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# u-substitution with exponential function

You can never get too much u-substitution practice. Created by Sal Khan.

Video transcript

Let's see if we can calculate the
definated roll from zero to one of x squared, times two, times two to the x to
the third power dx. And, like always I encourage you to pause
this video and see if you can figure this out, figure this
out on your own. So I'm assuming you've had a go at it and so there's a couple of interesting
things here. The first thing, at least in my brain
does, it says, well I'm used to taking derivatives and antiderivatives of
e to the x, not some other base to the x. So we know that the derivative with
respect to x of e to the x. Is e to the x or we can say the ante
derivirate of e to the x is equal to e to the x plus c. So, since I'm doing to the, something
raise to this particular situation from the
raise to the function of x, it seems like I might wanna put some, I might wanna change the base
here. But how do I do that? Well the way I would do that is re-express
2 in terms of e, so what would be 2 in terms of
e. Well 2 is equal to e is equal to e raised
to the power that you had to raise e to to
get to 2. Well what's the power that you have to
raise e to to get to 2? Well that's the natural log of 2. Once again, the natural log of 2 is the
exponent that you have to raise e to, to get to 2. So if you actually raise e to it, you're
going to get 2. So this is what 2 is, now what is 2 to the
x to the 3rd? Well, if we raise both sides of this to
the x to the 3rd power. If we raise both side of the x to the
third power, two to the x of the third is equal to, if I raise
something to an exponent and then raise that to an exponent it's going
to be equal to e to the x to the third times the natural log of two, times
the natural log of two. So that already seems pretty interesting. So let's, let's rewrite this. And actually, what I'm gonna do, let's just focus on the indefinite integral
first. See if we can figure that out, and then we
can apply. And then we can take, we can evaluate the
definite ones. So let's just, let's just think about
this. Let's think about the indefinite integral
of x squared. Times 2 to the x to the 3rd power, dx. So I really wanna find the anti-derivative
of this. Well, this is going to be the exact same
thing as the integral of. So I'll write my x squared, still. But instead of 2 to the x to the 3rd, I'm
gonna write all of this business. Let me just copy and paste that. We already established, this is the same
thing, as two to the ex to the third power. Copy, and paste, just like that. And then, let me just, close it with a dee
ex. So, I was able to get it, in terms of e as
a base. That makes me a little bit more comfortable, but it still seems pretty
complicated. But, you might be saying, well okay look, maybe use substitution could be at play
here because I had this, this kind of crazy expression,
x to the third times the natural log of 2. But what's the derivative of that? Well that's going to be 3x squared times
the natural log of 2. Or 3 times the natural log of two times x
squared. Well that's just a constant times x
squared. We already have an x squared here. And so maybe we can engineer this a little
bit to have the constant there as well. So let's think about that. So if we made this, if we defined this as
u, so if we said u is equal to x to the third times the natural
log of two, what is d,u going to be? Well do you is going to be, is going to be, well natural log of two is just a
constant. So it's going to be three x squared, times
the natural log of two. And we can actually just change the order
or multiplying a little bit. We could say that this is the same thing
as x squared. Time three natural log of two. Which is the same thing, just using
logarithm properties, as x squared times the natural log of 2 to the
third power. So this is equal to x squared time the natural log of
eight. So let's see. If this is u, where's du? And of course, we can't we can't forget
the dx. This is a dx right over here. dx, dx, dx, so where is the du? Well we have a dx. Let me circle things. So you have a dx here, you have a dx
there. You have an x squared here, you have an x
squared here. So really, all we need is. All we need here is the natural log of 8. So if we, ideally we would have a natural
log if 8 right over here. And we could put it there as long as we
also, we can multiply by a natural log of 8 as long as we also divide
by a natural log of 8. And so, we could, we could do it by, right
over here. We could write, we could divide by a
natural log of 8. But we know that the anti-derivative of
some constant times a function is the same thing as the constant times
the anti-derivative of that function. So we could just take that on the outside. So it's one over the natural log of eight. So let's writes this in terms of UNDU. This simplifies to one over. The natural log of eight times the
anti-derivative of e, e to the u, e to the u, that's the
u, du. This times this times that is du, du. And this is straightforward, we know what
this is going to be. This is going to be equal to, so let me
just write the 1 over natural log of 8 out here, 1 over natural log of 8, times,
times e to the u times e to the u. e to the u of course, if we're thinking
int terms of anti-derivative there'd be some constant out there and then we would just reverse the
substitution. We already know what us is so this is going to be equal to the anti-derivative
of this expression is 1 over the natural log of 8 times e to
the, instead of u, we know that u is. x to the third times a natural log of 2
and of course we can put a plus C there. Now going back to the original problem, we
just need to evaluate the anti derivative of this at
each of these points. So lets rewrite this. So given what we just figured out, so let
me copy and paste that. This is just going to be equal to, it's
going to be equal to the anti-derivative evaluated at one, minus
the anti-derivative evaluated at zero. We don't have to worry about the
constants, because those will cancel out. So we are going to get, we are going to
get one. Let me evaluate it first at one. So you're gonna get one over the natural
log of eight, times E to the one to the third
power. Which is just one times the natural log of
two. Natural log of two, that's it at evaluated
at one. And then we're gonna minus it evaluated at
zero. So it's going to be one over the natural
log of 8 times e to the well, when x is 0 this whole thing
is going to be 0. Well e to the 0 is just 1 and e to the natural log of 2, well that's just
going to be 2. We already established that early on, that
this is just going to be equal to 2.