| Start | Ok, so questions you should be able to do on the final. Make sure you can do something like this.
You should be able to do it without a calculator. |
| 1:06 | Ok, so approximate the area under the curve y=2x^2+3x+4 from x=2 to x=6 using n=4 left and right hand rectangles.
We'll work on it together at this point. So you get points for this setting it up, and you get points for doing the calculations. |
| 1:30 | Ok? So you get both. So I don't want you to get too excited about if you can't get the final number correct.
So um, let's draw a picture of what this looks like. We'll put this over here. It's some kind of upward parabola, the y-intercept is 4, and it goes up like that, and we're going from 2-6. |
| 2:00 | Using rectangles. So those are our left rectangles, thats an underestimate. These points are 3,4,5,6.
So the left end point rectangles will be evaluated at x=2, 3,4, 5. And the right end point rectangles will be at 3,4,5 and 6. And you literally just plug them into the equation and multiply them by width. And I'm very nice, I made the width of each of these rectangles 1, now on the exam maybe I'll make the width pi or something evil, but probably not. |
| 2:34 | I haven't done that so far, and theres no reason to think I'm doing that now.
So the width of each of these is 1, And you're going to want to do f(2)+f(3)+f(4)+f(5). 1 times, let's see. f(2)=18, f(3)=31, f(4)=48, and f(5)=69. That comes out to 166. |
| 3:40 | How did I do?
Ok, so that's the area underneath estimate, and it's less than the total area under the curve. Feel free to check my arithmetic, I'm doing it in my head while I'm talking so I could be getting them wrong. So for the right end point rectangles you do the same thing, except we're going to use 3,4,5,6 instead of 2,3,4,5. |
| 4:03 | So this is going to be 1[f(3)+f(4)+f(5)+f(6)]
Nobody has spoken up so I assume I have the number right.
So 3,4, and 5 I've already calculated. Thats 31+48+69. |
| 4:37 | And let's see f(6)=94.
So basically we took away 18 and added 94, so you're going to add 76 to the total so that's 242. So that's the over estimate. So that's if you do these rectangles. |
| 5:08 | So if I said what's your estimate, well, we would just average those 2 numbers. Now if we wanted to find the area exactly, we would have to evaluate this integral.
Which is from 2-6 2x^2+3x+4dx. So if we average that, let's see you get 204? |
| 5:39 | Yeah 204.
So the actual area is the integral of this which is 2x^3/3+3x^2/2+4x from 2-6. And then you plug in 6, you plug in 2 and you subtract |
| 6:03 | So let's see that's 202 and 2/3, I think that's correct I could be wrong.
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| 6:37 | As I said, I'm just sort of making that stuff up.
So we might ask you, find the left, find the right, find the actual. We don't really care about the averaging, you should be able to average 2 numbers together. Ok? So how'd we do on this? Good, alright, let's have you do some more integration. |
| 8:07 | Ok, theres 4 fun ones to start.
Ok, let's go over these. How do you know when to use u-substitution? Well first you try to do it without u-sub. So if you see a simple polynomial, you don't need to use u-substitution. Or even just plain old substitution. This is just 5x^4/4-8x^3/3+11x^2/2+4x+C. That's all. |
| 8:51 | Ok, you don't need u-sub to do e^x.
So thats e^2x/2, don't forget to divide by whatever is multiplied by x. |
| 9:05 | You don't do u-substitution when you just have something over x, because thats just a simple log.
And square root of x is just x^1/2. So the integral of that is x^(3/2)/(3/2) which you can leave over a fraction, you don't have to flip and multiply. You can if you want to but you don't have to. |
| 9:31 | You should be able to do that if you have to use numbers, if you did a definite integral where you have to plug in numbers.
So again, don't need u-sub for those. This one looks a little weird, you say agh, if I did u sub, what would I substitute? Cause the derivative of this is not this, or the other way around. But you could break this integral up into x^3/x +11x^2/x+7/x and now it's a simple integral. |
| 10:17 | So this is x^3/3+11x^2/2+7log(x)+C.
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| 10:34 | Remember I told you integrals are much more annoying than derivates.
This should be x^3/3=x^2, 11x^2/2 is 11x, and 7/x is 7/x. So each of these individually is over x. |
| 11:11 | Alright, good? The last one, first I would rewrite this as x^3/10-x^(-3/10)dx.
