Welcome back.
Well I'm now going to do just a bunch of integration by parts
problems, as many as I can do in ten minutes without
confusing you.
So let me just write the formula for integration by
parts, and if you ever forget it-- I mean, it doesn't hurt to
memorize it, but if you ever forget it-- you just really
have to just derive it from the product rule of
differentiation.
But it just says that if we have an integral of f of x
times g prime of x-- so if you see, within the integral, sign,
one function and then you see the derivative of another
function, and I think with practice-- integration by parts
is really a bit of an art.
It's not systematic-- that's g prime of x-- that equals f of
access times g of x-- this is how it's the product rule in
reverse-- minus the integral of the derivative of the first
function, f prime of x, times the second function.
And it kind of easy to memorize, because there's this
symmetry to the formula.
So let's see if we can apply this.
And really, once you know that you should use integration by
parts, I think you'll find that it's not that hard to do it.
The hard part is to recognize when you should use
integration by parts.
From my point of view, it's kind of my last resort, or once
you have a lot of practice, you might recognize, well, if
there's an e to the x in it, or if there's a trig function in
it, and I can't do the reverse chain rule or integration by
substitution, then integration by parts is probably my best
option, assuming I'm seeing this on an exam and
not in real life.
In real life, it might be an unsolvable integral, and
you'd have to use a computer or some other technique.
But if you're seeing it on an exam, you know it's a solvable
integral, and if you can't solve it any other way, it's
probably integration by parts.
But let's just do some problems.
Let's say I want to take the integral of x squared e to dx.
So if I saw this out of the blue, and I didn't know that
this was a presentation on the integration by parts, I would
first-- clearly this isn't a polynomial, so I can't just do
a simple polynomial antiderivative.
Then I would try to see, is there the derivative of
something, of one function, of kind of a composite function
here so I can do the reverse chain rule.
The derivative of x here is 1, so I can't do anything here.
So I use the chain rule.
And the way I think about the chain rule is I
want to simplify it.
So when I go into this term right here, I'm going to
have to pick up my f of x.
I have to pick my f of x out of probably one of these two
functions, so that the f prime of x is simpler.
And I need to pick my g prime of x, I would guess that either
x squared is going to be my g prime of x, or e to the x is
going to be my g prime of x, and I want to pick that so that
when I take the antiderivative of it, it's going
to be simpler.
Or at least, not more complicated.
I know that if I take the derivative of x squared,
that simplifies it.
And I also know that the-- and once again, this is one of my,
to me, a very mind blowing idea-- but that the
antiderivative of e to the x is e to the x.
So it's probably a good idea to say that f of x is equal to--
it's probably, let me change colors-- it's probably a good
idea to make f of x is equal to x squared, because later I'm
going to take the derivative of it, and the derivative of it
simplifies it, and it's probably a good idea to make g
prime of x e to the x, because later, I'm going to take the
antiderivative of it, and the antiderivative of e to
the x is e to the x.
It's not going to become any more complicated.
So if we assume what I'm doing is right, then what did we say?
Well here, we just multiply the two real functions, right?
So when I say the real functions, I mean not
the derivatives of e to one of them.
So f of x, we're saying f of x is x squared.
Let me try to stay color consistent.
And we said g of x-- now don't get confused-- we're saying g
prime of x-- let me write it in a corner down here-- we're
saying that g prime of x is equal to e to the x.
And of course, if g prime of x is equal to ex, then g of x
is also equal to e to the x.
So g of x-- I don't want you to think that I'm somehow
putting g prime of x here.
I've taken the antiderivative, it just happens to be
the same function.
And then from that, we subtract the integral, we take the
derivative of x squared, so you get 2x, and then times the
antiderivative of g prime of x.
Well, the g prime of x is e to the x, you take the
antiderivative, it's still e to the x.
Actually I should probably stay at least keep the colors kind
of consistent, so you know what I'm hopefully doing.
The example using e to the x might be a little complicated,
because it's hard to tell whether I've taken the
derivative or not.
And you can kind of keep reverting back to the top
formula if you get confused.
So it looks like I've simplified it a little bit.
This integral looks easier to solve than this integral.
But once again, when I look at this, I'm like, well,
how do I solve this?
I can't use the integration by substitution, because there's
not an embedded function and then I have the derivative of
it sitting right next to it, so.
Maybe I need to do integration by parts again.
So let's do that.
So let's say-- let me do it separately-- let's assume-- I
think you getting a little bit of the hang of it-- along the
same vein, that this is f of x, that this is f of x, and that
this is g prime of x now.
We're kind of doing integration by parts within
integration by parts.
So if that is the case, that this integral is going to
equal-- because we have the minus sign out front, this
minus sign out front-- this integral is going to equal f
of x times g of x where f of x is just 2x.
Our g of x, this is g prime of x now.
Remember, we're kind of doing a new problem within the
original big problem.
So this is g prime of x, but g of x is still just e of x.
I took the antiderivative of it.
And that's minus the integral of the derivative of
the first function.
That prime of x.
So that's just 2.
And then the antiderivative of the second function.
Well, that's easy.
The antiderivative of e to the x is just e to the x.
Interesting.
Now I think you see where we're going.
This is actually-- let me write out the whole thing.
Because this is-- x squared, e to the x.
Just so we don't lose track of our original problem.
Interesting.
Now I think we have an integral that is pretty
straightforward to solve.
Don't want to forget my dx's.
What's the integral of-- we could take this 2 out of
this, and I think it becomes pretty obvious-- what's the
integral of e to the x?
This is a scratch-out, this says dex.
Just so-- and it's a little messy.
I don't like this color.
Magenta.
Well the integral of e to the x, or the antiderivative of e
to the x is e to the x, right?
So let's write that.
So I'm going to rewrite everything we've done it.
So there's x squared e to the x minus 2xe to the x, and then
this minus, you distribute it so it becomes a plus, so then
it's plus 2-- I just took the minus, I multiplied it times
this minus, so I got a plus 2-- and then the antiderivative of
e to the x is just e to the x.
And then of course, we should never forget the plus c.
Pretty fancy, no?
We've figured out the antiderivative, the indefinite
integral of x squared e to the x is this big fancy thing.
I bet you, before listening to this video, you would never
imagine that you could tackle integration like this.
You could actually try x to the n times e to the x.
You can try x to the tenth times e to the x.
It actually turns out you'll just have to do this many many
many many times, but every time you do integration by parts,
that the exponent on the x-term just becomes smaller and
smaller and smaller until you get to something that's really
easy to integrate, and then you can do it, and you'll have kind
of this big long expression.
It might be tedious, it might be hairy, but at least you have
a tool kit-- or you have something in your tool kit--
that you can tackle integral problems like this.
I'll probably do one more video on integration by parts, just
because I think this is one of the harder concepts to really
grasp and feel comfortable with, and then I'll try to
do a bunch of examples.
Maybe not soon, but in the next couple of weeks, on
just a lot of integration.
And I'm going to mix it up so that hopefully you can get a
sense of how I try to figure out which of my integration
tools I should use for a certain problem when
I see the problem.
See you in the next presentation.