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HERBERT GROSS: Hi.
10
00:00:33,650 --> 00:00:37,290
Hopefully by now you have become
well experienced at
11
00:00:37,290 --> 00:00:39,430
computing limits
of double sums.
12
00:00:39,430 --> 00:00:41,770
You've learned to hate me a
little bit more because the
13
00:00:41,770 --> 00:00:45,190
work was quite complicated,
and hopefully you are
14
00:00:45,190 --> 00:00:47,290
convinced that there
must be an easier
15
00:00:47,290 --> 00:00:49,250
way to do this stuff.
16
00:00:49,250 --> 00:00:52,120
And fortunately, the
answer is there is
17
00:00:52,120 --> 00:00:54,640
an easier way sometimes.
18
00:00:54,640 --> 00:00:58,540
And the concept that we're going
to talk about today is
19
00:00:58,540 --> 00:01:01,890
again completely analogous to
what happened in calculus of a
20
00:01:01,890 --> 00:01:04,860
single variable where we showed
that the concept of
21
00:01:04,860 --> 00:01:08,050
definite integral existed
independently of being able to
22
00:01:08,050 --> 00:01:11,250
form derivatives, but in the
case where we knew a
23
00:01:11,250 --> 00:01:14,710
particular function whose
derivative would be a given
24
00:01:14,710 --> 00:01:18,300
thing, we were able to perform
the infinite sum-- or compute
25
00:01:18,300 --> 00:01:21,640
the infinite sum much more
readily than had we had to
26
00:01:21,640 --> 00:01:25,920
rely on the arithmetic
of infinite series.
27
00:01:25,920 --> 00:01:28,100
By the way-- and I'll mention
this later in a different
28
00:01:28,100 --> 00:01:29,380
context I hope--
29
00:01:29,380 --> 00:01:32,030
that the converse problem
was also true.
30
00:01:32,030 --> 00:01:37,910
In many cases one did not know
how to find the required
31
00:01:37,910 --> 00:01:40,920
function with the given
derivative, and in this case
32
00:01:40,920 --> 00:01:43,850
knowing how to find the area
by the limit process was
33
00:01:43,850 --> 00:01:48,370
equivalent to how we were then
able to invent new functions
34
00:01:48,370 --> 00:01:49,620
which had given derivatives.
35
00:01:49,620 --> 00:01:52,490
Essentially the difference
in point of view was the
36
00:01:52,490 --> 00:01:54,390
difference between what we
called the first fundamental
37
00:01:54,390 --> 00:01:56,940
theorem of integral calculus,
and the second fundamental
38
00:01:56,940 --> 00:01:58,450
theorem of integral calculus.
39
00:01:58,450 --> 00:02:01,350
As I said the same analogy will
hold here, and let's get
40
00:02:01,350 --> 00:02:03,440
into this now without
further ado.
41
00:02:03,440 --> 00:02:06,170
I call today's lesson the
fundamental theorem.
42
00:02:06,170 --> 00:02:08,600
And what I'm going to do now
is to pretend that we never
43
00:02:08,600 --> 00:02:10,800
had the lecture of last time.
44
00:02:10,800 --> 00:02:13,110
That we have never heard
of an infinite sum.
45
00:02:13,110 --> 00:02:17,490
I'm going to introduce what one
calls the anti-derivative
46
00:02:17,490 --> 00:02:20,130
of a function of two independent
variables, again
47
00:02:20,130 --> 00:02:22,710
keeping in mind the analogy
that for more than two
48
00:02:22,710 --> 00:02:25,480
independent variables a similar
treatment holds.
49
00:02:25,480 --> 00:02:28,790
And the exercises will include
problems that have more than
50
00:02:28,790 --> 00:02:30,150
two independent variables.
51
00:02:30,150 --> 00:02:33,320
The idea is that when I write
down this particular double
52
00:02:33,320 --> 00:02:39,040
integral, if I look at the
innermost integral, notice
53
00:02:39,040 --> 00:02:41,470
that x appears only
as a parameter.
54
00:02:41,470 --> 00:02:45,080
That the variable of integration
is y notice then,
55
00:02:45,080 --> 00:02:47,040
that x is being treated
as a constant.
56
00:02:47,040 --> 00:02:48,350
What this says is what?
57
00:02:48,350 --> 00:02:53,830
Pick a fixed value of x,
compute this integral,
58
00:02:53,830 --> 00:02:58,050
evaluate this as y goes from
g1 of x To g2 of x.
59
00:02:58,050 --> 00:03:01,400
The resulting function is a
function of x alone, and
60
00:03:01,400 --> 00:03:04,460
integrate that function
of x from a to b.
61
00:03:04,460 --> 00:03:07,620
And to do that for you in slow
motion so you see what happens
62
00:03:07,620 --> 00:03:10,210
here, what I'm essentially
saying is read this thing
63
00:03:10,210 --> 00:03:11,310
inside out.
64
00:03:11,310 --> 00:03:13,940
Imagine some brackets
placed here.
65
00:03:13,940 --> 00:03:16,240
Not imagine them, let's
put them in.
66
00:03:16,240 --> 00:03:18,070
And what we're saying
is look--
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00:03:18,070 --> 00:03:20,800
first of all, fix
a value of x.
68
00:03:20,800 --> 00:03:22,380
Hold x constant.
69
00:03:22,380 --> 00:03:27,700
And for that fixed value of x,
compute f of xydy from g1 of x
70
00:03:27,700 --> 00:03:29,280
to g2 of x.
71
00:03:29,280 --> 00:03:32,180
Compute that, remember you're
going to integrate that with
72
00:03:32,180 --> 00:03:32,820
respect to y.
