| Start | Down to the last 2 weeks, this week you will get one more homework assignment on paper and one more computer assignment, that will be the last one.
So you will not get one next week, because I like to wrap everything up before finals. I have no idea what your grade will be, until I get the final exam grades, the final exam numbers. And then I will put together the grades. Most of you are doing well, and paying attention. I see who comes to class on a regular basis, so I wouldn't worry too much. |
| 0:32 | You know, I just spoke to a student in my office, he was all worried because he said 'I'm not doing that well', different class.
Professor says don't worry you're doing fine, when we say don't worry you're doing fine, we know you're not failing. Ok? Unless you surprise us, in fact you're probably doing ok, but we can't tell you you're getting an A. Because then if I don't give you an A, you'll be upset. Cause you'll say "you said I'm getting an A" right? So I can't tell you how you're going to do, I can just say that most of the people in the course are doing reasonably well to very well, |
| 1:01 | some of you could do better but, keep up the good work and try to do well on the final.
It'll all come out ok, I can't promise you all A's, but very few of you are in danger of getting below a C, a few of you, some of you are really challenging that. And for many of you, the C is good enough, you'll get the C now and you'll leave. In fact, if I offered you a C and no final exam, you'd say yes. Ok? Which I understand, some of you want the A, which is also fine, you should try to do the best you can, |
| 1:30 | But don't stress too much about this course.
I will make practice problems for the final, I will do a practice final exam, even though I didn't do it with the midterms, I will make you a sample final exam, um, the final is 2 weeks from Wednesday, I think I may have to op, and there goes the microphone. Battery dead, do we have a replacement battery? No. The final exam thoughts scared off the microphone anyway what was I saying, practice final exam, oh room. I don't know yet what room the final exam will be in yet I have it in my computer, I will post it this week on blackboard. |
| 2:10 | It's not the same room as the midterms, [?] I will find out. Ok?
So there you go. Anyone have any questions? Can we get going alright. We're now going to do the more annoying topics in integration. So we learned how to take derivatives, and we did you know, the product rule, the quotient rule and the chain rule. |
| 2:32 | Now we got to integration. And there's lot of technique to integration, there are all sorts of integrals that are not obvious what the original derivative was.
So remember it's relatively easy to go from derivative to integral, it's very hard to go backwards because when you take the derivative you take something simple and make a mess. It's very hard to look at something, the integral, and figure out from that mess what the original function was. Working backwards is hard. |
| 3:04 | The good news, is in this course, we don't really cover any of that stuff. We leave almost all of it out because frankly that's an entire semester worth of stuff to cover.
And it's really not relevant to what you're doing, but you should learn how to do what's called substitution. Which is really working the chain rule backwards. So let's think again about what our 3 main forms of integrals are. You'll have the integral form of u^n and the integral of that is u^n+1/n+1, that's when you have something to a power. |
| 3:39 | So that's form number 1.
I'm using u for a specific reason. You can have an exponential, where you have e^u*constant and thats just e^u*constant/constant |
| 4:07 | And the third main type we get is you have a 1/u
So what we're going to go with substitution, is you're going to take an integral that doesn't immediately look like any of these,
and play with it to turn it in to something that looks like one of these. And once you can do that, it's very easy to integrate.
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| 4:33 | So you use this technique called substitution, where you take the integral and substitute some stuff into the integrand to make it easier to work with, once it's one of these 3 forms, you can integrate and then you can substitute back, for example, Suppose you had something that looked like that. (sqrt)(5x-3) |
| 5:00 | You say well, this is really (5x-3)^1/2
So it's something to the half, so we know that's the first type, it's a function to a power.
So how to do we get this to look like u^1/2? Well, you look at this and you say let's let u=5x-3 You can use any letter you want, u just seems to be the one traditional to use. |
| 5:34 | So if u is 5x-3, the derivative of u is 5.
Or if you cross multiply this, you could say du is 5dx. It's sort of traditional to break the derivative up like that. |
| 6:01 | Now you can look at your integral and you can say well I can substitute u for 5x-3, and I get u^1/2, but I haven't replaced dx.
Because if this is in terms of u, then I have to have a du, I can't have a dx. So du would be 5x, but I only have 1 dx, so I make that 1/5*du. So we divide both sides by 5. And now I can come to this integral, |
| 6:32 | And I can substitute, I can say well this is 5x-3 is now u^1/2,
And dx is now going to be 1/5 du.
Or I can move the 1/5 to the outside. So what I've done is I've sort of rewritten this as an integral of form 1. And made it into a power. It really is just by playing around with constants, it's really not that bad [?] but let's do a few of these so you get the hang of it. |
| 7:04 | Now if you integrate this, you get 1/5 of (u^3/2)/(3/2)
Which is 2/15*u^3/2,
And the you substitute back for u, which is 5x-3 and you get 2/15(5x-3)^3/2 + C.
|
| 7:44 | If you looked at this, the square root, it's not obvious that you get that 2/15 term, so when the u substitution does is it helps you figure out what that constant is.