Then we integrate, 3/10+1 is 13/10 so it's x^(13/10)/(13/10)-x(7/10)/(7/10)+C. Notice we didn't have to plug in to any of these. |
| 11:54 | Where would you see a definite integral other than just giving you one? Well, we could give you a word problem that looks something like this.
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| 12:22 | You could say something like, if marginal revenue, remember that's the derivative, marginal means derivative, marginal revenue for a manufacturer |
| 12:45 | is r'(x)=-x^2+18x+100, where x = units sold, find the total revenue for 10 units. Or 10,000 units.
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| 13:37 | What if I gave you something like that. You're a manufacturer, you have your marginal revenue equation, you figure out that it's -x^2+18x+100
Find the total revenue for 10,000 units, so what you're going to do is integrate from 0-10 the marginal revenue equation. That's all you have to do.
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| 14:03 | When you see the word marginal, think derivative, when you see the word total think integral.
And a clue is this is r'(x) and then you can go backwards and find r. And that will be r(10), because for 10,000 you plug in 10 not 10,000. So this is -x^3/3+9x^2+100x from 0-10 |
| 14:41 | Again, it's 9x^2 because it was 18x^2/2.
Then you plug in and you get -1000/3+900+1000. |
| 15:01 | Which is 4700/3-0, because when you plug in 0, these are all 0.
And you know, that's some number of dollars. Ok, I don't know what units we're making, are we making refrigerators, are we making I don't know, something that costs a penny? I don't know I just made a problem up. In the real world of course, you care about these things. |
| 15:36 | Ok, let's practice some other types of integrals.
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| 17:08 | There you go, that will entertain you for a few minutes.
Ok, so these are all u-substitution. How do you know these are u-substitution, otherwise you can't do them, You look and you say, I can't do I can do x^4 on it's own, I can't really do e^3x^5+1 on it's own, and since they're stuck together, |
| 17:30 | it's gotta be u-substitution because we haven't done other techniques. There are lots of different types of integration which we just don't cover.
So I would let u=3x^5+1 because that's the higher power term, du = 15x^4 dx, then I look at my integrand, and I don't see 15x^4, but I do see x^4. So, 1/15du=x^4dx. |
| 18:11 | Now substitute and you can write this as e^udu.
Now you integrate e^u is e^u+c, so it's (1/15)e^3x^5+1+C. |
| 18:42 | Ok? How'd we do on that one? Good? Ok, let's try the second one.
I look inside that cubed root and I see 1-x^2, I look outside I see x, and the derivative of 1-x^2 is something x. |
| 19:01 | So my substitution is I'll let u=1-x^2, and du is -2xdx
Therefore because I only have x, I'll do -1/2 du=x dx.
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| 19:33 | Substitute, this becomes -(1/2)u^(1/3)du.
So we took that messy integral and made it a nice integral. You integrate u^1/3, you get u^(4/3)/(4/3) And substitute back, and you can flip the fraction if you want, you get -3/8(1-x^2)^(4/3)+c. Alright? |
| 20:11 | Here I have 3x^2+1 in the denominator, and I have x in the numerator, So the other way around here you'd do the division like the other problem.
But here you can't really so let's let u=3x^2+1, and du=6x dx. |
| 20:32 | I only have xdx, so 1/6 du=x dx.
So I could rewrite this as (1/6)du/u. You can't have u/du, you can only have du/u. You can't have a du in the denominator, it makes things really complicated. |
| 21:03 | This becomes 1/6ln(u)+c, which is 1/6ln(3x^2+1)+C, you don't actually need the absolute value bars on this one because 3x^2+1 is always positive but if you left them in thats ok.
Alright last one. |
| 21:32 | Now notice here, this is one of those annoying ones x+3 and x are the same power.
So you have to do sort of a little trick. You let u=x+3, so du = dx, and x=u-3. So you can rewrite this as u^(1/2)(u-3)du. |
| 22:04 | Remember we did a few of these. Can I circle the answer? Sure.
Not really a circle, but you get the idea of a circle. So this becomes the integral of u^(3/2)-3u^(1/2)du, you get that by multiplying through. |
| 22:37 | That is,
u^(5/2)/(5/2)-3u^(3/2)/(3/2)
And then if you substitute back, it is 2/5(x+3)^(5/2)- 2/3(x+3)^(3/2)+c.
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| 23:28 | How'd we do? Ok? So that's everything.
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| 23:34 | Good luck on the final, we'll see you all Wednesday, I hope you enjoyed the class, and maybe I'll see some of you in a future course. Ok? Good luck.
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