73
00:03:32,820 --> 00:03:35,800
When you're all through with
this because the limits
74
00:03:35,800 --> 00:03:38,760
involve only x, the integrand
that you get--
75
00:03:38,760 --> 00:03:41,360
this bracketed expression will
now be a function of x alone,
76
00:03:41,360 --> 00:03:42,490
call it h of x.
77
00:03:42,490 --> 00:03:46,500
Integrate h of dx between a
and b, and the resulting
78
00:03:46,500 --> 00:03:49,370
expression assuming that you can
carry out the integration
79
00:03:49,370 --> 00:03:54,110
of course, will be a
particular number.
80
00:03:54,110 --> 00:03:57,710
That has nothing to do as I say,
with double sums so far.
81
00:03:57,710 --> 00:04:00,350
Though I suspect that you're
getting a bit suspicious.
82
00:04:00,350 --> 00:04:04,360
And in the same way the definite
integral was a sum,
83
00:04:04,360 --> 00:04:09,980
and not an anti-derivative the
notation of the definite
84
00:04:09,980 --> 00:04:12,040
integral looked enough
like the notation of
85
00:04:12,040 --> 00:04:13,730
the indefinite integral.
86
00:04:13,730 --> 00:04:15,970
So we began to suspect there was
going to be a connection
87
00:04:15,970 --> 00:04:16,890
between them.
88
00:04:16,890 --> 00:04:18,790
That connection is what's
going to be called the
89
00:04:18,790 --> 00:04:20,890
fundamental theorem in
our present case.
90
00:04:20,890 --> 00:04:24,890
But going on, let me just give
you an example showing you how
91
00:04:24,890 --> 00:04:28,400
we compute the anti-derivative.
92
00:04:28,400 --> 00:04:31,570
Let's take, for example interval
from 0 to 2, integral
93
00:04:31,570 --> 00:04:35,210
from 0 to x squared,
x cubed y dy dx.
94
00:04:35,210 --> 00:04:39,270
Again we go from inside to out,
the way we read this is
95
00:04:39,270 --> 00:04:41,960
that we're integrating
with respect to y.
96
00:04:41,960 --> 00:04:45,410
That means we're holding x
constant for a fixed value of
97
00:04:45,410 --> 00:04:49,320
x integrate this as y goes
from 0 to x squared.
98
00:04:49,320 --> 00:04:52,070
So treating x as a constant,
I integrate this
99
00:04:52,070 --> 00:04:53,240
with respect to y.
100
00:04:53,240 --> 00:04:56,280
The integral of y dy
is 1/2 y squared.
101
00:04:56,280 --> 00:05:00,250
I evaluate that as y goes
from o to x squared.
102
00:05:00,250 --> 00:05:01,670
That's crucial you see.
103
00:05:01,670 --> 00:05:03,940
It's y that goes from
0 to x squared.
104
00:05:03,940 --> 00:05:09,360
When I replace y by the upper
limit, I get 1/2 x cubed times
105
00:05:09,360 --> 00:05:11,830
the quantity x squared,
squared.
106
00:05:11,830 --> 00:05:14,530
When I replace y by
the lower limit 0,
107
00:05:14,530 --> 00:05:16,450
the integrand vanishes.
108
00:05:16,450 --> 00:05:19,700
So that what I wind up with
is a function of x alone.
109
00:05:19,700 --> 00:05:21,430
Specifically what
function is it?
110
00:05:21,430 --> 00:05:24,680
It's 1/2 x to the seventh,
and I now integrate
111
00:05:24,680 --> 00:05:26,710
that from 0 to 2.
112
00:05:26,710 --> 00:05:31,070
That gives me 1/8 x to the
eighth over 2, that's x to the
113
00:05:31,070 --> 00:05:32,350
eighth over 16.
114
00:05:32,350 --> 00:05:34,600
Evaluated between 0 and 2.
115
00:05:34,600 --> 00:05:38,210
2 to the eighth over 16 is 2
to the eighth over 2 to the
116
00:05:38,210 --> 00:05:41,530
fourth, which is 2 to the
fourth, which in turn is 16.
117
00:05:41,530 --> 00:05:42,620
So again know this--
118
00:05:42,620 --> 00:05:47,110
I can carry out this operation
purely manipulatively.
119
00:05:47,110 --> 00:05:52,660
It is truly the inverse of
taking the partial derivative.
120
00:05:52,660 --> 00:05:54,960
In the same way that taking
the partial derivative
121
00:05:54,960 --> 00:05:58,190
involved holding all the
variables but one constant,
122
00:05:58,190 --> 00:06:02,500
taking a possible integral, an
anti-derivative, involves
123
00:06:02,500 --> 00:06:03,870
simply doing what?
124
00:06:03,870 --> 00:06:04,610
Integrating--
125
00:06:04,610 --> 00:06:08,280
treating all the variables but
the given one as a constant.
126
00:06:08,280 --> 00:06:11,170
Which reduces you again to the
equivalent of what the
127
00:06:11,170 --> 00:06:14,700
anti-derivative meant in the
calculus of a single variable.
128
00:06:14,700 --> 00:06:16,430
Now here's where we come to the
129
00:06:16,430 --> 00:06:17,730
so-called fundamental theorem.
130
00:06:17,730 --> 00:06:21,890
Let me tie what we've talked
about today, and what we
131
00:06:21,890 --> 00:06:25,030
talked about last time let me
see if I can't tie those
132
00:06:25,030 --> 00:06:28,790
together for you in a completely
different way.
133
00:06:28,790 --> 00:06:32,160
Meaning I want you to see that
conceptually, today's lesson
134
00:06:32,160 --> 00:06:35,800
and last time's lesson are
completely different, but the
135
00:06:35,800 --> 00:06:37,480
punch line is there's
a remarkable
136
00:06:37,480 --> 00:06:39,730
connection between the two.
