[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

this. Right?

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]

Hello there. I am currently investigating whether or not Integration by Parts is the same thing as "the Reverse Product Rule" I was trying to figure out how to process the left side of the equation in this example: First Order Linear Differential Equations / Integrating Factors – Ex 2

…and realized I couldn't process the uv – ∫ v du (I.B.P.) in an easy way. I had to figure out by using wolfram to do: g(df/dx) + f (dg/dx) = d/dx( f * g )

by picking what is f, f' and g, g' in order to form d/dx( f * g ).

Therefore, I humbly request if you could make a video comparing I.B.P. and this other "Reverse Product Rule". Or maybe just illustrate a few examples of this reverse method?

Is this really the reverse product rule? I think this is the proof of the u-subs. I'm not sure.

This is integration by parts.

You don't even realize he is talking about integration by parts until the end. Now, that is some exceptional teaching skill.

bruh

he reminds me of david from love it or list it

This guy needs to go outside and do something with his life, stroking his dick to anti-derivatives is making him insane

Yeah……that didn't help at all.

didn't help at all