It's really almost chain rule backwards, because you we this. If you take the derivative, you get the 3/2, that comes up front, |
| 8:04 | (5x-3)^1/2 gives you the 5, so that 2/15 helps you get the constants right.
|
| 8:32 | So very similar to the other one. So we have (8x-7)^10, now of course you can multiply that out, so the derivative of that isn't very hard.
So if you want to do the integral, Well the problem is the 8x-7. You can't just say it's (8x-7)^11/11, because if you take the derivative of that, you won't get this. So you have to play with it a little bit. You can't just make this to the 11th over 11. Because as I said, if you do that, and you took the derivative, you won't get back here. You'll be off. |
| 9:01 | So the u substitution helps you figure out how to do that, how to get the constant right. So you let u = 8x-7,
And du/dx = 8.
Cross multiply and now you look at the integral, you say well ok, now I can substitute the 8x-7, and I can substitute for the dx, if I just divide by 8. |
| 9:40 | And now, I can take this integral, and make this u^10, because this is u, and this is 1/8du, or 1/8 (integral)u^10 du |
| 10:06 | Ok? So u^10 becomes u^11/11, Or, u^11/88, And then you substitute back, |
| 10:31 | Ok? Ok. Suppose I had,
That. (integral)x(sqrt)(6x^2-4).
So one clue was this is a higher power than x, 6x^2-4, so if I make this u, the derivative will be 12x, and you just have to play with it to make it look like x. If I made this x, the derivative is just 1, and I wouldn't have anything to sub in here. |
| 11:08 | So I'm gonna let u=6x^2-4.
And then the derivative of u is 12x, or if you cross multiply, du= 12dx. I'm going to start skipping that du dx step. So I look at the integral, |
| 11:36 | Yes, all I'm doing is I'm putting it over here. Because I want to substitute, I want to isolate du.
So in the integral remember we had this d term, you always wondered why that was there, now you know. Ok? It comes out of the derivative, so we have to replace everything in here that's in terms of x and replace it with something in terms of u. So I look at the integral and say well i can replace the 6x^2 -4, |
| 12:03 | And I've got x, dx, this is xdx, I just have to divide by 12, And now I can substitute, and I can say this becomes u^1/2, And xdx becomes 1/12du, or if I move the constant outside, |
| 12:30 | 1/12(integral)u^1/2 du.
How did I get 1/12? Here I have du=12x, so I divide both sides by 12, that becomes 1/12 dx. Ok? And once you have a constant inside the integral, it's always a good idea to move it out so it doesn't clutter up the in integral and you don't worry about it too much. You just figure out all the constants when you're done. This is 1/12, integral of u^1/2 is u^(3/2)/(3/2) |
| 13:08 | which becomes 1/18(u^3/2).
And then substitute back. (1/18)(6x^2-4)^3/2 + C. |
| 13:33 | Which, as I said, is not obvious when you start with something like that. We can get significantly messier, but we won't get to too difficult.
So let's have you guys practice one. |
| 14:02 | By the way, you could just multiply this whole thing out, and then it would just be a polynomial and you could integrate.
So I would let u = 5x^3 + 11, not to the 4th, just 5x^3+11 because if you put in the integral of u^4, you can work with it from there. We got it? Yeah? Some of you still look like you're struggling a little bit. Ok. |
| 14:36 | Ok, so one of the clues is when you have 2 terms in the integrand.
One of them is x to a power and the other one is x to a lower power and the x term to the higher power will be u. So let u = 5x^3+11, And then you take the derivative. So du will be 15x^2 dx. It's really du*dx, but you move the du to the other side. |
| 15:10 | Then I look at the integral and I say can I substitute? Well, this could be u, and x^2dx, well here I have x^2dx.
So I have to divide by 15, so 1/15 du = x^2 dx. Now if I take this integral, I can rewrite it as u^4*1/15du, or 1/15(integral)u^4 du, it can get more complicated, but not so far. |
| 15:48 | These are all the very simple ones so far. Generally we'er not going to give you hard ones because they involve trig.
And combinations of messy functions. But theres some, theres some tricky ones. I just don't know whether to bother with those or not. |
| 16:02 | I know you guys are voting don't bother but I'm just going to go ahead and see whether the book bothers or not.
I don't think it does, the idea of the course is to teach you some basic tools not to turn you into integrating machines. Alright so, the integral of u^4 is u^5/5 + C, well we don't really write the + C till the end. |
| 16:33 | We want this to be correct. So thats u^5/75,
and then you substitute back.
So thats (5x^3+11)^5/75+ C As I said, when you look at the function initially, you don't always say to yourself oh sure, that's going to be over 75. You have to sort of get that part right. |
| 17:02 | Why is that true? Let's take the derivative of this. Just to check.
Well, the derivative would be 5(5x^3+11)^4*15x^2 Ok? Because the derivative of something to the 5th is 5*something^4 times the derivative of the inside, and the over 75 just stays there, it's a constant. |
| 17:35 | And then if you simplify, well 5*15 =75, so they cancel so you get x^2(5x^3+11)^4.
Which is what we wanted to get. Ok? So as I said, this helps you work through the chain rule backwards. Ok, let's see how that one goes. (integral)(x+2)* (sqrt)(x^2+4x+1dx) |
| 18:04 | Alright, let's do this one.
