1
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Let's say we have a path in the xy plane.

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That's my y-axis, that is my x-axis, in my path

4
00:00:09,779 --> 00:00:11,759
will look like this.

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00:00:11,759 --> 00:00:16,280
Let's say it looks like that; trying to draw a bit of an

6
00:00:16,280 --> 00:00:19,480
arbitrary path, and let's say we go in a counter clockwise

7
00:00:19,480 --> 00:00:23,449
direction like that along our path.

8
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And we could call this path-- so we're going in a counter

9
00:00:26,179 --> 00:00:29,179
clockwise direction --we could call that path c.

10
00:00:29,179 --> 00:00:31,730
And let's say we also have a vector field.

11
00:00:31,730 --> 00:00:34,335
And our vector field is going to be a little unusual;

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I'll call it p.

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p of xy.

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It only has an i component, or all of its vectors are only

15
00:00:40,869 --> 00:00:43,239
multiples of the i-unit vector.

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So it's capital P of xy times the unit vector i.

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00:00:48,409 --> 00:00:51,919
There is no j component, so if you have to visualize this

18
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vector field, all of the vectors, they're all multiples

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00:00:57,880 --> 00:00:59,980
of the i-unit vector.

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Or they could be negative multiples, so they could

21
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also go in that direction.

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But they don't go diagonal or they don't go up.

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They all go left to right or right to left.

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That's what this vector field would look like.

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Now what I'm interested in doing is figuring out the line

26
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integral over a closed loop-- the closed loop c, or the

27
00:01:20,170 --> 00:01:27,710
closed path c --of p dot dr, which is just our standard

28
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kind of way of solving for a line integral.

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And we've seen what dr is in the past.

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dr is equal to dx times i plus dy times the j-unit vector.

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And you might say, isn't it dx, dt times dt?

32
00:01:45,250 --> 00:01:51,530
Let me write that: can't dr be written as dx, dt times

33
00:01:51,530 --> 00:01:56,780
dti plus dy, dt times dtj?

34
00:01:56,780 --> 00:01:59,280
And it could, but if you imagine these differentials

35
00:01:59,280 --> 00:02:01,820
could cancel out, and you're just left with the dx and

36
00:02:01,819 --> 00:02:03,489
a dy, and we've seen that multiple times.

37
00:02:03,489 --> 00:02:06,359
And I'm going to leave it in this form because hopefully, if

38
00:02:06,359 --> 00:02:09,389
we're careful, we won't have to deal with the third

39
00:02:09,389 --> 00:02:10,539
parameter, t.

40
00:02:10,539 --> 00:02:13,239
So let's just look at it in this form right here with

41
00:02:13,240 --> 00:02:16,120
just the dx's and the dy's.

42
00:02:16,120 --> 00:02:21,580
So this integral can be rewritten as the line integral,

43
00:02:21,580 --> 00:02:25,040
the curve c-- actually let me do it over down here.

44
00:02:25,039 --> 00:02:32,599
The line integral over the path of the curve c of p dot dr. So

45
00:02:32,599 --> 00:02:35,750
we take the product of each of the coefficients, let's say the

46
00:02:35,750 --> 00:02:39,610
coefficient of the i component, so we get p-- I'll do that in

47
00:02:39,610 --> 00:02:47,450
green, actually do that purple color --so we get p of xy times

48
00:02:47,449 --> 00:02:54,769
dx plus-- well there's no 0 times j times dy; 0 times dy id

49
00:02:54,770 --> 00:02:58,290
just going to be 0 --so this our line integral simplified

50
00:02:58,289 --> 00:02:59,329
to this right here.

51
00:02:59,330 --> 00:03:02,890
This is equal to this original integral up here, so we're

52
00:03:02,889 --> 00:03:06,109
literally just taking the line integral around this path.

53
00:03:06,110 --> 00:03:09,890
Now I said that we play our cards right, we're not going to

54
00:03:09,889 --> 00:03:12,089
have to deal with the third variable, t; that we might be

55
00:03:12,090 --> 00:03:15,229
able just solve this integral only in terms of x.

