DAVID JORDAN: Hello, and
welcome back to recitation.
So the problem I'd like
to work with you today
is this one here.
It's just to compute this
two-variable integral,
and the integrand that
we're going to be computing
is e to the u over u.
And what you might
notice right away
is that this inner integral,
it's an integral over u of e
to the u over u, and
this is not an integral
that we have a nice formula
for from one-variable calculus.
So I'm going to suggest that,
as you try to solve this,
you think about how can
you use the fact that this
is a multivariable
integral, maybe
swapping the order
of integration,
et cetera, to solve this.
So why don't you go ahead and
work this problem on your own.
Check back with me
in a few minutes,
and I'll see how I did it.
OK, welcome back.
So as I suggested,
I think what we
should do is we should see
what happens if we switch
the order of integration.
I don't know how to do
this inside integral,
and so maybe if we switch
the order of integration,
then something's
going to work out.
So in order to get
started doing that, we
need to draw the
region of integration,
so why don't we
do that over here.
So I'll just walk over here.
So we've got-- our variables
are t and u, so I've
drawn the t- and u-axes here.
And now, let's look at
the region of integration.
So t is running from 0 to 1/4.
So we'll just draw
1/4 about there.
And now the range for
u, the bottom range
is the square root
of t, so I'm going
to draw the curve u is
the square root of t.
It just looks like a
parabola on its side.
And then the top bound
is at u equals 1/2.
And notice that
when t is 1/4, that
means that u is 1/2, because u
is just the square root of t.
And so what we're
really interested in
is this region here,
the region between u
is the square root of
t and between u is 1/2,
so this is our region.
So let's rewrite the
integral by swapping
the order of integration,
so I'll do that here.
So now on the outside, we want
to put the range of u first.
So the range of u, we can
see on the graph here,
u ranges from 0 to 1/2, so
that's going to be easy.
And now t, so t is always
starting right here at t
equals 0, and it's always
ending at this curve, which
is t equals u squared.
So we have these
little integrals here.
And so our ranges for
t is going to be t
is running from 0 to u squared.
Then we have the same
integrand e to the u over u,
and now we have dt du.
All right.
Now we see that this
was a nice thing to do,
because look: The first
integral that we need to take
is an integral in
t, but our integrand
doesn't involve the
variable t, so this
is going to be a very
easy integral to take.
So I just take that integrand
and I just multiply it
by the constant t, so we just
have e to the u over u times t,
and then it's a
definite integral which
ranges from u squared to 0, du.
And so this is just
going to be-- 1/2
here-- this is really just
going to be u e to the u du,
all right?
So let me write that
up over here again.
So we're at integral from
u equals 0 to 1/2 of u
e to the u du.
And now we want to remember
the method of integration
by parts from
one-variable calculus.
So integration by
parts, you'll remember,
will tell us that the
integral of u e to the u
is going to be u e to
the u minus e to the u.
So that's just applying
integration by parts.
And then this is a
definite integral,
so we have a range 1/2 to 0.
Well, now, we can just plug
this in, so we get 1/2 e
to the 1/2 minus e to the
1/2 minus the quantity--
so we just get 0
minus e to the 0.
And so altogether,
we have-- let's see,
we have a negative e to the
1/2 and then we have a plus 1.
And negative e to
the 1/2 over 2,
because we had 1/2 and a minus
a whole, and then plus 1.
And that's our solution.