137
00:06:39,730 --> 00:06:42,580
Let's suppose I take the region
R, which is a very
138
00:06:42,580 --> 00:06:45,290
simple region that's bounded
above by the curve y
139
00:06:45,290 --> 00:06:46,890
equals g2 of x.
140
00:06:46,890 --> 00:06:50,410
It's bounded below by the curve
y equals g1 of x, and
141
00:06:50,410 --> 00:06:53,540
that these two curves happen
to intersect at values
142
00:06:53,540 --> 00:06:56,460
corresponding to x equals
a and x equals b.
143
00:06:56,460 --> 00:07:03,110
And what I would like to find is
the mass of the region R--
144
00:07:03,110 --> 00:07:08,590
if its density is some given
function f of xy.
145
00:07:08,590 --> 00:07:13,290
Now in terms of what we did
last time this involved a
146
00:07:13,290 --> 00:07:14,670
double sum.
147
00:07:14,670 --> 00:07:16,320
What double sum was it?
148
00:07:16,320 --> 00:07:19,340
We partitioned this
into rectangles.
149
00:07:19,340 --> 00:07:25,150
We picked a particular number c
sub ij comma d sub ij in the
150
00:07:25,150 --> 00:07:26,690
ij-th rectangle.
151
00:07:26,690 --> 00:07:32,090
We computed f of c sub
ij comma d sub ij
152
00:07:32,090 --> 00:07:34,690
times delta a sub ij.
153
00:07:34,690 --> 00:07:38,030
Added these all up the double
sum as I went from 1 to n and
154
00:07:38,030 --> 00:07:44,400
j went from 1 to n, and took the
limit as the maximum delta
155
00:07:44,400 --> 00:07:49,660
xi and delta yj approach 0, and
that limit if it existed--
156
00:07:49,660 --> 00:07:52,430
and it would exist if f was
piecewise continuous--
157
00:07:52,430 --> 00:07:55,090
That limit was denoted
by what?
158
00:07:55,090 --> 00:08:00,480
Double integral over the
region R f of xy da.
159
00:08:00,480 --> 00:08:03,990
And that da could either be
viewed as being dydx if you
160
00:08:03,990 --> 00:08:08,400
wanted, it could be dxdy, It
could also be in different
161
00:08:08,400 --> 00:08:10,160
coordinate systems, we're
not going to worry about
162
00:08:10,160 --> 00:08:11,630
that part just yet.
163
00:08:11,630 --> 00:08:13,850
We'll worry about that
in future lectures.
164
00:08:13,850 --> 00:08:17,820
Now on the other hand, let's
stop right here for a moment.
165
00:08:17,820 --> 00:08:20,680
Let's see how one might have
been tempted to tackle this
166
00:08:20,680 --> 00:08:23,250
problem from a completely
intuitive point of view.
167
00:08:23,250 --> 00:08:26,050
The so-called engineering
approach if you would never
168
00:08:26,050 --> 00:08:31,340
heard of the anti-derivative
but had anti-derivative for
169
00:08:31,340 --> 00:08:34,730
several variables, but had taken
part one of our course.
170
00:08:34,730 --> 00:08:36,870
The engineering approach would
be something like this-- you
171
00:08:36,870 --> 00:08:40,700
would say let's pick a very
small infinitesimal region
172
00:08:40,700 --> 00:08:44,400
here, in which we can assume
that the density is constant.
173
00:08:44,400 --> 00:08:47,930
So we'll pick a little region
that's a very, very small
174
00:08:47,930 --> 00:08:50,640
rectangle of dimension what?
175
00:08:50,640 --> 00:08:52,820
We'll say it's dy by dx.
176
00:08:52,820 --> 00:08:57,370
177
00:08:57,370 --> 00:09:00,640
and the density of this
particular rectangle can be
178
00:09:00,640 --> 00:09:03,365
assumed to be the constant
value f of xy.
179
00:09:03,365 --> 00:09:06,640
180
00:09:06,640 --> 00:09:10,070
Because that density is
constant, the mass of this
181
00:09:10,070 --> 00:09:15,050
little piece should be
f of xy times dydx.
182
00:09:15,050 --> 00:09:16,240
and then we say what?
183
00:09:16,240 --> 00:09:19,700
Let's add these all up.
184
00:09:19,700 --> 00:09:21,460
Let's go down to here
we say look, this
185
00:09:21,460 --> 00:09:23,080
is a typical element.
186
00:09:23,080 --> 00:09:25,720
And we'll add them all up
for all possible y's
187
00:09:25,720 --> 00:09:27,130
all possible x's.
188
00:09:27,130 --> 00:09:28,740
Now how would you do
this intuitively?
189
00:09:28,740 --> 00:09:32,070
You'd say well look, let
me hold x constant.
190
00:09:32,070 --> 00:09:36,130
Well, for a constant value of
x let's call it x sub 1.
191
00:09:36,130 --> 00:09:40,500
For a constant value of x,
notice that y can vary any
192
00:09:40,500 --> 00:09:44,230
place to be in the region R. For
that fixed value of x, y
193
00:09:44,230 --> 00:09:46,480
can be any place from
here to here.
194
00:09:46,480 --> 00:09:49,640
In other words y varies
continuously from g1
195
00:09:49,640 --> 00:09:52,660
of x1 to g2 of x1.
196
00:09:52,660 --> 00:09:55,180
Because x1 could have been x--
197
00:09:55,180 --> 00:09:58,030
for that given x, y varies
continuously from g1
198
00:09:58,030 --> 00:10:00,440
of x to g2 of x.
