| Start | There's one. That should keep you busy for a minute or two.
Alright that's probably long enough. Now the first thing you do is you have to realize how you're going to do this integral. You look at this and you say "well, the first trick I know is u-substitution". This can't be u-substitution because the derivative of the e term is not the cos term and the derivative of the cos term isn't the e term. So I'm guessing this is integration by parts. |
| 0:30 | So let's let u =e^2x, it doesn't matter which you pick, dv= cos(3x)dx
du= 2e^2xdx, v = sin(3x)/3 , so this is going to get messy with all the 2s and the 3s. So be careful.
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| 1:01 | Ok, so now we're gonna have (integral)(e^2x)(cos(3x))dx = u*v = (e^2x)(sin(3x)/3) - (2/3) (integral) (e^2x)(sin(3x)dx), 2/3 comes from this 2 and that 3.
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| 1:40 | Alright we've got to do it again. Let u = e^2x, du = 2e^2xdx dv=sin(3x)dx, v = -cos(3x)/3 |
| 2:07 | So that means that our integral, of (e^2x)(cos(3x))dx = (e^2x)(sin(3x)/3) - (2/3)[(-e^2x(cos(3x)/3))- (2/3)(integral)((e^2x)cos(3x)dx] |
| 2:57 | So far so good? Now clean it up. So this becomes e^2x*sin(3x)/3 + (2/9)(e^2x)cos(3x) - (4/9)(integral)(e^2x)cos(3x)dx |
| 3:31 | At this point you question the entire idea. Now that is equal to the integral of e^2x(cos(3x))dx
Ok, so you see the more generic one was just e^x sin(x), and e^x cos(x), it's just harder if you throw in constants.
We could make it really fun if we threw in stuff like pi, but we wouldn't do that to you. Alright so now you notice that you have 1 e^2x(cos(3x)) over here and (4/9) of one over here, so I add it to the other side and I will get |
| 4:06 | (13/9)(integral)(e^2x(cos(3x))dx = (e^2x*sin(3x))/3 + (2/9)(e^2x)cos(3x))+ C and then you can just multiply everything by (9/13).
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| 4:36 | Ok, so you get (9/13) of this thing.
Which you would not have to simplify. Ok, how'd we do on this one? Anyone get it? Good. You got close? You got close, that's ok. |
| 5:05 | Alright.
I'll give you guys a minute to copy that down, remember we have a video the video for this class will go up probably friday. Maybe tomorrow afternoon. OK? So one of the things you should do for the final is go through your midterms, go through the 2 practice tests for the midterms, I don't know if we'll have more practice for the final but we'll try. |
| 5:36 | Ok? So do that integral.
Alright that's long enough. So when you see these trig ones, you really want to look for using some trig identities to simplify things. So you have sins and cosines, but remember the derivative of sin is cos. So you kind of have a function in this derivative. The problem is you have too many cos or too many sin. |
| 6:00 | But one thing we could do is we could rewrite this as sin^5(x)cos^2(x)cos(x)dx, why would we want to do that?
We want to do that because now you have a single cos, which is the derivative of sin. If only you could find a way to get rid of the other cos. But then we remember our trigonometry, we remember our MAT 123. And you know this is the integral of (sin^5x)(1-sin^2x)(cos(x))dx |
| 6:45 | Now you can do u substitution.
So you let u = sin(x) and du= cos(x)dx So this is going to become (u^5)(1-u^2)du |
| 7:13 | Ok, that's a much easier integral. Distribute u^5, you get u^5-u^7 du which is u^6/6- u^8/8 + c |
| 7:35 | And then we substitute back and you get sin^6(x)/6 - sin^8(x)/8+C
You do that one ok? Yes? No? the trick is you have to see if you can take out 2 of the cos.
And by the way, you could have taken sin^5x and made it sin(x) and sin^4x and made that into (1-cos^2x))^2 and put everything in terms of cos, its just more work. |
| 8:12 | So far so good? I'll give you another one.
We're having way too much fun today. I'll put it over there, that'll be my starting spot. Ok more of these. |
| 8:31 | You have 11x^2+5x+5 in the numerator and (2x+1)(x^2+1) in the denominator, you can factor that.
Integrate it. Just to give you guys the first step, you're gonna want some a/(2x+1) + (bx+c)/(x+1) or you could switch the letters. |
| 9:00 | Alright when it get's this noisy, it means people are done. So let's start doing this. Alright so first, we have A/(2x+1) + (bx+c)/(x+1) and that equals this thing.