56
00:03:15,229 --> 00:03:16,699
And so let's see if we can do that.

57
00:03:16,699 --> 00:03:19,639
So let's look at our minimum and maximum x points.

58
00:03:19,639 --> 00:03:22,409
That looks like our minimum x point.

59
00:03:22,409 --> 00:03:24,250
Let's call that a.

60
00:03:24,250 --> 00:03:29,939
Let's call that our maximum x point; let's call that b.

61
00:03:29,939 --> 00:03:33,259
What we could do is we can break up this curve into

62
00:03:33,259 --> 00:03:36,500
two functions of x. y is functions of x.

63
00:03:36,500 --> 00:03:40,750
So this bottom one right here we could call as y1 of x.

64
00:03:40,750 --> 00:03:44,569
This is just a standard curve; you know when we were just

65
00:03:44,569 --> 00:03:47,939
dealing with standard calculus, this is just you can

66
00:03:47,939 --> 00:03:50,569
imagine this is f of x and it's a function of x.

67
00:03:50,569 --> 00:03:51,900
And this is y2 of x.

68
00:03:51,900 --> 00:03:55,159


69
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Just like that.

70
00:03:56,090 --> 00:04:01,469
So you can imagine two paths; one path defined by y1 of x--

71
00:04:01,469 --> 00:04:04,800
let me do that in a different color; magenta --one path

72
00:04:04,800 --> 00:04:09,340
defined by y1 of x as we go from x is equal to a to x is

73
00:04:09,340 --> 00:04:14,110
equal to b, and then another path defined by y2 of x as we

74
00:04:14,110 --> 00:04:19,230
go from x is equal to b to x is equal to a.

75
00:04:19,230 --> 00:04:21,330
That is our curve.

76
00:04:21,329 --> 00:04:24,550
So what we could do is, we could rewrite this integral--

77
00:04:24,550 --> 00:04:29,819
which is the same thing as that integral --as this is equal to

78
00:04:29,819 --> 00:04:33,790
the integral-- we'll first do this first path --of x going

79
00:04:33,790 --> 00:04:40,129
from a to b of p of x.

80
00:04:40,129 --> 00:04:43,670
And I could to say p of x and y, but we know along this

81
00:04:43,670 --> 00:04:47,439
path y is a function of x.

82
00:04:47,439 --> 00:04:51,290
So we say x and y1 of x.

83
00:04:51,290 --> 00:04:53,220
Wherever we see a y we substitute it with

84
00:04:53,220 --> 00:04:56,910
a y1 of x, dx.

85
00:04:56,910 --> 00:05:02,030
So that covers that first path; I'll do it in the same color.

86
00:05:02,029 --> 00:05:04,429
We could imagine this is c1.

87
00:05:04,430 --> 00:05:07,180
This is kind of the first half of our curve-- well it's not

88
00:05:07,180 --> 00:05:09,819
exactly the half --but that takes us right from that

89
00:05:09,819 --> 00:05:10,990
point to that point.

90
00:05:10,990 --> 00:05:13,430
And then we want to complete the circle.

91
00:05:13,430 --> 00:05:16,959
Maybe I'll do that, and I'll do that in yellow.

92
00:05:16,959 --> 00:05:19,389
That's going to be equal to-- sorry we're going to have to

93
00:05:19,389 --> 00:05:25,919
add these two --plus the integral from x is equal to b

94
00:05:25,920 --> 00:05:37,290
to x is equal to a of-- do it in that same color --of p of x.

95
00:05:37,290 --> 00:05:40,230
And now y is going to be y2 of x.

96
00:05:40,230 --> 00:05:42,180
Wherever you see a y, you can substitute with y2

97
00:05:42,180 --> 00:05:44,889
of x along this curve.

98
00:05:44,889 --> 00:05:49,349
y2 of x, dx.

99
00:05:49,350 --> 00:05:51,650
This is already getting interesting and you

100
00:05:51,649 --> 00:05:53,089
might already see where I'm going with this.

101
00:05:53,089 --> 00:05:56,129
So this is the curve c2.

