Hovedindhold

## Integrating with u-substitution

Aktuel tid:0:00Samlet varighed:5:11

# 𝘶-substitution intro

## Video udskrift

Let's say that we have
the indefinite integral, and the function is 3x
squared plus 2x times e to x to the third
plus x squared dx. So how would we go
about solving this? So first when you
look at it, it seems like a really
complicated integral. We have this polynomial right
over here being multiplied by this exponential expression,
and over here in the exponent, we essentially have
another polynomial. It seems kind of crazy. And the key intuition
here, the key insight is that you might want
to use a technique here called u-substitution. And I'll tell you
in a second how I would recognize that we
have to use u-substitution. And then over time,
you might even be able to do this type
of thing in your head. u-substitution is essentially
unwinding the chain rule. And the chain rule-- I'll go
in more depth in another video, where I really talk
about that intuition. But the way I would
think about it is, well, I have this crazy
exponent right over here. I have the x to the
third plus x squared, and this thing right
over here happens to be the derivative of x
to the third plus x squared. The derivative of x to
the third is 3x squared, derivative of x squared is
2x, which is a huge clue to me that I could use u-substitution. So what I do here is this
thing, or this little expression here, where I also see its
derivative being multiplied, I can set that equal to u. So I can say u is equal to x
to the third plus x squared. Now, what is going to
be the derivative of u with respect to x? du dx. Well, we've done
this multiple times. It's going to be
3x squared plus 2x. And now we can write this
in differential form. And du dx, this isn't really
a fraction of the differential of du divided by
differential of dx. It really is a form
of notation, but it is often useful to kind of
pretend that it is a fraction, and you could kind of view this
if you wanted to just get a du, if you just wanted to get a
differential form over here, how much does u change
for a given change in x? You could multiply
both sides times a dx. So both sides times a dx. And so if we were to pretend
that they were fractions, and it will give you the
correct differential form, you're going to
be left with du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to the
trouble of doing that? Well we see we have
a 3x squared plus 2x, and then it's being multiplied
by a dx right over here. I could rewrite this
original integral. I could rewrite this
as the integral of-- and let me do it in that color--
of 3x squared plus 2x times dx times e-- let me do that
in that other color-- times e to the x to the
third plus x squared. Now what's interesting
about this? Well the stuff that
I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up
here, x to the third plus x squared, that is what
I set u equal to. That is going to be equal to u. So I can rewrite
my entire integral, and now you might
recognize why this is going to simplify
things a good bit, it's going to be equal to--
and what I'm going to do is I'm going to
change the order. I'm going to put the
du, this entire du, I'm gonna stick it on
the other side here, so it looks like more
of the standard form that we're used to seeing
our indefinite integrals in. So we're going to have
our du times e to the u. And so what would the
antiderivative of this be in terms of u? Well, the derivative of
e to the u is e to the u. The antiderivative of e
to the u is e to the u. So it's going to be
equal to e to the u. Now, there is a
possibility that there was some type of a
constant factor here, so let me write that. So plus c. And now, to get
it in terms of x, we just have to
unsubstitute the u. We know what u is
equal to, so we could say that this is
going to be equal to e. Instead of writing u, we
could say u is x to the third plus x squared. And then we have our plus c. And we are done. We have found the
antiderivative. And I encourage you to take
the derivative of this, and I think you will find
yourself using the chain rule, and getting right back
to what we had over here.