So if you had u=x^2+4x+1, This would become u^1/2 which is good because that's what we want. So we have to turn x+2 to x, it's a du so let's take a derivative, and you get 2x+4 dx. The derivaitve of x^2+4x is 2x+4. |
| 18:40 | And you say well, it's almost x+2, if I just divide this by 2. So I have 1/2 du, that = x+2.
So now I can go to the integral, |
| 19:00 | and I can say this becomes the square root of u, or u^1/2, and this, becomes 1/2du.
1/2(integral)u^1/2 du. You all see what I did? Ok, so again, you look at the integrand, and you say I gotta turn 1 thing into u^ power and everything else has to become du. So if I let u=x^2+4x+1, and the derivative of that is 2x+4, I don't have 2x+4, but I have x+2, which is just half of 2x+4, so just divide this by 2. |
| 19:42 | And now 1/2du is x+2dx. Now I can go here and I can substitute everything.
This is just u^1/2, this becomes 1/2du. Alright, what's the integral of u^1/2? It's u^3/2/(3/2), or 1/3u^3/2 |
| 20:11 | And now you substitute back in.
So as I said, not obvious how you get here, how you go backwards. |
| 20:30 | You can imagine, you can get some very messy integrals in here that turn into something very simple, or, something very simple in here, that turns into something quite messy.
So um, we don't do trigonometry, but, oh the integral of sec^3(x), you get this long complicated thing, but when you take the derivative of it, it just becomes sec^3(x), but it's a long mess, so. It's not always obvious how it goes the other way, let's do another one. |
| 21:33 | So suppose we're going to do xe^x^2dx.
We could take the derivative of that, that's the product rule, but we want to do the integral. So remember what the integral of e is. The integral of e is just e really. Right? e^u is just e^u. So we just want to kind of turn this into e^u. So we have to get rid of that x somehow. So if we let u=x^2, the derivative of that is 2xdx, |
| 22:08 | Notice we have xdx, almost have 2xdx.
We have this, xdx, and that xdx. So when I move this constant, I can now rewrite the integral as this is e^u and this is 1/2du. I just put the 1/2 on the outside right away. |
| 22:44 | You all see what I did?
So again, we're going to let u=x^2, not e^x^2, just x^2, and the derivative of u is 2xdx, and then we need to substitute for xdx, so if I pull this 2 out, I get 1/2du is xdx. |
| 23:04 | Now, the integral of e^u is just e^u,
And now you substitute back. So it's just 1/2e^x^2+C.
Ok, we'll give you an entertaining one. |
| 23:51 | Don't be scared off, remember we want to make this into e^u
So if you let u=(sqrt)x, remember what the derivative of the square root of x is? 1/2(sqrt)x.
|
| 24:11 | Which is almost what we have.
We just have 1/(sqrt)x, we don't have the 2 in there. So then multiply this by 2, I get 2du = 1/(sqrt)x dx |
| 24:30 | Cause I want to have this,
I want to have dx/(sqrt)x which is this. So I keep the 2 on the other side.
So now this integral becomes 2(integral)e^udu because 2du is dx/(sqrt)x. |
| 25:01 | That's 2e^u+C, whoops that's 2e^u, sorry, which is 2e^(sqrt)x + C.
Alright we ready? So let's let u = 4-3x^2 du = -6xdx |
| 25:38 | So if I divide by -6,
That will get x all alone.
So I can substitute for 4-3x^2 and put in u, and I can put in -1/6 du for the xdx, so you get -1/6(integral)e^udu |
| 26:05 | So that's the key, is the chain rule helps you get rid of all the other stuff and turn it into a nice simple integral.
So again, xdx is here, and thats -1/6 du. Now integrate and you get -1/6 e^u And then substitute that, -1/6e^(4-3x^2)+C. |
| 26:50 | How about this one?
|
| 27:01 | Alright, so what do you think I'm gonna let u be on this one?
u is x. If u is x, I don't get anywhere, we can't do that one. Any other ideas? Let u= 1+e^x, well if u=1+e^x, du= e^x dx. Which is exactly what I need to substitute for. So it's this general idea that if you see a radical, make u the thing under the radical. |
| 27:35 | So this becomes (sqrt)u, this becomes du.
Integrate this, you get u^(3/2)/(3/2) Which is (2/3)u^(3/2) And then substitute back. |
| 28:06 | How do we feel about these so far? They're a little trickier.
As I said, there's a lot of different types of integrals, and all sorts of tricks to work them backwards. When calculus was being developed in the 17/1800's, obviously, people would come up with some and they'd say "I have no idea how to do this. And of course, people would throw their brains at it until someone came up with a clever idea and said "oh here you go." Then they'd generalize and say "always use this technique". |
| 28:36 | There we go, we conquered another type of integral.
So of these integrals are really tricky how you integrate them. Fortunately you guys won't have to deal with that, and, these days, computers can do all the integration you want. They can do numerical integration which is by rectangles though sort of. And then they can do actual [?]. We're going to do a lot more of this on Wednesday, so have a nice day. |