sweat it out okay everybody come again one more day