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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like us to
work on the following problem:

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So given the region
R shown below,

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which I'll mention
more of in a moment,

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compute the integral
double integral

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over R of the
quantity 4 x squared

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minus y squared to
the fourth dx dy,

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and I want you to do it
by changing variables to u

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is equal 2x minus y and
v is equal to 2x plus y.

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And the region,
defined down here,

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is just these
three line segments

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and the intersection
of the region bounded

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by these three line segments.

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So one of them is a portion
of the y-axis, one of the them

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is a portion of the line
defined by 2x minus y equals 0,

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and one of them is a
portion of the line defined

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by 2x plus y equals 2.

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So why don't you do
this, make sure you

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use this change of variables,
and then I'll come back,

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and I'll draw the accompanying
region in the u, v plane,

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and then we'll see
how I set it up.

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OK, welcome back.

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Well, the first
thing I'm going to do

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is I'm going to draw the
region in the u, v plane that

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is determined by this region in
the x, y plane and the change

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of variables.

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Now, as was mentioned in the
lecture, all of the changes

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are linear, and so I'm going
to be taking lines to lines.

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So what I'm going to do is
determine at each endpoint

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here where that endpoint goes
under the transformation,

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and then I will
connect the dots.

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So what do I get?

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Well, this first one is (0,
0) for x and y, and so that

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corresponds to x and y
are 0 here, so u is 0,

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and x and y are 0 here,
so it would be a 0,

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so that's the point (0,
0) on the u, v plane.

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Now, I want to
mention something here

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is that based on how
these lines were defined,

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this is telling you that all
along this line 2x minus y

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equals 0, that means u is equal
to 0 all along that segment.

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And if you'll notice also,
2x plus y is equal to 2

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here, so v is equal to 2
all along this segment.

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So we should expect whatever
u is, v's going to be 2 here,

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and whatever v is here,
u is going to be 0.

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OK, so I'm going to have
something with u, 0 initially,

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and then v, 2, I
can already expect,

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it's going to come up here
and move over that way.

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But let's just check.

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So this is the point (0,
2), 0 for x and 2 for y.

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Let's see what it is
in the u, v variables.

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So u in that case is negative
2 and v in that case is 2,

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and so I'm going to
go left 2 and up 2.

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This might not be drawn to
scale, but this is negative 2,

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and this is up 2
here, and so that's

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where this point goes
in the transformation.

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This point is 1/2 comma 1.

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You can actually
check what it is.

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But actually, as I
mentioned, you already

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know what has to happen, because
this segment is connected right

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here, and then we said all
along this line, v is 2,

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and all along this line, u is 0,
and so that actually carves out

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that rectangle.

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So if you were worried and
you weren't able to do it

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that way, what you
could actually do

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is say, well, where does
the point (1/2, 1) go?

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If I plug in 1/2
for x and 1 for y,

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I notice that I get u
equals 0 and v equals 2,

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and that's this point.

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And so now, I'm
looking at this region.

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Now, these two things
are not drawn to scale,

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because what I want to point
out is they're both triangles.

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It's easy to find the area
based on how they're sitting

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in two-dimensional space.

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So what we see
here is if I wanted

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to find the area
of this triangle,

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let's make this the base.

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The base is 2 and
the height is 1/2,

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so 1/2 of base times height
is-- that area is equal to 1/2.

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So the area of R is 1/2, and
what's the area of this one?

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The area this one is-- well,
I've got base 2 and height 2,

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and so it's 1/2 of 2 times 2,
and so that gives me area as 2.

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So here the area is
1/2, here the area is 2,

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and so notice that
I've multiplied

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by 4 to get from here to here.

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So I can anticipate
that I should

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have-- based on the principle
you saw in lecture, I

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should have something
like du dv is going

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to be equal to 4 times dx dy.

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That is what I expect to get.

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Now let's see if when
we do the Jacobian,

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we get the same thing.

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Let me double check, right?

