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You hopefully have a little
intuition now on what a double

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integral is or how we go about
figuring out the volume

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00:00:06,920 --> 00:00:07,490
under a surface.

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So let's actually compute it
and I think it'll all become

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a lot more concrete.

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So let's say I have the
surface, z, and it's a

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function of x and y.

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And it equals xy squared.

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It's a surface in
three-dimensional space.

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And I want to know the
volume between this

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surface and the xy-plane.

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And the domain in the xy-plane
that I care about is x is

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greater than or equal to 0,
and less than or equal to 2.

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And y is greater than or
equal to 0, and less

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than or equal to 1.

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Let's see what that looks
like just so we have a

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good visualization of it.

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So I graphed it here and
we can rotate it around.

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This is z equals xy squared.

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This is the bounding box,
right? x goes from 0 to

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2; y goes from 0 to 1.

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We literally want this-- you
could almost view it the

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volume-- well, not almost.

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Exactly view it as the
volume under this surface.

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Between this surface, the top
surface, and the xy-plane.

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And I'll rotate it around so
you can get a little bit better

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sense of the actual volume.

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Let me rotate.

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Now I should use the
mouse for this.

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So it's this space,
underneath here.

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It's like a makeshift
shelter or something.

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I could rotate it a little bit.

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Whatever's under this,
between the two surfaces--

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that's the volume.

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Whoops, I've flipped it.

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There you go.

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So that's the volume
that we care about.

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Let's figure out how to do and
we'll try to gather a little

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bit of the intuition
as we go along.

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So I'm going to draw a not as
impressive version of that

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graph, but I think it'll
do the job for now.

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Let me draw my axes.

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That's my x-axis, that's my
y-axis, and that's my z-axis.

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x, y, z.

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x is going from 0 to 2.

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Let's say that's 2.

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y is going from 0 to 1.

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So we're taking the volume
above this rectangle

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in the xy-plane.

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And then the surface, I'm going
to try my best to draw it.

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I'll draw it in a
different color.

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I'm looking at the picture.

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At this end it looks
something like this.

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And then it has a
straight line.

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Let's see if I can draw this
surface going down like that.

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And then if I was really
good I could shade it.

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It looks something like this.

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If I were to shade it,
the surface looks

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something like that.

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And this right here
is above this.

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This is the bottom left corner,
you can almost view it.

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So let me just say the yellow
is the top of the surface.

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That's the top of the surface.

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And then this is
under the surface.

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So we care about this
volume underneath here.

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Let me show you the
actual surface.

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So this volume underneath here.

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I think you get the idea.

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So how do we do that?

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Well, in the last example we
said, well, let's pick an

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arbitrary y and for that
y, let's figure out the

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area under the curve.

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So if we fix some y-- when you
actually do the problem, you

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don't have to think about it in
this much detail, but I want

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to give you the intuition.

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So if we pick just an
arbitrary y here.

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So on that y, we could think of
it-- if we have a fixed y, then

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the function of x and y you can
almost view it as a function

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of just x for that given y.

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And so, we're kind of figuring
out the value of this, of

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the area under this curve.

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You should view this as kind of
an up down curve for a given y.

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So if we know a y we can figure
out then-- for example, if y

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was 5, this function would
become z equals 25x.

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And then that's easy to
figure out the value

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of the curve under.

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So we'll make the value under
the curve as a function of y.

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We'll pretend like
it's just a constant.

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So let's do that.

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So if we have a dx
that's our change in x.

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And then our height of each of
our rectangles is going to be a

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function-- it's going to be z.

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The height is z, which is
a function of x and y.

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So we can take the integral.

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So the area of each of these is
going to be our function, xy

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squared-- I'll do it here
because I'll run out of space.

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xy squared times the
width, which is dx.

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And if we want the area of this
slice for a given y, we just

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integrate along the x-axis.

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We're going to integrate
from x is equal to 0

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to x is equal to 2.

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From x is equal to 0 to 2.

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Fair enough.

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Now, but we just don't want to
figure out the area under the

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curve at one slice, for one
particular y, we want to

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figure out the entire
area of the curve.

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So what we do is
we say, OK, fine.

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The area under the curve, not
the surface-- under this curve

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for a particular y,
is this expression.

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Well, what if I gave it
a little bit of depth?

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If I multiplied this area times
dy then it would give me a

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little bit of depth, right?

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We'd kind of have a
three-dimensional slice of the

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volume that we care about.

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I know it's hard to imagine.

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Let me bring that here.

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So if I had a slice here, we
just figured out the area of a

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slice and then I'm multiplying
it by dy to give it a

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little bit of depth.

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So you multiply it by dy to
give it a little bit of depth,

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and then if we want the entire
volume under the curve we add

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all the dy's together, take the
infinite sum of these

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infinitely small
volumes really now.

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And so we will integrate
from y is equal to 0

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to y is equal to 1.

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I know this graph is a little
hard to understand, but you

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might want to re-watch
the first video.

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I had a slightly easier
to understand surface.

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So now, how do we
evaluate this?

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Well, like we said,
you evaluate from the

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inside and go outward.

