| Start | Hope everyone is doing well, we're almost at the end.
We only have a little bit more to go, and then you're done, and as I said, this is your last math class. You will be very happy, you never have to do this again. But of course, the reality is sort of a math class, BUS 120 is sort of a math class, and marketing and business and technique is sort of math, so there's math coming. There's math-y stuff, but this is sort of the end of the algebra, calculus direction you have to go. |
| 0:30 | And you're not going to use this that much in the sense of you probably not really be doing these calculations when the time comes.
Well, some of you might. You might really like this. But you really want to get the concepts, so that's why I keep trying to [?] you. So unfortunately we're not going to do that today, we're going to focus mostly on technique. But, um, generally, you really want to know exponential behaves, and how regular functions behave by using the polynomials, and what a derivative really tells you, what an integral really tells you, that's really what you need to get out of the course. |
| 1:07 | That's really what you're trying to learn in business calculus, because the truth is most of what you do mathematically, will be involved with excel.
Which can do an enormous number of things. But this is sort of the last topic to cover before the final. Next week, we're going to review. So let's see. Tomorrow, I forget what time I set it for, but you will have a new paper homework and you will have a new my Math Lab. |
| 1:33 | Those are the last ones. You don't have to do anymore.
Then when it's all done, I will drop the lowest homework and the lowest math lab. Ok? So if you have perfect scores on your math labs up to now, feel free not to do the last one. Ok? So as a favor to all of you, we drop the lowest one. Ok? Thought that would make you happy, ok, now we have to do a head count for a second. |
| 2:13 | 49 of you here. So we're going to do this bonus thing for you. So on your final exam, write on the cover of the final exam "Merry Christmas" and you will get 5 bonus points on your final exam.
There are 50 of you. If 51 of your write that, none of you will get it. |
| 2:33 | Got it? So you cannot tell your friends, it's their fault for not being here.
If you remember the magic phrase, do I have to repeat that? There's 50 of you, you write merry christmas, you get 5 points, you don't write it, you don't get the 5 points, if I see 51 of those, I will count, none of you will get the points. Got it? Good. last year the magic word was 'bananas' so people drew pictures of bananas, that got them 6 points instead of 5, |
| 3:01 | But nothing else. Alright.
You should be rewarded for attending, as i was saying to these 2 guys here, you know, you're going to be in the business world, the worst habit to develop is not showing up for work. Ok? So there you go, now let's practice some more of these integrals. So this is the u-substitution stuff. How about... |
| 3:41 | We'll start with this one.
So you look at that, and what you do when you do the integral, is you're looking for a function, and you're looking for its derivative. |
| 4:01 | And the function becomes u, and the derivative of the function becomes du.
You go here and you say, well, the derivative of e^x is e^x, so the 5 is irrelevant, cause 5 is just a constant, so you take the derivative of 5, you get 0. So you're going to let u=5+e^x. Alright that's long enough, so why don't we let u = 5+e^x. How do I know I don't want u=e^x. |
| 4:30 | Well first of all, if I did that, then this would just turn into u/(5+u), which wouldn't really help, or one of them would be du, and that's not useful either you need to get rid of 5 too.
So du=e^x dx. Now we can substitute into the integral, we have e^x dx, and (5+e^x), so that's gonna be du/u. |
| 5:06 | The (5+e^x) = u, and e^x is the du. You can never have du in the dominator.
Ok? So du always has to be in the numerator if you have a rational expression, so that's one clue. What's the integral of du/u? ln|u| + C and then substitute back. |
| 5:33 | You don't really need the absolute value bars because (5+e^x) is always positive.
Remember e^x is always positive, yes? Uh, yeah, if you have this form, it'll always be log. In fact, in general if you see an integral with something over something, it'll end up being a log. Be careful, not 100% |
| 6:03 | People are always looking for rules, and one of the problems with math is it's not always as simple as there's a rule, but a lot of the time it is.
Let's give you guys a couple practice ones. Make sure you can do them. We're gonna do another flavor of this. |
| 6:36 | So this one will violate your rule there, Jessica.
That's not going to be a logarithm. |
| 7:07 | This I can do better, |
| 7:38 | There you go, there's 3 entertaining ones.
Alright, so how do you do these. So as I said, you're looking for a function and it's derivative. So you can rewrite the integral in one of 3 basic types. You want it to be a power integral, |
| 8:06 | A log integral
or e.
So you want to turn whatever you got into one of these three types. And once it's turned into 1 of those 3 types, it's easy to integrate. So to use u-substitution, you get one of those three forms, and then you can substitute back. Remember I put those up on Monday but for those of you who weren't here. |
| 8:39 | So, you look at this one and you say, well, how do I know which one to substitute? Well, first,
the derivative of x^3, is something x^2. So I'm going to want to go that way.