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This is area 2, this is area
1/2, so I have 1/2 times

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4 is going to give me 2.

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Yeah, that's what I should get.

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Oops!

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That sticks out a little.

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So now let's just
check our Jacobian

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and see if that is
indeed what we get.

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So I'm going to look at u sub
x, u sub y, v sub x, v sub y,

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right?

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So u sub x is 2, u
sub y is negative 1,

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so I get 2, negative 1.

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v sub x is 2, v sub y
is 1, so I get 2, 1.

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2 times 1 is 2, minus negative
2 gives me, indeed, 4.

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So I do get, in fact, du
dv is equal to 4 dx dy.

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So I know that I'm going to
have to-- because I'm changing

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variables, though,
from dx dy to du dv,

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I'm going to divide
by 4, obviously,

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when I do the substitution.

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The substitution will
be replacing just dx dy.

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So I just mention
that, but notice:

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Again, we get what
we expect to get.

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We got a 4 based on the
picture and we got a 4

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based on the Jacobian.

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And so now, we need to
finish up the process.

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We need to figure out how to
write this in terms of u and v

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and then figure out our
bounds in terms of u and v,

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and then we're done.

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So let me mention, one
thing you should notice

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is that u times v is
equal to exactly-- this

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is going to be a
difference of two squares,

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and it is precisely 4 x
squared minus y squared.

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If you need to multiply it out
to check, you can check it,

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but that's indeed what it is,
which is why this particular

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substitution is quite nice,
because that means this

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function I'm supposed to be
integrating is just u*v raised

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to the fourth.

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So I'm almost done.

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So let me write in the
pieces I know, and then we'll

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fill in the last two spots,
or four spots, which will be

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all the bounds on the integral.

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So I'm going to write
it here, give myself

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some space to write the bounds.

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So I'm replacing the 4 x squared
minus y squared by a u*v,

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so I'm going to get u times
v, and then the function,

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that is raised to the fourth
in the initial problem,

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so I raise that to the fourth.

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Then dx dy, as I
mentioned, will be replaced

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by a du dv divided by 4.

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And so I can just put this
over 4 and write du dv.

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I should be careful
which order I want to do.

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It doesn't really matter.

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I can do either one.

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du dv, OK?

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And so now I want to
know what u goes from

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and to and then what v
has to go from and to.

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So if I come over
here, you'll see

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it didn't matter,
because I could have

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picked either direction to go.

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But if I'm going
to go with u, I'm

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coming from whatever this
function is over to here,

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right?

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And this value here is easy.

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That's u equals 0.

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So the top bound for u is 0,
and the bottom bound for u,

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this is v is equal to minus
u, so the bottom bound for u

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is when u is equal
to negative v.

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So I'm running from-- because
I put the du on the inside,

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my first one is
running from minus v

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to-- what did I say-- 0, and
then the v-values from there

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go between 0 and 2.

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You can see this easily
from the picture.

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So they go between 0 and 2.

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And I am not going to finish it
off from here, because it would

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require not too much work.

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It's actually quite
simple, because it's just

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a polynomial and u and
a polynomial and v,

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but there are a lot
of powers and there

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will be big powers of 2.

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So suffice it to
say, at this point,

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we can evaluate this
integral and it's quite easy

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to evaluate.

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And the main point I
think we should make

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is how did it make it simpler?

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I mean, the initial problem,
if we come over here,

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this is annoying to
find an anti-derivative,

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but not impossible.

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But the really
annoying part is I

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would have to take this
region and split it

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into a bunch of pieces,
or at least two pieces,

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to evaluate the integral
in a reasonable way

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or my y-values would go from
this line here up to this line

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here.

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It could be complicated.

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So what we've really done is
we've simplified the region.

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That's the easiest thing.

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I mean, we also
simplified the function,

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but we've really simplified the
region we're integrating over.

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And so we only have-- one is a
function and one is a constant,

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and that's quite nice
to have on the inside.

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So I think that's
where I'll stop.