102
00:05:56,129 --> 00:06:00,029
too I think you appreciate if you take the union of c1 and

103
00:06:00,029 --> 00:06:02,439
c2, we've got our whole curve.

104
00:06:02,439 --> 00:06:05,689
So let's see if we can simplify this integral a little bit.

105
00:06:05,689 --> 00:06:07,709
Well one thing we want to do, we might want to make

106
00:06:07,709 --> 00:06:09,129
their end points the same.

107
00:06:09,129 --> 00:06:12,149
So if you swap a and b here, it just turns

108
00:06:12,149 --> 00:06:13,179
the integral negative.

109
00:06:13,180 --> 00:06:17,810
So you make this into a b, that into an a, and then make that

110
00:06:17,810 --> 00:06:20,290
plus sign into a minus sign.

111
00:06:20,290 --> 00:06:24,000
And now we can rewrite this whole thing as being equal to

112
00:06:24,000 --> 00:06:33,810
the integral from a to b of this thing, of p of x and y1 of

113
00:06:33,810 --> 00:06:44,290
x minus this thing, minus p of x and y2 of x, and then

114
00:06:44,290 --> 00:06:46,840
all of that times dx.

115
00:06:46,839 --> 00:06:48,310
I'll write it in a third color.

116
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117
00:06:51,480 --> 00:06:53,740
Now, I'm going to do something a little bit arbitrary, but I

118
00:06:53,740 --> 00:06:56,189
think you'll appreciate why I did this by the end of this

119
00:06:56,189 --> 00:06:58,000
video, and it's just a very simple operation.

120
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What I'm going to do is I'm going to swap these two.

121
00:06:59,579 --> 00:07:02,490
So I'm essentially going to multiply this whole thing by

122
00:07:02,490 --> 00:07:05,019
negative 1, or essentially multiply and divide

123
00:07:05,019 --> 00:07:05,909
by negative 1.

124
00:07:05,910 --> 00:07:08,310
So I can multiply this by negative 1 and then multiply

125
00:07:08,310 --> 00:07:10,704
the outside by negative 1, and I will not have changed the

126
00:07:10,704 --> 00:07:13,289
integral; I'm multiplying by negative 1 twice.

127
00:07:13,290 --> 00:07:16,810
So if I swap these two things, if I multiply the inside times

128
00:07:16,810 --> 00:07:20,250
negative 1, so this is going to be equal to-- do the outside

129
00:07:20,250 --> 00:07:22,769
of the integral, a to b.

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If I multiply the inside-- I'll do a dx out here --if I

131
00:07:26,459 --> 00:07:28,789
multiply the inside of the integral by negative 1,

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these two guys switch.

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So it becomes p of x of y2 of x.

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And then you're going to have minus p of x and y1 of x.

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My handwriting's getting a little messy.

136
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But I can't just multiply just the inside by minus 1.

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I don't want to change the integral, so I multiplied the

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inside by minus 1, let me multiply the outside

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by minus 1.

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And since I multiplied by minus 1 twice, these two

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things are equivalent.

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Or you could say this is the negative of that.

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Either way, I think you appreciate that I haven't

144
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changed the integral at all, numerically.

145
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I multiplied the inside and the outside by minus 1.

146
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And now the next step I'm going to do, it might look a little

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bit foreign to you, but I think you'll appreciate it.

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149
00:08:20,660 --> 00:08:23,460
It might be obvious to you if you've recently done

150
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some double integrals.

151
00:08:24,319 --> 00:08:30,269
So this thing can be rewritten as minus the integral from a to

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00:08:30,269 --> 00:08:38,659
b of-- and let me do a new color --of the function p of x,

153
00:08:38,659 --> 00:08:49,459
y evaluated at y2 of x minus-- and let me make it very clear;

154
00:08:49,460 --> 00:08:54,970
this is y is equal to y2 of x --minus this function evaluated

155
00:08:54,970 --> 00:08:59,519
at y is equal to y1 of x.

156
00:08:59,519 --> 00:09:03,819
And of course all of that times dx.

157
00:09:03,820 --> 00:09:08,379
This statement and what we saw right here-- this

158
00:09:08,379 --> 00:09:14,090
statement right here --are completely identical.