The second is, this is to the 5th. So if I can put this as u^5, it would be easy to integrate. Or in this case u^-5, but same thing. So let's let u = 1+x^3. |
| 9:04 | So du = 3x^2 dx.
And 1/3 du, well actually you don't even need the 1/3, never mind. We've 3x^2 dx. So I can turn that into du, and this, will be u^5. So I'll have du/u^5, which is the same thing as u^-5*du. |
| 9:37 | Which is the first type.
Why do we use the letter u? Why not. You can use any letter you want, it's just traditional to use u. So the integral of u^-5 is u^-4/-4, which is (-1/4u^4) |
| 10:03 | And now substitute back.
u = 1+x^3, so it becomes (-1/(4(1+x^3)^4))+ C. So as I said, what you're doing is you're using u-substitution to take a messy integral and turn it into something simple. So the simple integration and then plug back in the messy part. |
| 10:30 | How do we feel about that one? Little better? Alright.
So, we'll go here, so again, the derivative of something x^2 is going to be something x, so let's let u=1+24x^2. du = 48x dx. |
| 11:01 | And I look at the integrand, and say well, I have 12 dx, which is kind of like 48dx, it's just multiplied by 4.
So 1/4 du = 12x. So I can rewrite this as 1/4*u^1/5*du. Why to the 1/5? Because it's a 5th root. |
| 11:33 | So the 1+24x^2 becomes the u to the 5th root, and then the 12x becomes 1/4du. And I always recommend that you move the constant outside the integral.
It's easier to work with. Move it out of the integrand, that way, when you get messier integrals, it'll be less cluttered, and will just be the part you have to integrate. |
| 12:00 | Alright, what's the integral of u^1/5? It's u^(6/5)/(6/5)
Which is 5/24 u^(6/5).
And then substitute back. And you get, 5/24(1+24x^2)^(6/5)+C. |
| 12:36 | Remember that constant is very important. It looks like it's not important, you go yeah yeah, + C. But you need that when you actually work with these things and figure out what the constant is.
Yes? Why did I take a quarter of du? Because I looked at the original integral, right? And if I wanted to substitute it, this could be u^1/5. This isn't quite du, because du is 48x. |
| 13:00 | I only have 12x. So if I divide this by 4, then I get 12x, and then I can substitute in. Ok?
Anybody else? No? Why is du 48x? Because it's the derivative of this. So what we're really doing is we're saying du/dx = 48x and then we just move the dx over there. The du and the dx you can think of as separate things. |
| 13:35 | This is Leibniz way. Calculus was simultaneously by Newton and by Leibniz, Newton in England and Leibniz in I guess Germany,
and um, this is his way of doing it, which turned out to be a better way. More efficient, yes?
Well, ok, we took 3x^2 dx, and we replaced it with du. So this, becomes du. |
| 14:05 | So what's going to happen is everything in the original integral has to be replaced with things that are in terms of u. You can't have any x's left.
Yes? I'd have to see what you're doing, but it's probably wrong. Ok? I hate to say it that way. Well, I write out every step. But when you turn the integral, the goal is to turn it into something that looks like this. |
| 14:36 | Because that's a power like the first one, and then you can integrate it and then you can substitute back.
Which you don't want to substitute back until you've done your integration. Cause the idea is to have an easy integral. Ok, let's do another one now. Let's let u equal the denominator, because it's a higher power. |
| 15:03 | du, well what's the derivative of x^2+x-1? 2x+1 dx.
Well, how would you simplify that? Right, if you try to factor the bottom, it doesn't really factor easily, you'd have to do quadratic formula. |
| 15:31 | So you're not going to get anywhere with that. Right, obviously, when you have integrals, the first thing you should be looking to do is simplify. Before you do anything,
is the thing inside the integral in the simplest possible form? Generally, that's the way I'll give it to you, maybe once and a while I'll cross you up, but not too much.
I tried to give a bunch of these for you guys to practice on the my math lab. Plus I will put up some stuff from my book on u-substitution which people find helpful and I recommend working through, it's about 7 pages. |
| 16:00 | It's in my study guide, and people say it really helps them learn u-substitution so sure. Ignore the trigonometry ones obviously.
Um, ok, so, notice this is from x=1, to x=4, but we're going to do u substitution for u, so first we're just going to ignore the limits of substitution and just take this integral. |
| 16:31 | And do the substitution. So x^2+x-1 is u, and 2x+1 dx is du.
Ok, that integral is ln|u|, no + C, we don't need it. So this becomes the ln|x^2+x-1| and remember there was a definite integral. |
| 17:00 | So now we have to go from x=1, to x=4.
So if you plug in, you get ln|19|-ln|1|, ln(1) is 0, so this just equals ln|19|. |
| 17:31 | How do we feel about these? Still a little confused? Ok, practice them some more, we're going to do another type,
another u-substitution type. And we'll just keep practicing until you guys get good at them.