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It's taking a partial
derivative in reverse.

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So we're integrating here with
respect to x, so we can treat

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y just like a constant.

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Like it's like the number
5 or something like that.

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So it really doesn't
change the integral.

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So what's the antiderivative
of xy squared?

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Well, the antiderivative of
xy squared-- I want to make

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sure I'm color consistent.

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Well, the antiderivative
of x is x to the 1/2--

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sorry. x squared over 2.

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And then y squared is
just a constant, right?

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And then we don't have to
worry about plus c since

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this is a definite integral.

151
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And we're going to
evaluate that at 2 and 0.

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And then we still have
the outside integral

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with respect to y.

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So once we figure that out
we're going to integrate it

155
00:07:25,190 --> 00:07:29,800
from 0 to 1 with respect to dy.

156
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Now what does this evaluate?

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00:07:31,470 --> 00:07:32,912
We put a 2 in here.

158
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If you put a 2 in there
you get 2 squared over 2.

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160
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That's just 4 over 2.

161
00:07:41,740 --> 00:07:43,565
So it's 2 y squared.

162
00:07:43,565 --> 00:07:47,670


163
00:07:47,670 --> 00:07:51,210
Minus 0 squared over
2 times y squared.

164
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Well, that's just
going to be 0.

165
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So it's minus 0.

166
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I won't write that down because
hopefully that's a little

167
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bit of second nature to you.

168
00:07:56,190 --> 00:07:58,510
We just evaluated this
at the 2 endpoints and

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I'm short for space.

170
00:08:00,660 --> 00:08:03,710
So this evaluated at 2y
squared and now we evaluate

171
00:08:03,710 --> 00:08:05,580
the outside integral.

172
00:08:05,580 --> 00:08:08,910
0, 1 dy.

173
00:08:08,910 --> 00:08:10,230
And this is an important
thing to realize.

174
00:08:10,230 --> 00:08:13,120
When we evaluated this
inside integral, remember

175
00:08:13,120 --> 00:08:13,820
what we were doing?

176
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We were trying to figure out
for a given y, what the

177
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area of this surface was.

178
00:08:19,180 --> 00:08:23,070
Well, not this surface, the
area under the surface

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00:08:23,070 --> 00:08:24,380
for a given y.

180
00:08:24,380 --> 00:08:27,190
For a given y that surface
kind of turns into a curve.

181
00:08:27,190 --> 00:08:30,110
And we tried to figure out
the area under that curve

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00:08:30,110 --> 00:08:33,540
in the traditional sense.

183
00:08:33,540 --> 00:08:36,870
This ended up being
a function of y.

184
00:08:36,870 --> 00:08:40,500
And that makes sense because
depending on which y we pick

185
00:08:40,500 --> 00:08:44,390
we're going to get a
different area here.

186
00:08:44,390 --> 00:08:47,810
Obviously, depending on which y
we pick, the area-- kind of a

187
00:08:47,810 --> 00:08:52,620
wall dropped straight down--
that area's going to change.

188
00:08:52,620 --> 00:08:55,760
So we got a function of y when
we evaluated this and now we

189
00:08:55,760 --> 00:08:58,330
just integrate with respect to
y and this is just plain old

190
00:08:58,330 --> 00:09:00,810
vanilla definite integration.

191
00:09:00,810 --> 00:09:03,350
What's the antiderivative
of 2y squared?

192
00:09:03,350 --> 00:09:08,140
Well, that equals 2 times
y to the third over 3,

193
00:09:08,140 --> 00:09:11,510
or 2/3 y to the third.

194
00:09:11,510 --> 00:09:14,740
We're going to evaluate
that at 1 and 0, which

195
00:09:14,740 --> 00:09:16,100
is equal to-- let's see.

196
00:09:16,100 --> 00:09:17,480
1 to the third times 2/3.

197
00:09:17,480 --> 00:09:18,870
That's 2/3.

198
00:09:18,870 --> 00:09:20,460
Minus 0 to the third times 2/3.

199
00:09:20,460 --> 00:09:21,580
Well, that's just 0.

200
00:09:21,580 --> 00:09:25,270
So it equals 2/3.

201
00:09:25,270 --> 00:09:29,620
If our units were meters
these would be 2/3 meters

202
00:09:29,620 --> 00:09:31,230
cubed or cubic meters.

203
00:09:31,230 --> 00:09:32,280
But there you go.

204
00:09:32,280 --> 00:09:34,890
That's how you evaluate
a double integral.

205
00:09:34,890 --> 00:09:36,450
There really isn't
a new skill here.

206
00:09:36,450 --> 00:09:38,650
You just have to make sure to
keep track of the variables.

207
00:09:38,650 --> 00:09:39,760
Treat them constant.

208
00:09:39,760 --> 00:09:41,620
They need to be treated
constant, and then treat them

209
00:09:41,620 --> 00:09:44,710
as a variable of integration
when it's appropriate.

210
00:09:44,710 --> 00:09:49,090
Anyway, I will see you
in the next video.

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00:09:49,090 --> 00:09:49,900