If you multiply through by the denominator, you'll end up with a(x^2+1)+(bx+c)(2x+1) = (11x^2+8x+5). |
| 9:52 | Now do some algebra. Foil out the left side and you're gonna get ax^2+a+2bx^2+bx+2cx+c = (11x^2+8x+5) |
| 10:20 | Ok, so that becomes x^2(a+2b)+x(b+2c)+(a+c)=(11x^2+8x+5).
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| 10:42 | See and the cover up that the teachers taught you in school starts to fail in these problems.
sometimes. So this tells us that a+2b =11, b+2c=8, |
| 11:00 | and a+c = 5, alright we could just sort of guess.
maybe we'll come up with the right numbers. Or the simplest thing for these is get rid of one of the variables. Put it in the equation for the other 2 so you get down to 2 equations and 2 variables. For example here you could say c=5-a. and now you have a+2b=11 and b+2(5-a)=8 |
| 11:38 | So that's a+2b=11 and b-2a=-2
So now you gotta solve that one. So you can either do something clever like multiply this equation by 2 and add them, or you can do substitution again.
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| 12:08 | I'm gonna multiply it by 2, that's pretty easy. So this becomes 2a+4b=22 and then -2a +b =-2, so now when I add them I get 5b=20, so b=4.
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| 12:34 | Now that we know b =4, we can find the other 2 things. So a+2b=11, so a=3, b=4 and c=2. How'd we do on that?
Alright so that means we can take our integral and rewrite it as (3/2x+1)dx + (4x+2)/(x^2+1)dx |
| 13:10 | or 3(integral)(dx/2x+1) + 4(integral)(xdx/(x^2+1)) + 2(integral)(dx/(x^2+1)) A lot of work.
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| 13:37 | Ok, so far so good? These 2 are straight forward this one takes a tiny bit of work.
so let u = x^2+1, du =2xdx so (1/2)du = dx |
| 14:03 | That little integral becomes 4/2(integral) (du/u)I told you guys before I would memorize how to do this integral.
That will become (1/2)ln|x^2+1| So we are going to get 3(ln|2x+1|)/2 + 2(ln|x^2+1|) + 2arctan(x)+C, how'd we do? Yay? |
| 14:49 | Yes? Which last part? Sure, let u= x^2+1, du=2x, when you substitute here you have xdx, so I need 1/2 of du.
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| 15:05 | So I get 4*(1/2)(du/u), 4*(1/2) is 2, so I recommend, because we might have one of these on the final, we might not I don't really remember.
But if you have the integral of xdx/(x^2+1) just remember that that's a 1/2ln(x^2+1), you don't need absolute value bars because x^2+1 is always positive. Alright? |
| 15:35 | Let's give you guys something else to work on.
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| 17:43 | So we have the region between y=25-x^2 and y = 25-5x
find the area of R, find the volume when R is revolved about the y axis when using shells, and about the x axis washers.
find the area of the region bounded by y = 25-x^2 and y=25-5x. |
| 18:05 | Let's draw a picture. I'm pretty sure that we will actually give you a picture on the exam,
but your region looks something like that. This is the piece of the parabola that will matter. Right this actually sort of keeps going like that.
This is 25-x^2 and this is 25-5x, ok did everyone draw a picture? |
| 18:35 | OK the area,
Simply equals the area from here to here
of our slices. So where do they intercept? Well, this is 25, this is (0,25),
and this is (5,0)
so we're going to integrate from x=0 to x=5, top curve - bottom curvedx.
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| 19:15 | I'm not really that interested in you doing the actual integration but, since we're here, the integral from 0-5, of 5x-x^2 dx. How am I doing? So far so good?
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| 19:35 | That equals 5x^2/2 -x^3/3 from 0-5 and that equals 125/6.
Right did everyone get 125/6? Or 20 and 5/6? 20.833? Or I give up. |
| 20:01 | Any of those? The last one? ok.
That's find volume. So we want to go around the y axis, we're going to get a cylindrical shell because, you take a slice like here, and we're gonna get that shell shape. The height of the shell will be the slice, going from the top of the curve to the bottom of the curve. Radius of the shell is just x. |
| 20:36 | X determines where you're taking the slice. You're gonna go out x, and then you slice it.
So the volume will be 2*pi(integral) from 0-5, of x[(25-x^2)-(25-x)]dx |
| 21:04 | That equals 2pi*(integral) from 0-5 of (5x^2-x^3)dx
So far so good?