As I said in last class, and as I keep emphasizing, there's lots and lots of integrals, that are quite difficult to do, or not possible at all. |
| 18:03 | And the only way to do them is with approximation. Basically the Riemann sum concept, there are some other techniques, that you'll never get to in a class like this.
And you just cannot integrate them, they cannot be solved, so, all you can do is a numerical approximation. There's all sorts of stuff out there in the real world where there is no actual solution. So the approximation techniques are actually very important. |
| 18:32 | And computers and calculators can do the approximation technique to zillions of decimal places, so you don't have to worry about how close you get.
It's like pi. You can use pi to a trillion decimal places. You know, it's not actually a pi, but it's certainly not something you need to worry about. Alright, what if I gave you something like |
| 19:03 | x*(sqrt)(x+1)dx. How would you integrate that?
You say well, here's my problem. If I let u = x, that doesn't do anything, I'm just swapping u for x. Right, this would be u and this would be u+1, I'm not any better off than before. If I let u=x+1, the derivative of x+1 is not x, it's 1. So how do you do u substitution on something like this? |
| 19:32 | It doesn't make any sense right? If I let u = x+1, du =1dx. So I can substitute this, this can become du,
and this would become (sqrt)u but what about this?
You say, ah, well another thing you could do, is if u = x+1, then u-1=x. So you can rewrite this now as (u-1)(u)^(1/2)du. |
| 20:06 | So how do I know to do this trick? The power of x is the same.
Remember I said when power is 1 different, then you'll be able to get the derivative. So 24x^2, 12x. The derivative of x^2 is x. But here, this is x and this is x. So the derivative of 1 doesn't give you the other. Now why would I do this substitution, it doesn't seem like it's in much better shape. Well now you can distribute the u^1/2. |
| 20:31 | And you get u^3/2-u^1/2 du.
You say, well why didn't you distribute the x here? Well, then you would have gotten the (sqrt)(x^3+x^2) which doesn't help you in any way. So what you could do is rewrite the integral in a way where now, it's easy to integrate. The integral of u^3/2 is u^(5/2)/(5/2)-u^(3/2)/(3/2) |
| 21:06 | That's the same as (2/5)u^(5/2)-(2/3)u^(3/2)
And then you just put x+1 back.
So that becomes (2/5)(x+1)^(5/2)-(2/3)(1+x)^(3/2) +C. |
| 21:44 | So let's do a couple of these, make sure we get the hang of it.
I always wonder, who was the first person to think of this. Because the first time I saw an integral like that I had no idea what to do. |
| 22:05 | Suppose I gave you,
Wait, wait, don't pay attention to that one.
|
| 22:38 | Ok, suppose I gave you something like that.
So you say, let's let u =x^3+2, and u-2=x^3, and du = 3x^2 dx. You say ok, well this is u. |
| 23:03 | But wait, I don't have an x^3 and I don't have a 3x^2. What do I do. I have x^3*x^2, which is x^5. So, I could let
this, what kind of integrand is that?
So you take the x^5 and you break it up into x^3 and x^2. |
| 23:36 | That's not fun, I wouldn't have thought of that.
Ok? So, (sqrt)(x^3+2) becomes u^1/2, x^3 becomes (u-2), And x^2 dx, becomes 1/3 du. Remember the 1/3 is on the outside. |
| 24:06 | So again, this becomes (sqrt)u, x^3 becomes u-2, x^2dx becomes the 1/3du.
That's a really tricky one. Now, distribute. |
| 24:41 | And now it's relatively easy to integrate, in fact it's the same integral as this one.
u^(3/2)-u^(1/2). You get 1/3[(2/5)u^(5/2)-(4/3)u^(3/2)] |
| 25:11 | The the answer is (2/15)(x^3+2)^(5/2)-(4/9)(x^3+2)^(3/2) +C.
So remember what I told you, simple looking integrals, turn out to be messy solutions. |
| 25:34 | Right? Let's go back and do one of the slightly easier ones.
|
| 26:11 | Alright so, u=x+3,
so this will become u^5. du=dx. Because the derivative of du is 1, right? 1dx.
And then let u-3=x. |
| 26:34 | This then becomes the integral of (u-3)(u)^5 du.
By the way, the other thing you could do is multiply this out to the 5th, distribute the x so you just have a big polynomial. Certainly valid. But if that was to the 50th, it would take you a while. So you want to have a better technique just in case. Distribute the u^5, and you get u^6-3u^5 du. |
| 27:17 | And that is u^7/7- 3u^6/6
and substitute back, (x+3)^7/7-(x+3)^6/2 + C.
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| 27:44 | Got it? Ok, we're going to do some other stuff on Friday, see you Friday.
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