[MUSIC]. 
I'm mostly selling you on the idea of 
u-substitution, as a way to find 
anti-derivatives. 
But anti-differentiation is just a means 
to an end. 
The real goal, at this point in the 
course, is evaluating definite integrals. 
Well, here's an example of a definite 
integral. 
Let's evaluate the integral of 2x times x 
squared plus 1 to the 3rd power dx. 
As x goes from 0 to 2. 
We can do it with u-substitution. 
The substitution that I want to make is u 
equals x squared plus 1, and in that case 
du, what's the derivative of this, is 2x 
dx. 
And that's great 'cuz I've got a 2x dx 
right there. 
So, this integration problem becomes the 
integral x goes from 0 to 2. 
but what is the integrand now. 
It's u cube du and I know the 
anti-derivative of u cube is u to the 
forth over 4 and I want to make sure that 
I evaluate this when x equals to 0. 
And 2. 
Now we replace u by x squared plus 1. 
So this is x squared plus 1 to the 4th, 
over 4. 
And I want to evaluate at 0 and 2. 
Now we plug in x equals 2, and x equals 
0, and take the difference. 
Okay, well when I plug in 2, I get 2 
squared plus 1 to the 4th over 4. 
And when I plug in 0, I get 0 squared 
plus 1 to the 4th over 4. 
What's 2 squared plus 1? 
That's 5 to the 4th over 4 and that's 1 
to the 4th, which is just 1. 
A quarter. 
And, now I got think about what's 5 to 
the 4th? 
Well that's 25 times 25. 
Tha'ts 625 over 4 minus 1 over 4. 
And, now I can combine these into a 
single fraction, that's 624 over 4. 
And that I can simplify a bit. 
That's 156. 
We did it. 
But I could've finished this problem off 
in a slightly different but equivalent 
way. 
Let's back up, I'll get rid of this, and 
let's suppose that I didn't go down this 
path but I just stopped here. 
The problem is that my endpoints are in 
terms of x but my integrand now is in 
terms of u. 
So I'm just going to change those 
endpoints. 
So instead I'll see that, that 
integration problem is the same as the 
integral of u cubed du. 
And when x is equal to 0, u is 1 and when 
x is equal to 2, u is 2 squared plus 1, 
which is 5. 
So if I change the endpoint to be in term 
of u, then I don't have to go back to x. 
So again I know the anti-derivative, it's 
u to the 4th over 4 and I'm evaluating 
it. 
At 5 and 1 in term of u and taking the 
difference so when I plug in the 5, I 
just get 5 to the 4th over 4 and when I 
plug in 1 I just get 1 to the 4th over 4. 
And 5 to the 4th is 25 squared which is 
625 over 4. 
And this is now the same as before. 
Minus a quarter and, just like before, 
this ends up being 156. 
Let's summarize these two different 
approaches. 
The first time I went through this 
problem, I found the antiderivative in 
terms of x, alright? 
I wanted to integrate this and I found an 
antidirivitive and I just evaluated it at 
b and a and took the difference. 
In the second method, instead of finding 
the antidirivitive in terms of x, I 
change the endpoints to make the 
endpoints be in terms of u. 
So, I took this original problem and 
after making the substitution u equals g 
of x, then I rewrote the endpoints to go 
from g of a to g of b in terms of u. 
And then I found an antiderivative of f 
prime u du and now f of u and I evaluated 
that at g of b and g of a and took the 
difference. 
At the end of it I'm doing the same 
calculation. 
In both cases I'm calculating, you know, 
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 
I"m setting it up slightly differently. 
In the first case I'm finding the 
anti-derivative in terms of x, and in the 
second case, I'm changing the bounds on 
the integral to be in terms of u.