159
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And then if we assume that a partial derivative of capital P

160
00:09:20,000 --> 00:09:24,590
with respect to y exists, hopefully you'll realize-- and

161
00:09:24,590 --> 00:09:26,870
I'll focus on this a little bit because I don't want to

162
00:09:26,870 --> 00:09:27,700
confuse you on this step.

163
00:09:27,700 --> 00:09:28,879
Let me write the outside of this integral.

164
00:09:28,879 --> 00:09:32,179
So this is going to be equal to-- and this is kind of a neat

165
00:09:32,179 --> 00:09:35,149
outcome, and we're starting to build up to a very neat

166
00:09:35,149 --> 00:09:38,490
outcome, which will probably have to take the next video to

167
00:09:38,490 --> 00:09:41,210
do --so we do the outside dx.

168
00:09:41,210 --> 00:09:43,860


169
00:09:43,860 --> 00:09:48,830
If we assume that capital P has a partial derivative, this

170
00:09:48,830 --> 00:09:51,410
right here is the exact same thing.

171
00:09:51,409 --> 00:09:54,549
This right here is the exact same thing as the partial

172
00:09:54,549 --> 00:10:01,859
derivative of P with respect to y, dy, the antiderivative of

173
00:10:01,860 --> 00:10:10,009
that from y1 of x to y2 of x.

174
00:10:10,009 --> 00:10:13,039
I want to make you feel comfortable that these two

175
00:10:13,039 --> 00:10:14,740
things are equivalent.

176
00:10:14,740 --> 00:10:16,799
And to realize they're equivalent, you'll probably

177
00:10:16,799 --> 00:10:19,309
just have to start here and then go to that.

178
00:10:19,309 --> 00:10:21,539
We're used to seeing this; we're used to seeing a double

179
00:10:21,539 --> 00:10:25,099
integral like this, and then the very first step we say, OK

180
00:10:25,100 --> 00:10:29,500
to solve this double integral we start on the inside integral

181
00:10:29,500 --> 00:10:33,210
right there, and we say, OK let's take the antiderivative

182
00:10:33,210 --> 00:10:34,870
of this with respect to y.

183
00:10:34,870 --> 00:10:37,419
So if you take the antiderivative of the partial

184
00:10:37,419 --> 00:10:41,829
of p with respect to y, you're going to end up with p.

185
00:10:41,830 --> 00:10:44,330
And since this is a definite integral, the boundaries are

186
00:10:44,330 --> 00:10:47,420
going to be in terms of x, you're going to evaluate that

187
00:10:47,419 --> 00:10:51,019
from y is equal to y2 of x, and you're going to subtract from

188
00:10:51,019 --> 00:10:53,210
that y is equal to y1 of x.

189
00:10:53,210 --> 00:10:55,200
Normally we start with something like this, and we

190
00:10:55,200 --> 00:10:56,610
go to something like this.

191
00:10:56,610 --> 00:10:59,620
This is kind of unusual that we started, we kind of solved, we

192
00:10:59,620 --> 00:11:02,029
started with the solution of the definite integral, and then

193
00:11:02,029 --> 00:11:05,149
we slowly built back to the definite integral.

194
00:11:05,149 --> 00:11:08,399
So hopefully you realize that this is true, that this is

195
00:11:08,399 --> 00:11:10,110
just we're kind of going in a reverse direction

196
00:11:10,110 --> 00:11:11,019
than we normally do.

197
00:11:11,019 --> 00:11:13,750
And if you do realize that, then we've just established

198
00:11:13,750 --> 00:11:14,850
a pretty neat outcome.

199
00:11:14,850 --> 00:11:18,879
Because what is this right here?

200
00:11:18,879 --> 00:11:21,570
Let me go back, let me see if I can fit everything.

201
00:11:21,570 --> 00:11:25,745
I have some function-- and I'm assuming that the partial of

202
00:11:25,745 --> 00:11:29,789
P with respect to y exists --but I have some function

203
00:11:29,789 --> 00:11:31,509
defined over the xy plane.

