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[MUSIC]. 

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I'm mostly selling you on the idea of 
u-substitution, as a way to find 

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anti-derivatives. 
But anti-differentiation is just a means 

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to an end. 
The real goal, at this point in the 

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course, is evaluating definite integrals. 
Well, here's an example of a definite 

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integral. 
Let's evaluate the integral of 2x times x 

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squared plus 1 to the 3rd power dx. 
As x goes from 0 to 2. 

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We can do it with u-substitution. 
The substitution that I want to make is u 

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equals x squared plus 1, and in that case 
du, what's the derivative of this, is 2x 

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dx. 
And that's great 'cuz I've got a 2x dx 

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right there. 
So, this integration problem becomes the 

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integral x goes from 0 to 2. 
but what is the integrand now. 

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It's u cube du and I know the 
anti-derivative of u cube is u to the 

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forth over 4 and I want to make sure that 
I evaluate this when x equals to 0. 

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And 2. 
Now we replace u by x squared plus 1. 

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So this is x squared plus 1 to the 4th, 
over 4. 

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And I want to evaluate at 0 and 2. 
Now we plug in x equals 2, and x equals 

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0, and take the difference. 
Okay, well when I plug in 2, I get 2 

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squared plus 1 to the 4th over 4. 
And when I plug in 0, I get 0 squared 

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plus 1 to the 4th over 4. 
What's 2 squared plus 1? 

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That's 5 to the 4th over 4 and that's 1 
to the 4th, which is just 1. 

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A quarter. 
And, now I got think about what's 5 to 

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the 4th? 
Well that's 25 times 25. 

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Tha'ts 625 over 4 minus 1 over 4. 
And, now I can combine these into a 

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single fraction, that's 624 over 4. 
And that I can simplify a bit. 

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That's 156. 
We did it. 

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But I could've finished this problem off 
in a slightly different but equivalent 

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way. 
Let's back up, I'll get rid of this, and 

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let's suppose that I didn't go down this 
path but I just stopped here. 

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The problem is that my endpoints are in 
terms of x but my integrand now is in 

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terms of u. 
So I'm just going to change those 

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endpoints. 
So instead I'll see that, that 

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integration problem is the same as the 
integral of u cubed du. 

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And when x is equal to 0, u is 1 and when 
x is equal to 2, u is 2 squared plus 1, 

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which is 5. 
So if I change the endpoint to be in term 

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of u, then I don't have to go back to x. 
So again I know the anti-derivative, it's 

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u to the 4th over 4 and I'm evaluating 
it. 

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At 5 and 1 in term of u and taking the 
difference so when I plug in the 5, I 

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just get 5 to the 4th over 4 and when I 
plug in 1 I just get 1 to the 4th over 4. 

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And 5 to the 4th is 25 squared which is 
625 over 4. 

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And this is now the same as before. 
Minus a quarter and, just like before, 

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this ends up being 156. 
Let's summarize these two different 

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approaches. 
The first time I went through this 

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problem, I found the antiderivative in 
terms of x, alright? 

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I wanted to integrate this and I found an 
antidirivitive and I just evaluated it at 

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b and a and took the difference. 
In the second method, instead of finding 

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the antidirivitive in terms of x, I 
change the endpoints to make the 

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endpoints be in terms of u. 
So, I took this original problem and 

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after making the substitution u equals g 
of x, then I rewrote the endpoints to go 

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from g of a to g of b in terms of u. 
And then I found an antiderivative of f 

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prime u du and now f of u and I evaluated 
that at g of b and g of a and took the 

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difference. 
At the end of it I'm doing the same 

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calculation. 
In both cases I'm calculating, you know, 

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f of g of b, f of g of b, and I'm 
subtracting f of g of a, f of g of a, but 

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I"m setting it up slightly differently. 
In the first case I'm finding the 

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anti-derivative in terms of x, and in the 
second case, I'm changing the bounds on 

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the integral to be in terms of u.