[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.