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DAVID JORDAN: Hello, and
welcome back to recitation.

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So the problem I'd like
to work with you today

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is this one here.

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It's just to compute this
two-variable integral,

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and the integrand that
we're going to be computing

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is e to the u over u.

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And what you might
notice right away

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is that this inner integral,
it's an integral over u of e

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to the u over u, and
this is not an integral

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that we have a nice formula
for from one-variable calculus.

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So I'm going to suggest that,
as you try to solve this,

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you think about how can
you use the fact that this

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is a multivariable
integral, maybe

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swapping the order
of integration,

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et cetera, to solve this.

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So why don't you go ahead and
work this problem on your own.

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Check back with me
in a few minutes,

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and I'll see how I did it.

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OK, welcome back.

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So as I suggested,
I think what we

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should do is we should see
what happens if we switch

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the order of integration.

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I don't know how to do
this inside integral,

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and so maybe if we switch
the order of integration,

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then something's
going to work out.

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So in order to get
started doing that, we

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need to draw the
region of integration,

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so why don't we
do that over here.

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So I'll just walk over here.

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So we've got-- our variables
are t and u, so I've

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drawn the t- and u-axes here.

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And now, let's look at
the region of integration.

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So t is running from 0 to 1/4.

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So we'll just draw
1/4 about there.

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And now the range for
u, the bottom range

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is the square root
of t, so I'm going

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to draw the curve u is
the square root of t.

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It just looks like a
parabola on its side.

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And then the top bound
is at u equals 1/2.

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And notice that
when t is 1/4, that

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means that u is 1/2, because u
is just the square root of t.

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And so what we're
really interested in

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is this region here,
the region between u

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is the square root of
t and between u is 1/2,

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so this is our region.

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So let's rewrite the
integral by swapping

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the order of integration,
so I'll do that here.

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So now on the outside, we want
to put the range of u first.

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So the range of u, we can
see on the graph here,

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u ranges from 0 to 1/2, so
that's going to be easy.

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And now t, so t is always
starting right here at t

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equals 0, and it's always
ending at this curve, which

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is t equals u squared.

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So we have these
little integrals here.

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And so our ranges for
t is going to be t

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is running from 0 to u squared.

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Then we have the same
integrand e to the u over u,

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and now we have dt du.

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All right.

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Now we see that this
was a nice thing to do,

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because look: The first
integral that we need to take

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is an integral in
t, but our integrand

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doesn't involve the
variable t, so this

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is going to be a very
easy integral to take.

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So I just take that integrand
and I just multiply it

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by the constant t, so we just
have e to the u over u times t,

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and then it's a
definite integral which

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ranges from u squared to 0, du.

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And so this is just
going to be-- 1/2

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here-- this is really just
going to be u e to the u du,

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all right?

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So let me write that
up over here again.

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So we're at integral from
u equals 0 to 1/2 of u

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e to the u du.

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And now we want to remember
the method of integration

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by parts from
one-variable calculus.

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So integration by
parts, you'll remember,

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will tell us that the
integral of u e to the u

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is going to be u e to
the u minus e to the u.

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So that's just applying
integration by parts.

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And then this is a
definite integral,

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so we have a range 1/2 to 0.

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Well, now, we can just plug
this in, so we get 1/2 e

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to the 1/2 minus e to the
1/2 minus the quantity--

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so we just get 0
minus e to the 0.

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And so altogether,
we have-- let's see,

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we have a negative e to the
1/2 and then we have a plus 1.

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And negative e to
the 1/2 over 2,

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because we had 1/2 and a minus
a whole, and then plus 1.

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And that's our solution.