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PROFESSOR: Hi.

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Welcome back to recitation.

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Today we're going to
do another example

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of a definite integral.

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This time, it's going to be one
that you're going to need

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to do some substitution for.

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You're not going to know the
antiderivative right off the

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top of your head, I think.

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So, I've got it on the
board behind me.

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So it's compute the integral
from minus 2 to 2 of the

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function x squared cosine
of the quantity x

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cubed over 8 dx.

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So, this is a kind of
weird a integrand.

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I roughly, you know, very, very
rough graph of kind of

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what it looks like.

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So this is sort of the
area of this region.

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It's a positive function
over this interval.

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So it's the area of this
region that I'm

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asking you to compute.

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So, all right, so definite
integral.

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Why don't you pause the video,
take a couple minutes, work

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this out yourself, come back,
we can work it out together.

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All right, welcome back.

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So, from looking at this
integrand here, it seems to me

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that I don't know immediately
off the top of my head what

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its antiderivative is.

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So, I can't just sort of apply
the fundamental theorem of

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calculus directly without
a little bit of work.

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So there are two different
ways that I can go about

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computing this then,
about computing

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this definite integral.

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One is, that I can forget for
the minute that it's a

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definite integral and compute
the antiderivative and then

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use the fundamental theorem
of calculus.

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So, let's try that way first and
then we'll do it a second

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way as well.

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So, method one is to compute
the antiderivative.

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So we compute the antiderivative
of x squared

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cosine of x cubed over 8 dx.

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So here, it's not a definite
integral anymore.

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Now I'm just looking at
the antiderivative.

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And so, OK, so either by
intelligent guessing or that--

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I forget if that's what
Professor Jerison called it--

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or by just a substitution
method.

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So, a good thing to substitute
here, is we see inside this

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function we have this x cubed
part and then outside we have

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an x squared part.

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So x squared dx, that's
closely related to the

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differential of x cubed.

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So we might try the
substitution.

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We might try the substitution
u equals--

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well we can maybe get rid of all
of that at once-- so make

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u equals x cubed over 8.

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So in that case, du is equal
to 3/8 x squared dx.

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And so this integral is equal
to the integral of--

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well so, x squared
dx, we have that.

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So we need to multiply
and divide by 3/8.

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So this is, 8/3 cosine
of u du.

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All right, and so now,
OK, so the 8/3,

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that's just a constant.

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That pulls out in front.

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That's not hard.

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So then the antiderivative
of cosine u du, that's

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something we know.

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So this is 8/3 of minus sine--

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sorry not minus; I was doing the
derivative instead of the

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antiderivative.

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Just 8/3 sine u.

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OK.

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And good.

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So that's the anti plus the
constant, which we don't need

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because we're going to plug
this in the fundamental

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theorem of calculus
in a minute.

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So it's 8/3 sine u.

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And, OK, so, but that's
not good.

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I need it back in
terms of my x's.

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And so I have to back
substitute.

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So this is 8/3 sine
of x cubed over 8.

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All right, now I've got an
antiderivative of this

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expression.

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So now I can take this
antiderivative and apply

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fundamental theorem of
calculus, Right?

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So, the integral from minus 2 to
2 of x squared cosine of x

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cubed over 8 dx is equal to--

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well it's equal to this
antiderivative--

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8/3 sine x cubed over 8 taken
between x equals minus 2

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and x equals 2.

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So this is equal to 8/3 times
sine of 1 minus 8/3 times sine

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of minus 1.

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OK, and sine of minus x is
equal to minus sine x.

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So we could rewrite this as
16/3 times sine of 1.

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So that's the answer.

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That's the value of
this integral.

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And we did this first method
by first computing the

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antiderivative and then using
it with the fundamental

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theorem of calculus.

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The second way we can do it is
the one that we just learned

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more recently in lecture.

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Which is to use substitution
directly in

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the definite integral.

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So let me come over
here and do that.

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So, method two.

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So now, when we do it
this way, mostly

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this looks the same.

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So we start off, we don't go
back to antiderivative.

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We start off with the
definite integral.

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So that's one difference.

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We can use the exact
same substitution.

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So we can use exactly the
substitution that we used

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before, which was u equals
x cubed over 8.

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That doesn't change.

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And we still have the same du is
equal to 3x squared over 8,

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but what does change--
sorry, dx.

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Good.

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You need a differential always
to be equal to a differential.

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So what does change
is that this--

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I need to change the bounds.

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Right?

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If I'm keeping it as definite
integrals all the time, so I'm

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changing here from an integral
with respect to x to an

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integral with respect to u.

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And so that means that when I
write down the second integral

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with respect to u, the bounds
that I write down have to be

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the bounds for u.

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Not for x.

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So when x is equal to minus
2, I need to know what the

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corresponding value of u is.

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So in this case, when x is equal
to minus 2, we see that

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the lower bound here--

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when x is equal to minus 2--

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the lower bound on u becomes
u equals, well, minus

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2 cubed over 8.

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That's minus 1.

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And when x is equal to 2, the
upper bound, u becomes

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2 cubed over 8.

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So u becomes 1.

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So we have this extra step when
we do it this way, of

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changing the bounds.

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OK, so I do that.

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So now I get the integral.

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So I'm going to write u equals
minus 1 here just to remind

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myself that I made
this change to 1.

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OK, and now the inside
transforms exactly the same.

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So it's 8/3 times cosine u du.

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OK, and now again, this is
something for which I already

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know the antiderivative.

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So I apply the fundamental
theorem of calculus here.

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So I get the integral is equal
to 8/3 sine of u, where u goes

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between minus 1 and 1.

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So the other thing that's
changed here is that now I

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don't have to change this
back in terms of x's.

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I have the values that I'm going
to plug in here already

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in terms of u.

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So I don't have to switch
back in terms of x's.

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And so I just plug these
values in and take the

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difference.

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So this is equal to 8/3 sine of
1 minus 8/3 sine of minus

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1, which as we said before
is 16/3 times sine of 1.

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OK, so we have these two
different methods.

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They're very, very, very,
similar in how you apply them.

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The key differences are, that
when you do it, the

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substitution in terms of
definite integrals, you have

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to change your bounds
of integration.

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And what you get for changing
your bounds of integration is

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at the end you don't have to
switch back from u's into x's.

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Whereas, if you do it by first
computing the antiderivative,

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well, you don't have any bounds
of integration when you

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compute the antiderivative so
you don't have to change them.

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But once you computed the
antiderivative, then you have

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to go and make your
back substitution.

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Put everything back in terms
of x's before going on to

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compute the definite integral
that you started with.

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So the those are the two
different ways you can do

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substitution in definite
integrals and

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I'll end with that.

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