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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

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today is this one here.

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

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integrand that we're going
to be computing is e

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to the u over u.

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And what you might notice right
away is that this inner

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integral, it's an integral over
u of e to the u over u,

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and this is not an integral that
we have a nice formula

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for from one-variable
calculus.

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

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think about how can you use
the fact that this is a

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multivariable integral, maybe
swapping the order of

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integration, 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, and I'll

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see how I did it.

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OK, welcome back.

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So as I suggested, I think what
we should do is we should

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see what happens if we switch
the order of integration.

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

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if we switch the order of
integration, then something's

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going to work out.

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So in order to get started doing
that, we need to draw

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the region of integration,
so why don't we

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do that over here.

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So I'll just walk over here.

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So we've got--

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our variables are t and
u, so I've drawn the

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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 is the square

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root of t, so I'm going to
draw the curve u is the

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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 means that u is 1/2,

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because u is just the
square root of t.

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And so what we're really
interested in is this region

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here, the region between u is
the square root of t and

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between u is 1/2, so
this is our region.

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So let's rewrite the integral
by swapping the order of

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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, u ranges

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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 is

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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 is running from

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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, because

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look: The first integral that we
need to take is an integral

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in t, but our integrand doesn't
involve the variable

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t, so this 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 by the

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

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then it's a definite integral
which ranges from u

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squared to 0 du.

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And so this is just
going to be--

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1/2 here--

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this is really just going to be
ue to the u du, 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

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of ue to the u du.

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And now we want to remember the
method of integration by

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parts from one-variable
calculus.

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So integration by parts, you'll
remember, will tell us

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that the integral of ue to the
u is going to be ue to the u

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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, so we have a

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range 1/2 to 0.

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

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1/2 minus e to the 1/2 minus the
quantity, so we just get 0

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minus e to the 0.

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And so altogether, we have--

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let's see, we have a negative
e to the 1/2 and then

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we have a plus 1.

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And negative e to the 1/2 over
2, because we had 1/2 and a

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minus a whole, and
then plus 1.

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And that's our solution.

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