199
00:10:00,440 --> 00:10:03,030
And notice that x could have
been chosen if we want to be
200
00:10:03,030 --> 00:10:06,470
in the region R to be any
place from a to b.
201
00:10:06,470 --> 00:10:10,970
And so mechanically, we might
say let's just say here is the
202
00:10:10,970 --> 00:10:15,120
mass of a small element, and
we'll add these all up so that
203
00:10:15,120 --> 00:10:19,040
for a fixed x, y goes from
g1 of x to g2 of x.
204
00:10:19,040 --> 00:10:21,750
And x could be anywhere
from a to b.
205
00:10:21,750 --> 00:10:25,270
And it appears that
this is a truism.
206
00:10:25,270 --> 00:10:27,220
This is not a truism.
207
00:10:27,220 --> 00:10:29,410
It's a rather remarkable
result.
208
00:10:29,410 --> 00:10:31,750
It is analogous to what happened
in calculus of a
209
00:10:31,750 --> 00:10:33,350
single variable.
210
00:10:33,350 --> 00:10:36,440
Notice that when we had the
definite integral from a to b,
211
00:10:36,440 --> 00:10:39,100
f of x dx That was
an infinite sum.
212
00:10:39,100 --> 00:10:44,720
It was summation f of c sub k,
delta x sub k, as k went from
213
00:10:44,720 --> 00:10:48,370
1 to n taking the limit as
the maximum delta x sub k
214
00:10:48,370 --> 00:10:49,750
approached 0.
215
00:10:49,750 --> 00:10:53,710
It turned out that if we knew
a function capital f whose
216
00:10:53,710 --> 00:10:58,430
derivative was little f, then
this sum could be evaluated by
217
00:10:58,430 --> 00:11:02,350
computing capital f of b
minus capital f of a.
218
00:11:02,350 --> 00:11:05,100
Again the idea being if you
knew a function whose
219
00:11:05,100 --> 00:11:07,970
derivative was f, that gave
you an easy way to
220
00:11:07,970 --> 00:11:09,460
evaluate the sum.
221
00:11:09,460 --> 00:11:12,950
On the other hand, if you didn't
know a function whose
222
00:11:12,950 --> 00:11:16,700
derivative was little f,
evaluating the sum as a limit
223
00:11:16,700 --> 00:11:21,620
gave you a way of being able to
find the anti-derivative.
224
00:11:21,620 --> 00:11:25,220
A particular case in point you
may recall was trying to
225
00:11:25,220 --> 00:11:29,140
handle the problem e to the
minus x squared dx.
226
00:11:29,140 --> 00:11:31,480
As x goes from 0 to 1.
227
00:11:31,480 --> 00:11:35,580
This certainly existed as an
area, but we did not know in
228
00:11:35,580 --> 00:11:37,620
fact, there is no elementary
function.
229
00:11:37,620 --> 00:11:39,950
We invented one called the
error function whose
230
00:11:39,950 --> 00:11:43,000
derivative is e to the
minus x squared.
231
00:11:43,000 --> 00:11:44,510
So the idea is--
232
00:11:44,510 --> 00:11:46,230
and this is the key point--
233
00:11:46,230 --> 00:11:49,940
this is what becomes known as
the fundamental theorem for
234
00:11:49,940 --> 00:11:52,000
several variables.
235
00:11:52,000 --> 00:11:56,690
That there is a connection
between a double infinite sum,
236
00:11:56,690 --> 00:12:00,220
and a double anti-derivative.
237
00:12:00,220 --> 00:12:02,560
That this particular expression
here does not
238
00:12:02,560 --> 00:12:06,010
involve knowing anything about
partial derivatives.
239
00:12:06,010 --> 00:12:09,310
This particular expression here
does not involve being it
240
00:12:09,310 --> 00:12:11,970
any knowledge of knowing
partial sums.
241
00:12:11,970 --> 00:12:14,500
This simply says what?
242
00:12:14,500 --> 00:12:16,470
Integrate this thing twice.
243
00:12:16,470 --> 00:12:20,270
Once holding x constant and
letting y go from g1 of x to
244
00:12:20,270 --> 00:12:22,920
g2 of x, and integrating
with respect to x.
245
00:12:22,920 --> 00:12:28,450
And the amazing result is that
if f is continuous, this limit
246
00:12:28,450 --> 00:12:32,630
exists, and in particular, if
we happen to know how to
247
00:12:32,630 --> 00:12:37,290
actually carry out these
repeated or iterated
248
00:12:37,290 --> 00:12:41,740
integrations, we can compute
this complicated sum simply by
249
00:12:41,740 --> 00:12:44,950
carrying out this
anti-derivative.
250
00:12:44,950 --> 00:12:48,160
And I think the best way to
emphasize that to you is to
251
00:12:48,160 --> 00:12:53,080
repeat the punchline to the
homework exercises of last
252
00:12:53,080 --> 00:12:54,000
assignment.
253
00:12:54,000 --> 00:12:57,570
You may recall that last time
we were dealing with the
254
00:12:57,570 --> 00:13:02,110
square whose vertices were
0 0, 1 0, 1 1, and 0 1.
255
00:13:02,110 --> 00:13:06,170
The density of the square at
the point x comma y was x
256
00:13:06,170 --> 00:13:07,820
squared plus y squared.
257
00:13:07,820 --> 00:13:13,340
And I asked you as a homework
problem to compute the mass of
258
00:13:13,340 --> 00:13:16,430
this plate R, to compute
it exactly.
259
00:13:16,430 --> 00:13:21,400
And we said OK, by
definition of--
260
00:13:21,400 --> 00:13:23,230
Remember this thing here now
with the R under here
261
00:13:23,230 --> 00:13:26,440
indicates a limit
of a double sum.
