WEBVTT
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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 of a definite integral.
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This time, it's going
to be one that you're
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going to need to do
some substitution for.
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You're not going to know
the antiderivative right
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off the 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
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of the function x squared
cosine of the quantity
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x cubed over 8 dx.
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So, this is a kind
of a weird integrand.
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Roughly, you know,
very, very rough graph
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of kind of 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 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,
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work this out
yourself, come back, we
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can work it out together.
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All right, welcome back.
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So, from looking at
this integrand here,
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it seems to me that I
don't know immediately
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off the top of my head
what its antiderivative is.
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So, I can't just sort of
apply the fundamental theorem
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of calculus directly without
a little bit of work.
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So there are two
different ways that I
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can go about 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
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that it's a definite integral
and compute the antiderivative
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and then 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 way
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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
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guessing or that--
I forget if that's
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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,
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is we see inside this function
we have this x cubed part
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and then outside we
have an x squared part.
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So x squared dx,
that's closely related
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to the 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-- well we can maybe
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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-- well so,
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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
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u du, that's something we know.
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So this is 8/3 of minus
sine-- sorry not minus;
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I was doing the derivative
instead of the 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
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we don't need
because we're going
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to plug this in the
fundamental theorem of calculus
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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
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of this expression.
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So now I can take
this antiderivative
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and apply fundamental
theorem of calculus, Right?
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So, the integral
from minus 2 to 2
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of x squared cosine of x cubed
over 8 dx is equal to-- well
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it's equal to this
antiderivative-- 8/3 sine x
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cubed over 8 taken between x
equals minus 2 and x equals 2.
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So this is equal to 8/3 times
sine of 1 minus 8/3 times
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sine 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
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the antiderivative
and then using it
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with the fundamental
theorem of calculus.
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The second way we
can do it is the one
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that we just learned
more recently in lecture.
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Which is to use
substitution directly
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in 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 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
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that we used 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 3 x 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-- I need
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to change the bounds.
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Right?
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If I'm keeping it as definite
integrals all the time,
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so I'm changing here from
an integral with respect
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to x to an integral
with respect to u.
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And so that means
that when I write down
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the second integral
with respect to u,
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the bounds that I write down
have to be 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
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to know what the
corresponding value of u is.
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So in this case, when
x is equal to minus 2,
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we see that the
lower bound here--
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when x is equal to minus
2-- the lower bound on u
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becomes u equals, well,
minus 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,
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u becomes 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,
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of 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
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here just to remind 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,
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where u goes 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 difference.
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So this is equal to 8/3 sine
of 1 minus 8/3 sine of minus 1,
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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
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you do it, the substitution in
terms of definite integrals,
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you have 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
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when you compute the
antiderivative so you
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don't have to change them.
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But once you computed
the antiderivative,
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then you have to go and
make your back substitution.
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Put everything back in
terms of x's before going on
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to compute the definite
integral that you started with.
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So the those are the
two different ways
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you can do substitution
in definite integrals
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and I'll end with that.