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Studying for a test? Prepare with these 12 lessons on Antiderivatives and the fundamental theorem of calculus.

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# Integration by parts: ∫x²⋅𝑒ˣdx

Video transcript

Let's see if we can
take the antiderivative of x squared times
e to the x dx. Now, the key is to recognize
when you can at least attempt to use integration by parts. And it might be a
little bit obvious, because this video is
about integration by parts. But the clue that
integration by parts may be applicable
is to say, look, I've got a function
that's the product of two other functions-- in this
case, x squared and e to the x. And integration by
parts can be useful is if I can take the
derivative of one of them and it becomes simpler. And if I take the
antiderivative of the other one, it becomes no more complicated. So in this case, if I were to
take the antiderivative of x squared, it does become simpler. It becomes 2x. And if I take the
antiderivative of e to the x, it doesn't become
any more complicated. So let's assign f of x
to be equal to x squared. And we want that one
to be the one where if I take the derivative
it becomes simpler. Because I'm going to have
to take the derivative of f of x right over here in the
integration by parts formula. And let's assign g prime of
x to be equal to e to the x. Because later, I'm going to
have to take its antiderivative, and the antiderivative of e to
the x is still just e to the x. So let me write this down. So we are saying that f of x--
I'll do it right over here-- f of x is equal to x squared,
in which case f prime of x is going to be equal to 2x. And I'm not worrying about
the constants right now. We'll just add a
constant at the end to make sure that
our antiderivative is in the most general form. And then g prime of x is
equal to e to the x, which means its
antiderivative, g of x, is still equal to e to the x. And now we're ready to apply
the right-hand side right over here. So all this thing
right over here is going to be equal to f
of x, which is x squared-- and let me put it right
underneath-- x squared times g of x, which is e
to the x, minus. Let me do that in
that yellow color. I want to make the
colors match up-- minus the antiderivative
of f prime of x. Well, f prime of x is
2x, times g of x. g of x is e to the x dx. So you might say,
hey, Sal, we're left with another
antiderivative, another indefinite
integral right over here. How do we solve this one? And as you might
guess, the key might be integration by parts again. And we're making progress. This right over here is a
simpler expression than this. Notice we were able to reduce
the degree of this x squared. It now is just a 2x. And what we can do to
simplify this a little bit. Since 2 is just a
scalar-- it's a constant, it's multiplying the
function-- we can take that out of the integral sign. So let's take it this way. So let me rewrite it this way. We can only do that
with a constant that's multiplying the function. So let me put the
2 right out here. And so now what
we're concerned about is finding the integral--
let me write it right over here-- the
integral of xe to the x dx. And this is another
integration by parts problem. And so let's again apply the
same principles of integration by parts. What, when I take
its derivative, is going to get simpler? Well, x is going to get simpler
when I take its derivative. So now, for the purposes
of integration by parts, let's redefine f of x
to be equal to just x. And then we can still have g
prime of x equaling e to the x. And so in this case, let
me write it all down. f of x is equal to x. f prime of x is equal to 1. g prime of x is
equal to e to the x. g of x, just the
antiderivative of this, is equal to e to the x. So let's apply integration
by parts again. So this is going to be equal
to f of x times g of x. Now f of x is x. g of x is e to the x, minus
the antiderivative of f prime of x-- well, that's just 1--
times g of x-- e to the x. It's just 1 times e to the x dx. And remember, all
I'm doing right now-- you might have lost
track of things-- I'm just focused on
this antiderivative. That antiderivative is
that antiderivative there. If we can figure
out what it is, we can then substitute back
into our original expression. Now, you might appreciate
integration by parts. What does this right
over here simplify to? What is the antiderivative
of 1 times e to the x dx? Or what is the antiderivative
of 1 times e to the x? Well, it's just the
antiderivative of e to the x, which is
just e to the x. So this simplifies
to x times e to the x minus the antiderivative of
e to the x, which is just e to the x, so minus e to the x. And then we can take this
and substitute it back. This is the
antiderivative of this. So we can substitute
it back up here to figure out the antiderivative
of our original expression. So the antiderivative of
our original expression-- we're getting really
close-- is going to be equal to-- I'm going
to use different colors so we can keep track of things. It's going to be equal
to x squared times e to the x minus 2 times
all of this business. So minus 2 times-- well,
this antiderivative we just figured out is this. Minus 2 times xe to
the x minus e to the x. And if we want, now is a
good time to put our plus C. And of course, we
can simplify this. This is equal to x squared. I like to keep the same colors. This is equal to x
squared e to the x. You distribute the negative 2. You get minus 2xe to
the x plus 2e to the x, and then finally, plus
C. And we're done. We figured out
the antiderivative of what looked like a kind of
hairy-looking expression using integration by parts twice.