262
00:13:26,440 --> 00:13:28,720
That this mass was
by definition
263
00:13:28,720 --> 00:13:30,290
this particular result.
264
00:13:30,290 --> 00:13:34,680
And notice that last time we
showed, in terms of double
265
00:13:34,680 --> 00:13:36,790
sums, that this came
out to be 2/3.
266
00:13:36,790 --> 00:13:40,780
Let me show you in terms of
new theory how we can find
267
00:13:40,780 --> 00:13:44,550
this much more conveniently
with no sweat so to speak.
268
00:13:44,550 --> 00:13:47,510
What I'm going to do now
is the following--
269
00:13:47,510 --> 00:13:49,470
The same analogy
I used before.
270
00:13:49,470 --> 00:13:52,120
I look at my region here,
which I'll call
271
00:13:52,120 --> 00:13:53,580
a, and I say what?
272
00:13:53,580 --> 00:13:56,030
For a fixed value of x.
273
00:13:56,030 --> 00:13:58,400
For a fixed value of x.
274
00:13:58,400 --> 00:14:02,470
Notice that y varies
continuously from 0 to 1.
275
00:14:02,470 --> 00:14:05,330
See y varies continuously
from 0 to 1.
276
00:14:05,330 --> 00:14:08,840
And then x in turn could have
been chosen to be any place
277
00:14:08,840 --> 00:14:09,700
from 0 to 1.
278
00:14:09,700 --> 00:14:12,830
That as we run through all these
strips added up from 0
279
00:14:12,830 --> 00:14:17,930
to 1, that covers our region R.
So what we're saying is all
280
00:14:17,930 --> 00:14:22,110
right, a little element
of our mass will be x
281
00:14:22,110 --> 00:14:24,260
squared plus y squared.
282
00:14:24,260 --> 00:14:26,870
dydx, we'll add these
all up from 0 to
283
00:14:26,870 --> 00:14:28,930
1, holding x constant.
284
00:14:28,930 --> 00:14:33,140
Then add them all up again
as x goes from 0 to 1.
285
00:14:33,140 --> 00:14:35,470
Again, this is the iterated
integral.
286
00:14:35,470 --> 00:14:37,990
How do we evaluate this?
287
00:14:37,990 --> 00:14:41,400
Well we treat x as a constant,
integrate this
288
00:14:41,400 --> 00:14:43,680
with respect to y.
289
00:14:43,680 --> 00:14:45,890
If we're treating x as a
constant, this then will come
290
00:14:45,890 --> 00:14:49,480
out to be x squared y, plus
1/3 y cubed, and we now
291
00:14:49,480 --> 00:14:53,580
evaluate that as y
goes from 0 to 1.
292
00:14:53,580 --> 00:14:55,940
Leaving these details out
because they are essentially
293
00:14:55,940 --> 00:14:58,640
the calculus of a single
variable all over again, this
294
00:14:58,640 --> 00:15:01,890
turns out to be the integral
from 0 to 1 x squared plus 1/3
295
00:15:01,890 --> 00:15:06,810
the x, which in turn is 1/3 x
cubed, plus 1/3 x evaluated as
296
00:15:06,810 --> 00:15:08,390
x goes from 0 to 1.
297
00:15:08,390 --> 00:15:10,870
The upper limit gives me a
third plus a third, the
298
00:15:10,870 --> 00:15:12,000
lower limit's 0.
299
00:15:12,000 --> 00:15:19,450
The mass here is 2/3, which
checks with the result that we
300
00:15:19,450 --> 00:15:21,930
got last time.
301
00:15:21,930 --> 00:15:24,340
Again, I'm not going to go
through the proof of the
302
00:15:24,340 --> 00:15:27,930
fundamental theorem, I don't
think it's that crucial.
303
00:15:27,930 --> 00:15:32,680
It's available in textbooks on
advanced calculus some of the
304
00:15:32,680 --> 00:15:36,920
exercises may possibly give
you hints as to how these
305
00:15:36,920 --> 00:15:38,170
results come about.
306
00:15:38,170 --> 00:15:41,150
But by and large, I'm more
interested now in you seeing
307
00:15:41,150 --> 00:15:44,430
the overview, and at this stage
of the game letting you
308
00:15:44,430 --> 00:15:48,040
get whatever specific
theoretical details you desire
309
00:15:48,040 --> 00:15:50,030
on your own.
310
00:15:50,030 --> 00:15:53,950
At any rate, let's continue
on with examples.
311
00:15:53,950 --> 00:15:57,890
I think a very nice counterpart
to example one and
312
00:15:57,890 --> 00:16:00,170
to refresh your memories without
looking back at it,
313
00:16:00,170 --> 00:16:06,370
example one asked us to compute
this anti-derivative.
314
00:16:06,370 --> 00:16:09,460
And what I'd like to do now is
to emphasize the fundamental
315
00:16:09,460 --> 00:16:12,500
theorem by wording this
a different way.
316
00:16:12,500 --> 00:16:17,420
What I want to do now is to
describe the plate R. If its
317
00:16:17,420 --> 00:16:22,840
mass is given by the double
integral over the region R row
318
00:16:22,840 --> 00:16:27,150
of x, y, da and that turns out
to be integral from 0 to 3, 0
319
00:16:27,150 --> 00:16:30,070
to x squared, x cubed y, dydx.
320
00:16:30,070 --> 00:16:36,900
What is the region R if this
is how it's mass is given?
321
00:16:36,900 --> 00:16:40,810
The first thing I want you to
observe is that this part here
322
00:16:40,810 --> 00:16:45,110
is identified with the density
part of the problem.