204
00:11:31,509 --> 00:11:33,490
You know, you could imagine we're dealing in

205
00:11:33,490 --> 00:11:35,389
three dimensions now.

206
00:11:35,389 --> 00:11:37,980
We'll draw a little bit neater.

207
00:11:37,980 --> 00:11:46,110
So that's y, that's x, that's z, so this, you could imagine,

208
00:11:46,110 --> 00:11:48,560
is some surface; it just happens to be the partial

209
00:11:48,559 --> 00:11:50,349
of P with respect to x.

210
00:11:50,350 --> 00:11:55,790
So it's some surface on the xy plane like that.

211
00:11:55,789 --> 00:11:57,169
And what are we doing?

212
00:11:57,169 --> 00:12:01,309
We're taking the double integral under that surface,

213
00:12:01,309 --> 00:12:02,250
around this region.

214
00:12:02,250 --> 00:12:04,580
The region's boundaries in terms of y are defined

215
00:12:04,580 --> 00:12:06,180
by y2 and y1 of x.

216
00:12:06,179 --> 00:12:08,969


217
00:12:08,970 --> 00:12:11,160
So you literally have that curve.

218
00:12:11,159 --> 00:12:16,500
That's y2 on top, y1 on the bottom.

219
00:12:16,500 --> 00:12:18,590
And so we're essentially taking the volume above.

220
00:12:18,590 --> 00:12:22,780


221
00:12:22,779 --> 00:12:27,189
So if you imagine with the base is-- the whole floor of this is

222
00:12:27,190 --> 00:12:32,060
going to be the area inside of this curve, and then the height

223
00:12:32,059 --> 00:12:36,069
is going to be the function partial of P with respect to y.

224
00:12:36,070 --> 00:12:40,490


225
00:12:40,490 --> 00:12:42,700
It's going to be a little hard for me to draw, but this is

226
00:12:42,700 --> 00:12:44,670
essentially some type of a volume, if you want to

227
00:12:44,669 --> 00:12:45,610
visualize it that way.

228
00:12:45,610 --> 00:12:52,779
But the really neat outcome here is if you call this region

229
00:12:52,779 --> 00:12:56,600
r, we've just simplified this line integral.

230
00:12:56,600 --> 00:12:57,580
And this was a special one.

231
00:12:57,580 --> 00:13:01,270
It only had an x-component, the vector field, but we've just

232
00:13:01,269 --> 00:13:06,309
simplified this line integral to being equivalent to-- maybe

233
00:13:06,309 --> 00:13:09,079
I should write this line integral because that's what's

234
00:13:09,080 --> 00:13:10,830
the really neat outcome.

235
00:13:10,830 --> 00:13:13,129
We've just established that this thing right here, which

236
00:13:13,129 --> 00:13:16,970
is the same as our original one, so let me write that.

237
00:13:16,970 --> 00:13:22,570
The closed line integral around the curve c of p of xy, dx,

238
00:13:22,570 --> 00:13:25,300
we've just established that that's the same thing as the

239
00:13:25,299 --> 00:13:31,979
double integral over the region r-- this is the region

240
00:13:31,980 --> 00:13:40,332
r --of the partial of P with respect to y.

241
00:13:40,332 --> 00:13:47,990


242
00:13:47,990 --> 00:13:52,120
And we could write dy, dx, or we could write da, whatever you

243
00:13:52,120 --> 00:13:53,810
want to write, but this is the double integral

244
00:13:53,809 --> 00:13:55,309
over that region.

245
00:13:55,309 --> 00:13:59,559
The neat thing here is using a vector field that only had an

246
00:13:59,559 --> 00:14:02,849
x-component, we were able to connect its line integral to

247
00:14:02,850 --> 00:14:04,730
the double integral over region-- oh, and I forgot

248
00:14:04,730 --> 00:14:06,000
something very important.

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We had a negative sign out here.

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So this was a minus sign out here.

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Or we could even put the minus in here, but I think you

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get the general idea.

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In the next video, I'm going to do the same exact thing with

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the vector field that only has vectors in the y-direction.

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And then we'll connect the two and we'll end up

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with Green's theorem.

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