WEBVTT

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Sal: Let's see if we can
calculate the definite

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integral from zero to
one of x squared times

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two to the x to the third power d x.

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Like always I encourage
you to pause this video

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and see if you can figure
this out on your own.

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I'm assuming you've had a go at it.

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There's a couple of
interesting things here.

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The first thing, at least
that my brain does, it says,

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"I'm used to taking derivatives
and anti-derivatives

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of e to the x, not some
other base to the x."

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We know that the derivative
with respect to x of e to the x

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is e to the x, or we could
say that the anti-derivative

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of e to the x is equal
to e to the x plus c.

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Since I'm dealing with
something raised to,

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this particular situation,
something raised to

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a function of x, it seems
like I might want to put,

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I might want to change the base
here, but how do I do that?

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The way I would do that is
re-express two in terms of e.

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What would be two in terms of e?

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Two is equal to e, is
equal to e raised to the

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power that you need to
raise e to to get to two.

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What's the power that you have

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to raise two to to get to two?

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Well that's the natural log of two.

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Once again the natural
log of two is the exponent

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that you have to raise e to to get to two.

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If you actually raise e to
it you're going to get two.

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This is what two is.

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Now what is two to the x to the third?

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Well if we raise both sides
of this to the x to the third

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power, we raise both sides
to the x to the third power,

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two to the x to the third
is equal to, if I raise

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something to an exponent and
then raise that to an exponent,

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it's going to be equal to
e to the x to the third,

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x to the third, times
the natural log of two,

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times the natural log of two.

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That already seems pretty interesting.

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Let's rewrite this, and
actually what I'm going to do,

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let's just focus on the
indefinite integral first,

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see if we can figure that out.

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Then we can apply, then we can take,

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we can evaluate the definite ones.

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Let's just think about
this, let's think about

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the indefinite integral of x squared times

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two to the x to the third power d x.

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I really want to find the
anti-derivative of this.

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Well this is going to be
the exact same thing as

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the integral of, I'll
write my x squared still,

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but instead of two to the x to the third

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I'm going to write all of this business.

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Let me just copy and paste that.

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We already established
that this is the same

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thing as two to the x to the third power.

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Copy and paste, just like that.

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Then let me close it with a d x.

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I was able to get it in
terms of e as a base.

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That makes me a little
bit more comfortable

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but it still seems pretty complicated.

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You might be saying, "Okay, look.

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"Maybe u substitution
could be at play here."

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Because I have this crazy
expression, x to the third times

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the natural log of two, but
what's the derivative of that?

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Well that's going to be three x

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squared times the natural log of two,

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or three times the natural
log of two times x squared.

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That's just a constant times x squared.

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We already have a x squared
here so maybe we can engineer

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this a little bit to have
the constant there as well.

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Let's think about that.

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If we made this, if we defined this as u,

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if we said u is equal
to x to the third times

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the natural log of two,
what is du going to be?

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du is going to be, it's
going to be, well natural

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log of two is just a
constant so it's going to be

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three x squared times
the natural log of two.

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We could actually just change the order

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we're multiplying a little bit.

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We could say that this
is the same thing as

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x squared times three natural log of two,

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which is the same thing just
using logarithm properties,

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as x squared times the natural
log of two to the third power.

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Three natural log of two is the same thing

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as the natural log of
two to the third power.

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This is equal to x squared
times the natural log of eight.

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Let's see, if this is u, where is du?

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Oh, and of course we can't forget the dx.

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This is a dx right over here, dx, dx, dx.

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Where is the du? Well we have
a dx. Let me circle things.

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You have a dx here, you have a dx there.

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You have an x squared here,
you have an x squared here.

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So really all we need is,

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all we need here is the
natural log of eight.

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Ideally we would have the
natural log of eight right over

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here, and we could put it
there as long as we also,

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we could multiply by
the natural log of eight

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as long as we also divide
by a natural log of eight.

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We can do it like right over here,

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we could divide by natural log of eight.

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But we know that the
anti-derivative of some constant

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times a function is the
same thing as a constant

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times the anti-derivative
of that function.

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We could just take that on the outside.

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It's one over the natural log of eight.

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Let's write this in terms of u and du.

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This simplifies to one over the natural

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log of eight times the anti-derivative of

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e to the u, e to the u, that's the u, du.

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This times this times that is du, du.

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And this is straightforward,

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we know what this is going to be.

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This is going to be equal
to, let me just write

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the one over natural
log of eight out here,

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one over natural log of
eight times e to the u,

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and of course if we're
thinking in terms of just

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anti-derivative there would
be some constant out there.

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Then we would just
reverse the substitution.

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We already know what u is.

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This is going to be equal
to, the anti-derivative of

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this expression is one over
the natural log of eight

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times e to the, instead
of u, we know that u is

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x to the third times
the natural log of two.

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And of course we could put a plus c there.

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Now, going back to the original problem.

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We just need to evaluate
the anti-derivative

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of this at each of these points.

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Let's rewrite that.

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Given what we just figured out,
let me copy and paste that.

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This is just going to be equal to,

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it's going to be equal to
the anti-derivative evaluated

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at one minus the anti-derivative
evaluated at zero.

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We don't have to worry about the

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constants because those will cancel out.

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So we are going to get,
we are going to get one--

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Let me evaluate it first at one.

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You're going to get one
over the natural log of

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eight times e to the
one to the third power,

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which is just one, times
the natural log of two,

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natural log of two,
that's evaluated at one.

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Then we're going to have
minus it evaluated it at zero.

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It's going to be one
over the natural log of

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eight times e to the, well when x is zero

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this whole thing is going to be zero.

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Well e to the zero is just
one, and e to the natural

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log of two, well that's
just going to be two,

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we already established that early on,

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this is just going to be equal to two.

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We are left with two over the
natural log of eight minus

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one over the natural log of
eight, which is just going

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to be equal to one over
the natural log of eight.

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And we are, and we are done.