323
00:16:45,110 --> 00:16:47,810
And that these limits of
integration determine the
324
00:16:47,810 --> 00:16:51,370
region R. For example, what
this says is if you hold x
325
00:16:51,370 --> 00:16:55,760
constant, y varies from
0 to x squared.
326
00:16:55,760 --> 00:16:57,300
Let's see what that means.
327
00:16:57,300 --> 00:17:00,980
If you hold x constant,
y varies from 0.
328
00:17:00,980 --> 00:17:03,390
Well y equals 0 is the x-axis.
329
00:17:03,390 --> 00:17:05,619
To y equals x squared--
330
00:17:05,619 --> 00:17:07,240
that's this particular
parabola--
331
00:17:07,240 --> 00:17:12,930
what this says is for a fixed
value of x, say x equals x1 to
332
00:17:12,930 --> 00:17:17,180
be in the region R, you can be
any place from the x-axis
333
00:17:17,180 --> 00:17:21,560
along this strip up to
y equals x1 squared.
334
00:17:21,560 --> 00:17:23,440
So your strip is like this.
335
00:17:23,440 --> 00:17:27,010
Then we're told that in turn
x could be chosen to be any
336
00:17:27,010 --> 00:17:28,730
place from 0 to 2.
337
00:17:28,730 --> 00:17:31,980
So now what we know is that
this strip could have been
338
00:17:31,980 --> 00:17:35,080
chosen for any value of
x between 0 and 2.
339
00:17:35,080 --> 00:17:38,230
And what that tells us
therefore, is that the shape
340
00:17:38,230 --> 00:17:42,320
of our region R is the curve--
341
00:17:42,320 --> 00:17:44,870
the region that's bounded above
by the curved y equals x
342
00:17:44,870 --> 00:17:50,660
squared, below by the x-axis,
and on the right by the
343
00:17:50,660 --> 00:17:52,940
line x equals 2.
344
00:17:52,940 --> 00:17:57,230
This is our region R. You see
the region R is determined by
345
00:17:57,230 --> 00:18:02,000
the limits of integration, and
the density of R is given by
346
00:18:02,000 --> 00:18:06,296
row of xy equals x cubed y
at the point x comma y.
347
00:18:06,296 --> 00:18:07,546
OK?
348
00:18:07,546 --> 00:18:10,150
349
00:18:10,150 --> 00:18:13,700
We'd mentioned before that why
do you have to write dydx?
350
00:18:13,700 --> 00:18:15,740
Couldn't you have
written dxdy?
351
00:18:15,740 --> 00:18:19,420
What I thought I'd like to do
for my next example for you is
352
00:18:19,420 --> 00:18:21,960
to show you how one inverts--
353
00:18:21,960 --> 00:18:22,830
or changes--
354
00:18:22,830 --> 00:18:24,920
the order of integration.
355
00:18:24,920 --> 00:18:26,950
For example, given the
same integral--
356
00:18:26,950 --> 00:18:28,360
the same region R--
357
00:18:28,360 --> 00:18:31,850
suppose we now want to express
the mass in the form double
358
00:18:31,850 --> 00:18:34,720
integral x cubed y dxdy?
359
00:18:34,720 --> 00:18:37,830
You see the region R is still
the same as it was before.
360
00:18:37,830 --> 00:18:41,590
But now you see what we want to
do, how would we read this?
361
00:18:41,590 --> 00:18:43,375
This says we're integrating
with respect to x.
362
00:18:43,375 --> 00:18:44,730
x is varying.
363
00:18:44,730 --> 00:18:49,030
So this says for a fixed value
of y, integrate this thing.
364
00:18:49,030 --> 00:18:52,100
See for a fixed value
of y integrate this.
365
00:18:52,100 --> 00:18:54,910
Evaluated between the
appropriate limits of x.
366
00:18:54,910 --> 00:18:56,060
In terms of y.
367
00:18:56,060 --> 00:18:58,230
And then integrate this
with respect to y.
368
00:18:58,230 --> 00:19:01,450
Well, what this says is let's
pick a fixed value of y.
369
00:19:01,450 --> 00:19:03,490
Let's say y equals y1.
370
00:19:03,490 --> 00:19:07,890
For that fixed value of y notice
that the curve y equals
371
00:19:07,890 --> 00:19:12,440
x squared for x positive in
inverted form has the form x
372
00:19:12,440 --> 00:19:15,050
equals the square root of y.
373
00:19:15,050 --> 00:19:20,090
So for a fixed value of y,
notice that x varies from the
374
00:19:20,090 --> 00:19:25,340
square root of y1 up
to x equals 2.
375
00:19:25,340 --> 00:19:29,010
See x goes from the square root
of y to 2, for a fixed
376
00:19:29,010 --> 00:19:31,250
value of y.
377
00:19:31,250 --> 00:19:33,350
Now where can y be?
378
00:19:33,350 --> 00:19:34,840
This is our limits here.
379
00:19:34,840 --> 00:19:38,290
To be in the region R, y could
have been chosen as low as
380
00:19:38,290 --> 00:19:42,210
this, or as high as this.
381
00:19:42,210 --> 00:19:46,660
In other words y varies
continuously from 0 to four.
382
00:19:46,660 --> 00:19:49,570
And again, because I don't
want to have our lecture
383
00:19:49,570 --> 00:19:54,490
obscured by computational
detail, I simply urge you to--
384
00:19:54,490 --> 00:19:55,580
on your own--
385
00:19:55,580 --> 00:19:58,830
compute this double integral,
compute the double integral
386
00:19:58,830 --> 00:20:02,720
that we obtained in example
number three, and show that
387
00:20:02,720 --> 00:20:04,150
the answers are the same.
