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[MUSIC].
I've mostly been selling you on the idea of u-substitution as a method for finding
antiderivatives, for evaluating indefinite integrals.
Let's think about what u-substitution says just for definite integrals.
So here's the setup. I've got the integral x goes from a to b
of some function f, not the derivative of f, just f g of x times g prime of x dx.
Now, how could I approach a problem like this so I can make a substitution?
U equals g of x so that du is g prime of x dx and, in that case, this integration
problem becomes the integral of f of u u is g of x times this is du.
But now I've gotta change the bounds of integration.
When is x is equal to a, u is g of a, and when x is equal to b, u is g of b.
Assuming that those two integrals exist to justify the equalty of those two
integrals. It's just the chain rule and the
fundamental theorem of calculus. And yet, this statement, the equality of
this integral and this integral, that's just a statement about two definite
integrals. I don't need to know the fundamental
theorem of calculus in order to make this assertion.
This raises a fascinating question, u substitution is making a claim about
definite integrals. And yet the naive way to justify that
claim goes through the fundamental theorem of calculus and the chain rule.
Is it possible to see why the statement about u substitution for definite
integral should be true without resorting back to anti-differentiation?
Well, to gain some insight into this, here's a random graph that I made of some
function that's nearly constant. But the function I'm calling f.
And this, area, in here is the interval from 1 to 10 of f of udu.
And that's this area in here. Now I want to imagine what happens when I
do the substitution. What happens when I replace.
U with x squared plus 1, and that changes du, right.
That's that formula that we saw a moment ago.
If I'm integrating f of u du, I could alternatively integrate f of g of x times
du g prime of xdx, and I can imagine that in the graph, transforming from the u
coordinates to these x coordinates. Well the u-coordinates are more stretched
out, right. When x is 0, u is 1.
When x is 1, u is 2. When is x is 2, u is 5.
And when x is 3, u is 10. So if I redraw this picture in the,
x-coordinates, it's as if I've taken this picture and squished it in, right, and
when I squish it in. Then I'm going to get a picture more like
this. And to calculate the area, there, I'd be
calculating this integral. The integral from 0 to 3, this is x, of f
of what u is equal to in terms of x, du, which is 2x dx.
And the whole gimmick here is just that I'm trading the width in this picture for
the height in this picture. And the factor that relates the width
here, the du's and the height here, well that's exactly this factor here, right?
That's du but written in terms of dx.