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In all of the double integrals we've done so

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far, the boundaries on x and y were fixed.

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Now we'll see what happens when the boundaries on

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x and y are variables.

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So let's say I have the same surface, and I'm not going to

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draw it the way it looks, I'll just kind of draw

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it figuratively.

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But the problem we're actually going to do is z, and this is

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the exact same one we've been doing all along.

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The point of here isn't to show you how to integrate, the point

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of here is to show you how to visualize and think

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about these problems.

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And frankly, in double integral problems the hardest part is

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figuring out the boundaries.

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Once you do that, the integration is pretty

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straightforward.

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It's really not any harder then single variable integration.

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So let's say that's our surface: z is equal

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to xy squared.

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Let me draw the axes again.

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So that's my x-axis.

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That's my z-axis.

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That's my y-axis.

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x, y, and z.

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And you saw what this graph looked like several videos ago.

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I took out the whole grapher and we rotated and things.

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I'm not going to draw the graph the way it looks; I'm just

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going to brought fairly abstractly as just an

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abstract surface.

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Because the point here it's really to figure out the

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boundaries of integration.

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Before I actually even draw the surface, I'm going

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to draw the boundary.

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The first time we did this problem we said, OK, x goes

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from 0 to 2, y goes from 0 to 1, and then we figured out the

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volume above that bounded domain.

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Now let's do something else.

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Let's say that x goes from 0 to 1.

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And let's say that the volume that we want to figure out

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under the surface, it's not from a fixed y to

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an upper-bound y.

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I'll show you: it's actually a curve.

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So this is all on the xy plane, everything I'm drawing here.

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And this curve, we could view it two ways: we could say y is

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a function of x, y is equal to x squared.

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Or we could write is equal to square root of y.

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We don't have to write plus or minus or anything like that

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because we're in the first quadrant.

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So this is the area above which we want to

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figure out the volume.

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Let me, yeah, it doesn't hurt to color it in just so we

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can really hone in on what we care about.

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So that's the area above which we want to

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figure out the volume.

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You could kind of say, that's our bounded domain.

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And so x goes from 0 to 1, and then this point

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is going to be what?

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That point's going to be 1 comma 1, right?

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1 is equal to 1 squared, 1 is equal to the square root of 1.

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So this point is y is equal to 1.

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And then I'm not going to draw this surface exactly.

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I'm just trying to give you a sense of what the volume of

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the figure we're trying to calculate is.

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If this is just some arbitrary surface-- let me do it in a

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different color --so this is the top.

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This line is going vertical in the z-direction.

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Actually, I could draw it like this, like it's a curve.

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And then this curve back here is going to be like a wall.

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And maybe I'll paint this side of the wall just so you can see

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what it kind of looks like.

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Trying my best.

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Think you get an idea.

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Let me make it a little darker; this is actually more of an

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exercise in art than in math, in many ways.

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You get the idea.

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And then the boundary here is like this.

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And this top isn't flat, you know, it could

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be curved surface.

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I do a little like that, but it's a curved surface.

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And we know in the example we're about to do that the

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surface right here is z is equal to x squared.

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So we want to figure out the volume under this.

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So how do we do it?

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Well, let's think about it.

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We could actually use the intuition that I just gave you.

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We're essentially just going to take a da, which is a little

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small square down here, and that little area, that's the

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same thing as the dx-- let me use a darker color --as a dx

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times a dy, and then we just have to multiply it times f of

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xy, which is this, for each area, and then

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some them all up.

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And then we could take a sum in the x-direction first

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or the y-direction first.

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Now before doing that, just to make sure that you have

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the intuition because the boundaries are the hard part,

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let me just draw our xy plane.

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So let me rotate it up like that.

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I'm just going to draw our xy plane.

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Because that's what matters.

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Because the hard part here is just figuring out our

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bounds of integration.

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So the curve is just y is equal to x squared, look

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something like that.

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This is the point y is equal to 1.

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This is y-axis, this is the x-axis, this is the

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point x is equal to 1.

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That's not an x, that's a 1.

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This is the x.

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Anyway, so we want to figure out, how do we sum up this dx

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times dy, or this da, along this domain?

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So let's draw it.

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Let's visually draw it and it doesn't hurt to do this when

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you actually have to do the problem because this

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frankly is the hard part.

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A lot of calculus teachers will just have you set up the

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integral and then say, OK, well the rest is easy.

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Or the rest is Calc 1.

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OK, so this area, this area here is the same thing

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as this area here.

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So its base is dx and its height is dy.

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And then you could imagine that we're looking at

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this thing from above.

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So the surface is up here some place and we're looking

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straight down on it, and so this is just this area.

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So let's say we wanted to take the integral with

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respect to x first.

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So we want to sum up, so if we want the volume above this

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column, first of all, is this area times dx, dy, right?

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So let's write the volume above that column.

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It's going to be the value of the function, the height at

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that point, which is xy squared times dx, dy.

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This expression gives us the volume above this area, or

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this column right here.

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And let's say we want the sum in the x direction first.

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So we want to sum that dx, sum one here, sum here,

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et cetera, et cetera.

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So we're going to sum in the x-direction.

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So my question to you is, what is our lower

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bound of integration?

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Well, we're kind of holding our y constant, right?

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And so if we go to the left, if we go lower and lower x's we

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kind of bump into the curve here.

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So the lower bound of integration is

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actually the curve.

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And what is this curve if we were to write x

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is a function of y?

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This curve is y is equal to x squared, or x is equal

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to the square root of y.

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So if we're integrating with respect to x for a fixed y

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right here-- we're integrating in the horizontal direction

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first --our lower bound is x is equal to the square root of y.

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That's interesting.

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I think it's the first time you've probably seen a

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variable bound integral.

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But it makes sense because for this row that we're adding up

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right here, the upper bound is easy.

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The upper bound is x is equal to 1.

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The upper bound is x is equal to 1, but the lower bound is

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x is equal to the square root of y.

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Because you go back like, oh, I bump into the curve.

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And what's the curve?

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Well the curve is x is equal to the square root of y because we

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don't know which y we picked.

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Fair enough.

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So once we've figured out the volume-- so that'll give us the

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volume above this rectangle right here --and then we

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want to add up the dy's.

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And remember, there's a whole volume above what

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I'm drawing right here.

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I'm just drawing this part in the xy plane.

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So what we've done just now, this expression, as it's

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written right now, figures out the volume above

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that rectangle.

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Now if we want to figure out the entire volume of the solid,

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we integrate along the y-axis.

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Or we add up all the dy's.

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This was a dy right here, not a dx.

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My dx's and dy's look too similar.

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So now what is the lower bound on the y-axis if I'm summing

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up these rectangles?

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Well, the lower bound is y is equal to 0.

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So we're going to go from y is equal to 0 to what--

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what is the upper bound?

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--to y is equal to 1.

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And there you have it.

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Let me rewrite that integral.

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So the double integral is going to be from x is equal to square

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root of y to x is equal to 1, xy squared, dx, dy.

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And then the y bound, y goes from 0 to y to 1.

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I've just realized I've run out of time.

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In the next video we'll evaluate this, and then we'll

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do it in the other order.

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See you soon.

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