388
00:20:04,150 --> 00:20:06,540
What I do want you
to observe is how
389
00:20:06,540 --> 00:20:09,080
different the limits look.
390
00:20:09,080 --> 00:20:11,440
Notice that in the other
integral it was from 0 to 2
391
00:20:11,440 --> 00:20:14,010
outside, now it's from 0 to 4.
392
00:20:14,010 --> 00:20:16,325
Notice also on the inside
integral it
393
00:20:16,325 --> 00:20:18,180
was from 0 to x squared.
394
00:20:18,180 --> 00:20:20,570
Now it's from the square
root of y to 2.
395
00:20:20,570 --> 00:20:23,710
There is no mechanical way of
doing this, at least in the
396
00:20:23,710 --> 00:20:25,790
two dimensional case,
we can see from the
397
00:20:25,790 --> 00:20:27,110
picture what's happening.
398
00:20:27,110 --> 00:20:30,290
In the multi-dimensional case,
we have to resort to the
399
00:20:30,290 --> 00:20:33,730
theory of inverse functions,
and this becomes at best a
400
00:20:33,730 --> 00:20:35,260
very messy procedure.
401
00:20:35,260 --> 00:20:38,070
Of course the answer is if it's
such a messy procedure
402
00:20:38,070 --> 00:20:39,790
why bother with it?
403
00:20:39,790 --> 00:20:42,140
And so in terms of another
example that I would like to
404
00:20:42,140 --> 00:20:47,680
show you, let me give you an
illustration in which it may
405
00:20:47,680 --> 00:20:51,870
be possible to find the
required answer if we
406
00:20:51,870 --> 00:20:55,060
integrated one order, but not
if we integrate with respect
407
00:20:55,060 --> 00:20:56,060
to the other order.
408
00:20:56,060 --> 00:21:01,780
Let me evaluate the double
integral 0 to 2, x to 2, e to
409
00:21:01,780 --> 00:21:04,900
the y squared, dydx.
410
00:21:04,900 --> 00:21:07,660
Let me see if I can at least
write down what this thing
411
00:21:07,660 --> 00:21:10,490
means geometrically before
I even begin.
412
00:21:10,490 --> 00:21:14,480
Notice that I can think of
this as a plate R, whose
413
00:21:14,480 --> 00:21:18,530
density at the point x comma y
is e to the y squared, and
414
00:21:18,530 --> 00:21:21,250
what is the shape
of this plate?
415
00:21:21,250 --> 00:21:22,740
Integrate with respect
to y first.
416
00:21:22,740 --> 00:21:28,350
For a fixed value of x, y goes
from y equals x-- well that's
417
00:21:28,350 --> 00:21:29,790
this line here--
418
00:21:29,790 --> 00:21:31,310
to y equals 2.
419
00:21:31,310 --> 00:21:32,830
That's this line here.
420
00:21:32,830 --> 00:21:35,920
So for a fixed x, y varies
continuously
421
00:21:35,920 --> 00:21:37,280
from here to here.
422
00:21:37,280 --> 00:21:43,920
x can be any place from 0 to 2
that happens to be the point 2
423
00:21:43,920 --> 00:21:47,310
comma 2 where these two
lines intersect.
424
00:21:47,310 --> 00:21:50,870
So the region R is this
rectangular region here.
425
00:21:50,870 --> 00:21:53,710
OK, this is our region R. And
what we're saying is find the
426
00:21:53,710 --> 00:21:57,400
mass of the region R if its
density at the point x comma y
427
00:21:57,400 --> 00:21:59,170
is e to the y squared.
428
00:21:59,170 --> 00:22:01,410
Now, notice that this density
exists even if I've never
429
00:22:01,410 --> 00:22:03,190
heard of the anti-derivative.
430
00:22:03,190 --> 00:22:07,270
If however I elect to use the
fundamental theorem, and I say
431
00:22:07,270 --> 00:22:11,290
OK, what I'll do is
I'll compute this.
432
00:22:11,290 --> 00:22:13,660
Notice I'm back to
an old bugaboo.
433
00:22:13,660 --> 00:22:17,940
I don't know how to integrate
e to the y squared dy, other
434
00:22:17,940 --> 00:22:19,340
than by an approximation.
435
00:22:19,340 --> 00:22:22,010
In terms of elementary
functions, there is no
436
00:22:22,010 --> 00:22:24,210
function whose derivative
with respect to y
437
00:22:24,210 --> 00:22:26,240
is e to the y squared.
438
00:22:26,240 --> 00:22:29,100
So what I do is I elect
to change the order of
439
00:22:29,100 --> 00:22:30,490
integration.
440
00:22:30,490 --> 00:22:31,630
Why is that?
441
00:22:31,630 --> 00:22:35,380
Well because if I change the
order of integration sure I'm
442
00:22:35,380 --> 00:22:38,290
integrating e to the y squared,
but now with respect
443
00:22:38,290 --> 00:22:41,860
to x, which means that e to the
y squared is a constant.
444
00:22:41,860 --> 00:22:43,660
This is trivial to integrate.
445
00:22:43,660 --> 00:22:47,200
It's just going to be x times
e to the y squared.
446
00:22:47,200 --> 00:22:49,000
The problem is what?
447
00:22:49,000 --> 00:22:52,500
If I'm going to change the order
here, I must make sure
448
00:22:52,500 --> 00:22:56,260
that I take care of the limits
of integration accordingly.
449
00:22:56,260 --> 00:22:57,530
Now what this says is what?
450
00:22:57,530 --> 00:23:00,240
I'm going to integrate with
respect to x first. x is going
451
00:23:00,240 --> 00:23:02,900
to vary, so I'm treating
y as a constant.