That's coming, you have to be patient. |
| 21:30 | Now, that integral, is 2*pi*5x^3/3-x^4/4 from 0-5.
1350pi/12, which can be reduced. Right? |
| 22:02 | something like that 160?
Something like that, I'm wrong a lot. So far so good? Alright, C. Now I want to draw around the x axis. So I'm gonna redraw this picture for a second. Now if I'm drawing around the x axis, |
| 22:30 | I'm not gonna get the same answer. So now if I slice this way I will get a washer.
So the volume is pi*r^2, so the volume is gonna be pi*(integral) from 0-5 of the top curve ^2 - bottom curve ^2, so (25-x^2)^2-(25-5x)^2 |
| 23:11 | So I don't really think we'll ask you to evaluate this because these get annoying, but 625-50x^2+x^4-625+250x-25x^2dx |
| 23:38 | which is pi(integral) from 0-5 of (x^4-75x^2+250x, did you guys get that?
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| 24:08 | Well if I wanted to change in terms of x to in terms of y, so for example, if I wanted to do shells around the x axis,
then I'd have to slice it this way.
If I am slicing that way I'm gonna get things in terms of y, and have to convert the equation from x to y. Similarly, if I was going around the y axis and I wanted to do washers, I'd have that slice. |
| 24:32 | So if I'm going to slice horizontally it has to be in terms of y. If I'm going to slice vertically it has to be in terms of x. Notice how I spent a lot of time.
Right, let's give you guys another kind of problem. There's still time. |
| 25:50 | OK, approximate the area under y=x^2+3 from x=1 to x=5, using n=4 right handed rectangles. Then write but do not evaluate the Riemann sum.
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| 26:04 | Alright, this one, this should be easier. We'll see if that's true.
No? No, oh my gosh folks, you have to remember this stuff! Ok, you have a week. Well you have a week right. Well, just under a week. The exam is at 8 in the morning, it's 5 in the afternoon, so some of you will be studying at 10:30, so you still have 162 hours. |
| 26:42 | Subtract 2 hours a day for sleep, which I do not suggest.
Alright let's get this one done. |
| 27:03 | So first of all, you should know how to find the answer, ok? Because the answer is just the integral from 1-5 of (x^2+3)dx which is x^3/3+3x from 1-5
which is (125/3 +15)-(1/3 +3) which is 170/3 -10/3 = 160/3, so when you do a problem like this, do the integral first so you know what you're doing.
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| 27:46 | So let's do n=4 right hand rectangles.
So x^2+3 is something boring like that, we're going from 1-5, |
| 28:08 | We're gonna go from 1-5
We're gonna use right hand rectangles.
So the width of each of these rectangles is 1. You go from 1-5 that's a distance of 4, so the width is (5-1)/4 = 1 |
| 28:36 | The height is f(2), f(3), f(4) and f(5) so the area is going to be,
1[f(2)+f(3)+f(4)+f(5)] and we're going to over state here, so we're gonna come out with something bigger than 160/3.
|
| 29:09 | So what is that, well let's see, f(2) is 2^2+3 = 7, f(3) is 3^2+3 f(4) is 4^2+3 f(5) is 5^2+3 I already know how this comes out, |
| 29:38 | 66. Which is bigger than 160/3 so we were able to execute part a.
No because we forgot it. Ok, well you got a week to remember. A week minus about 6 hours. It's not like you have anything else to study. The only thing you take is math right? Good. |
| 30:00 | Alright, now to do the Riemann sum, all you want to do is say well how wide is each one of these? Well instead of being 4/4 which is 1, this is going to be 4/n.
And you're going to start at 1 and you're going to add 4/n each time. Sorry, so you're gonna do 4/n[f(1+4/n)+f(1+2*(4/n))+f(1+3(4/n))+f(1+n(4/n))] |
| 30:53 | So if you were making that fancy sigma thing, it's going to be from i=1 to n of 4/n, of (1+(4i/n))^2 +3 and you need the limit as n goes to infinity. So again this comes from x^2+3, you essentially take that first |
| 31:29 | insertion into the parentheses and stick and i in there. Because each time, you're taking this 4/n and multiplying it by a bigger and bigger number, starting at 1 and going up to n.
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| 31:47 | So when we say set up the Riemann sum, I am looking for this
Ok folks? Alright, look for stuff on blackboard, good luck you guys were a great class, have a wonderful summer if I don't see you.
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