[MUSIC] We’ve already learned about the
chain rule in reverse. That technique was u substitution.
I want to anti-differentiate a function evaluated at g times the derivative of g.
I can make a substitution and reduce that just down to anti-differenting f of this
new variable u. Now we’re going to run the product rule
in reverse. What is the product rule?
Well the product rule tells us how to differentiate the product of two
functions. So here’s two functions, f and g, and if
I want to differentiate their product, using the product rule, right.
The derivative of the product is the derivative of the first, times the second
plus the first, times the derivative of the second.
Now, integrate both sides. So then I get that anti derivative of the
derivative of f times g, is anti derivative of, what’s this by the product
rule? Alright, it’s derivative of the first,
times the second, plus the first, times the derivative of the second.
But the antiderivative of the derivative is just the original function.
So let’s write that down. That tells me that an antiderivative of
the derivative of f times g plus f times the derivative of g is this.
Which is just f of x times g of x, and I’ll include a constant.
That’s the integral of a sum, so its the sum of the intergrates.
So integrate f prime of x g of x dx plus integral of f of x, g prime of x dx and
we get f of x times g of x plus a constant.
And that’s one very symmetric way of writing down the product rule in reverse.
But we can rearrange it a bit more. I’ll subtract this integral from both
sides, so the left hand side is just this integral.
So the integral of f of x, g prime of x dx is equal to, well heres what we got on
the righthand side f of x times g of x. But I’m going to subtract this.
Subtracting f prime of x g of x dx. Now look at what this is saying.
It’s saying that I can do this integration problem if I can do this
integration problem. And how do these two integration problems
differ? Well here I’ve got a function times a
derivative and here I’ve got the derivative of f and an antiderivative of
g is an antiderivative of g prime. So I can replace this integration problem
with another integration problem. Where I’ve differentiated part of the
integrand, and anti-differentiated another piece of the integrand.
It’ll be a bit easier to see what’s going on if I make some substitutions.
Let’s set u equal to f of x, and dv equal to g prime xdx.
And in that case d u is f prime x d x. And what’s an anti-derivative of this.
Well one of them is just g of x. So I can use these substitutions to
rewrite what I’ve got up here. This integral is u times dv.
And it’s equal to u times v, minus the integral of g is v.
And f prime dx is du. So now I’ve got the integral of udv is uv
minus the integral of vdu. This is maybe why it makes sense to call
this Integration by parts. So I can integrate udv provided I can
integrated vdu. It’s this trading game.
I’m trading this integration problem for this integration problem.
But now one part is differentiated and another part of the inner grand is
antidifferentiated. Maybe that’ll make things better.
With u substitution, we had to come up with a single u.
In contrast, when you’re doing integration by parts, when using this
formula. You not only have to pick a u, but you’ve
got to pick a dv so that you can write your integrand as udv.
This makes parts a bit harder to apply the u substitution.
I’ve gotta find both a u and a dv. But any time you’re willing to
differentiate part of the integrand at the price of antidifferentiating the
other part of the integrand. Well if that’s something you’re willing
to do, parts will do that for you. [BLANK_AUDIO]