452
00:23:02,900 --> 00:23:06,090
For that constant value of y,
notice that the b in the
453
00:23:06,090 --> 00:23:11,830
region R for that constant
value of y our
454
00:23:11,830 --> 00:23:14,310
x varies from what?
455
00:23:14,310 --> 00:23:19,310
x varies from 0 to the
line y equals x, or
456
00:23:19,310 --> 00:23:21,050
accordingly x equals y.
457
00:23:21,050 --> 00:23:27,420
So for that fixed value of
y, x varies from x equals
458
00:23:27,420 --> 00:23:29,120
0 to x equal y.
459
00:23:29,120 --> 00:23:33,680
And correspondingly, y can be
chosen to be any fixed number
460
00:23:33,680 --> 00:23:36,720
between 0 and 2, and that
gives me this particular
461
00:23:36,720 --> 00:23:37,840
double integral.
462
00:23:37,840 --> 00:23:39,860
And by the way, from
this point on the
463
00:23:39,860 --> 00:23:41,720
rest is child's play.
464
00:23:41,720 --> 00:23:45,180
Because when I integrate this
you see this integrand is xe
465
00:23:45,180 --> 00:23:46,430
to the y squared.
466
00:23:46,430 --> 00:23:53,400
467
00:23:53,400 --> 00:23:56,760
When x is equal to y, this
becomes ye to the y squared.
468
00:23:56,760 --> 00:23:59,340
When x is 0 the lower
limit drops out.
469
00:23:59,340 --> 00:24:03,520
So what I want to integrate
now is 0 to 2 ye to the y
470
00:24:03,520 --> 00:24:04,750
squared dy.
471
00:24:04,750 --> 00:24:07,890
But this is beautiful for me,
because the factor of y in
472
00:24:07,890 --> 00:24:11,080
here is exactly what I need
to be able to handle this.
473
00:24:11,080 --> 00:24:14,930
In other words, this is nothing
more than 1/2 e to the
474
00:24:14,930 --> 00:24:18,340
y squared evaluated
between 0 and 2.
475
00:24:18,340 --> 00:24:21,810
Replacing y by two gives
me 1/2 e to the fourth.
476
00:24:21,810 --> 00:24:25,070
Replacing y by 0 remember
e to the 0 is one.
477
00:24:25,070 --> 00:24:29,130
Gives me minus 1/2 because I'm
subtracting the lower limit.
478
00:24:29,130 --> 00:24:33,540
The answer to evaluating this
integral, which was impossible
479
00:24:33,540 --> 00:24:36,190
to do in this form because of
the fact that there was no
480
00:24:36,190 --> 00:24:38,350
elementary function whose
derivative is e to the y
481
00:24:38,350 --> 00:24:41,280
squared turns out
quite nicely.
482
00:24:41,280 --> 00:24:45,710
To be given by e to the
fourth minus 1 over 2.
483
00:24:45,710 --> 00:24:48,260
To summarize today's lecture,
to show you what the
484
00:24:48,260 --> 00:24:51,550
fundamental theorem really
means, all we're saying is
485
00:24:51,550 --> 00:24:55,110
that we often compute the
infinite double sum--
486
00:24:55,110 --> 00:24:59,500
double integral r f
of x, y da by an
487
00:24:59,500 --> 00:25:01,560
appropriate iterated integral.
488
00:25:01,560 --> 00:25:02,930
That's what we mean is what?
489
00:25:02,930 --> 00:25:04,340
The anti-derivative.
490
00:25:04,340 --> 00:25:07,850
Integrating the anti-derivative
successively.
491
00:25:07,850 --> 00:25:09,270
Integrated integral.
492
00:25:09,270 --> 00:25:10,040
See?
493
00:25:10,040 --> 00:25:12,820
Iterated integral.
494
00:25:12,820 --> 00:25:18,870
And conversely, the gist of this
whole thing is that we
495
00:25:18,870 --> 00:25:22,060
now have two entirely different
topics that are
496
00:25:22,060 --> 00:25:26,770
related by a fantastic unifying
thread that allows us
497
00:25:26,770 --> 00:25:30,560
to solve one of the problems
in terms of the other, and
498
00:25:30,560 --> 00:25:31,320
vice versa.
499
00:25:31,320 --> 00:25:35,430
You see again the analogy
being complete with what
500
00:25:35,430 --> 00:25:39,420
happened in the calculus
of a single variable.
501
00:25:39,420 --> 00:25:44,070
That we can evaluate double sums
which was the aim of the
502
00:25:44,070 --> 00:25:48,960
previous lecture, by means of
an iterated integral, we can
503
00:25:48,960 --> 00:25:52,950
evaluate iterated integrals
sometimes by appropriately
504
00:25:52,950 --> 00:25:59,040
knowing how to find the limit of
an appropriate double sum.
505
00:25:59,040 --> 00:26:03,250
And again we shall make use
of this in the exercises.
506
00:26:03,250 --> 00:26:06,790
We will go into this in more
detail from other aspects, and
507
00:26:06,790 --> 00:26:08,400
other points of view
next time.
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00:26:08,400 --> 00:26:10,200
At any rate, until next
time, goodbye.
509
00:26:10,200 --> 00:26:13,230
510
00:26:13,230 --> 00:26:16,420
Funding for the publication of
this video was provided by
511
00:26:16,420 --> 00:26:20,480
Gabriella and PaulRosenbaum
foundation.
512
00:26:20,480 --> 00:26:24,650
Help OCW continue to provide
free and open access to MIT
513
00:26:24,650 --> 00:26:32,360
courses by making a donation
at ocw.mit.edu/donate.
514
00:26:32,360 --> 00:26:33,642