Iterated Integrals

Uploaded by TheIntegralCALC on 06.05.2011

Hi, everyone! Welcome back to Today we’re going to be doing an iterated
integral problem. And the one that we’re going to be working with is the function f(xy)
= 2xy – 3y^2 and we’ve been asked to find the integrated integral over the region (-1,1)
by (-2,2) and these coordinates here define a rectangular region.
When we’re talking about integrated integrals, we are finding the area of a rectangular region
defined by the coordinates here in r. The way that we set-up our iterated integral are
the following. It doesn’t really matter whether we integrate with respect to x first
or y first. We can do it either way but we usually like to do x first unless doing so
is significantly more difficult than integrating with respect to y first instead. So when we’re
looking at r here, keep in mind that r is basically (x)(y). These first set of coordinates
are always your x coordinates and y here with your y coordinates. So what you want to do
is set up your integral like this. We’ll do the first one and we’re kind of working
from the outside in. We’re working backwards. We want to do the coordinates here for y first.
We’re going to do from -2 to 2, then we do our inter-integral with our x coordinates
here, -1 and 1. Then we put our function inside, 2xy – 3y^2. And then because we have our
x coordinates there, dx goes first and then dy. We’re making an integral sandwich and
the information about y is the bread of our sandwich and the information about x is the
lettuce or something. But here, this integral corresponds to dy and this integral corresponds
tp dx. So we’re only going to be looking at this part first. We’re going to finish
this part and then when we’re done, we’ll do the outer integral. You always want to
work from the inside out. Working from the inside out, we’re going
to be looking at this inner integral right here and we’re going to be integrating with
respect to dx first, which means we’re going to be holding y as a constant. When we do
that, we treat y as a constant just like it were a constant number like 2 or 3 or something
like that. Let’s go ahead and integrate term by term. remember that we’re just going
to leave this outer integral alone for now and we’ll integrate term by term. So the
integral of 2xy with respect to x, which means we’re treating x as the variable and holding
the y as the constant. Basically, we’ve got 2y right here as a coefficient on our
x term. This 2y, you could put, like, 3 in place of that. It’s the coefficient on the
x term because y is a constant. So in order to integrate this, we’ll add 1 to the exponent
to get a 2 here and then divide by 2. That’s how we integrate. We’ll end up with 2/2
x^2y. We just moved the y to the end again because we always want to have our variables
in order, x and then y. To integrate -3y^2, we realize that x is the variable we’re
dealing with, we’re integrating with respect to x and we’re holding y constant. y is
again a number like 2 or 3 which means that this whole term here, this -3y^2 is a constant.
It’s as if therefore saying -4 instead of saying -3y^2. That whole thing is a constant.
There’s no x variable involved. If we had -4, the integral of that will be -4x. We will
just tack on the x variable so similarly in this case, we’re just going to add an x
variable to this term and we’ll get -3xy^2. We stick that x in there and then we’re
going to evaluate this on the range here, 1 to -1 because that’s what this integral
tells us here and then we’ve got this dy out in the end.
Before we evaluate on this range, let’s go ahead and simplify. Obviously, 2/2 cancels
so we’ve just got (x^2)y – 3xy^2 on the range -1 to 1 and then dy out here. Now, evaluating,
we’ve got -2 to 2, we’ll plug in our top number first 1 for x, remember because we’re
taking the integral with respect to x here so we’ve got (1^2)y – 3(1)y^2 and then
from that, we will subtract what we get when we plug in the bottom number, -1. We’ll
plug in -1 for x. (-1^2)y – 3(-1)y^2. And then we’ll put dy out here because that’s
going to correspond to our second integral. We will now simplify what we’ve got inside
here. We’ve got -2 to 2 and then this 1^2 is going to go away. It’s redundant so we’re
going to say y and looks like we’ve got -3y^2 and then here, this -1^2 is just going
to become 1 so that will be redundant and notice that we’ve just got –y then. This
end term here, we’ve got a -3 and -1 so this is actually going to be a plus 3y^2 because
we’ve got this minus sign in the middle, it’s going to end up being a minus 3y^2
and then dy. Now, one more simplification step before we move on to our y integral.
We’ve got y – y so that’s going to go away. -3y^2 – 3y^2 is a -6y^2 dy. And we
finally simplified down to the simplest form of the integral with respect to x.
That part’s now done and now we can integrate with respect to y. Doing so, we’ll add 1
to the exponent to get 3, which means we’re going to be dividing -6 by 3. This is going
to end up being -2y^3 and we’ll be evaluating that on the range -2 to 2. Plugging these in here we’ll get -2(2^3) minus what
we get when we plug in -2 so -2(-2^3) and if we simplify this, we’ll get -2(8) which
will be a -16 - -2(-8) = -16 – 16. And of course, -16 – 16 = -32.
So that’s it. That’s our final answer. We can just provide the answer -32. I hope
that video helped you guys and I will see you in the next one. Bye!