1
00:00:00,000 --> 00:00:00,040
2
00:00:00,040 --> 00:00:01,940
FEMALE SPEAKER: The following
content is provided under a
3
00:00:01,940 --> 00:00:03,690
Creative Commons License.
4
00:00:03,690 --> 00:00:06,630
Your support will help MIT
OpenCourseWare continue to
5
00:00:06,630 --> 00:00:09,990
offer high quality educational
resources for free.
6
00:00:09,990 --> 00:00:12,830
To make a donation or to view
additional materials from
7
00:00:12,830 --> 00:00:16,760
hundreds of MIT courses, visit
MIT OpenCourseWare at
8
00:00:16,760 --> 00:00:18,010
ocw.mit.edu.
9
00:00:18,010 --> 00:00:31,286
10
00:00:31,286 --> 00:00:32,290
PROFESSOR: Hi.
11
00:00:32,290 --> 00:00:35,520
Technically speaking, I suppose,
if time permitted, we
12
00:00:35,520 --> 00:00:39,030
should spend, or shall we say,
give equal time to integral
13
00:00:39,030 --> 00:00:42,180
calculus, the same time that
we gave to differential
14
00:00:42,180 --> 00:00:46,080
calculus, starting from scratch,
building ideas slowly
15
00:00:46,080 --> 00:00:46,750
and the like.
16
00:00:46,750 --> 00:00:49,710
But there are two logistic
reasons for not doing this,
17
00:00:49,710 --> 00:00:53,390
one of which is that the basic
definitions would be the same
18
00:00:53,390 --> 00:00:55,620
regardless of which branch
we started with.
19
00:00:55,620 --> 00:00:58,550
The second one is, of course,
that time becomes preciously
20
00:00:58,550 --> 00:01:02,010
tight, and we have to make the
best of what we can here.
21
00:01:02,010 --> 00:01:06,750
So what we're going to do now
is try to show, in terms of
22
00:01:06,750 --> 00:01:10,700
hindsight being better than
foresight, a motivation as to
23
00:01:10,700 --> 00:01:14,970
how one would have invented
differential calculus had it
24
00:01:14,970 --> 00:01:18,550
been motivated by the existing
integral calculus.
25
00:01:18,550 --> 00:01:22,070
In other words, what we want
to do today is to show the
26
00:01:22,070 --> 00:01:24,580
beautiful interplay between
differential
27
00:01:24,580 --> 00:01:26,290
and integral calculus.
28
00:01:26,290 --> 00:01:29,350
Since we have said in our last
lecture that we would like to
29
00:01:29,350 --> 00:01:32,830
begin with a study of integral
calculus, assuming that
30
00:01:32,830 --> 00:01:35,990
differential calculus had never
been invented, I think
31
00:01:35,990 --> 00:01:39,720
that we're obligated to try to
show how, then, differential
32
00:01:39,720 --> 00:01:42,860
calculus could have been
invented in the environment of
33
00:01:42,860 --> 00:01:44,510
the studying of areas.
34
00:01:44,510 --> 00:01:48,030
For this reason, I have called
today's lecture "The Marriage
35
00:01:48,030 --> 00:01:51,910
of Differential and Integral
Calculus." And essentially,
36
00:01:51,910 --> 00:01:54,560
the idea goes something
like this.
37
00:01:54,560 --> 00:01:57,370
Last time, we were talking
about finding
38
00:01:57,370 --> 00:01:59,240
areas under a curve.
39
00:01:59,240 --> 00:02:02,170
And to find the area under the
curve, we saw that we could do
40
00:02:02,170 --> 00:02:04,450
this as the limit of
a certain sum.
41
00:02:04,450 --> 00:02:07,630
In other words, an infinite sum
having a particular limit.
42
00:02:07,630 --> 00:02:11,450
Now, to motivate the concept of
rate of change, one could
43
00:02:11,450 --> 00:02:13,500
ask the following question.
44
00:02:13,500 --> 00:02:16,710
A curve like 'y' equals 'f of
x', and a starting point, 'x'
45
00:02:16,710 --> 00:02:17,830
equals 'a'.
46
00:02:17,830 --> 00:02:21,980
As I move out along the x-axis
from 'a', I could study the
47
00:02:21,980 --> 00:02:26,670
area under the curve, say, of
the curve, 'y' equals 'f of
48
00:02:26,670 --> 00:02:30,600
x', from 'x' equals 'a'
to some value 'x1'.
49
00:02:30,600 --> 00:02:34,280
And the question I could then
ask is, I wonder how fast this
50
00:02:34,280 --> 00:02:38,040
area is changing at the instant
that 'x' equals 'x1'.
51
00:02:38,040 --> 00:02:41,860
In other words, how fast is the
area changing under the
52
00:02:41,860 --> 00:02:47,150
curve as 'x' moves out along the
positive x-axis this way?
53
00:02:47,150 --> 00:02:48,160
You see?
54
00:02:48,160 --> 00:02:50,960
Now, the idea is this.
55
00:02:50,960 --> 00:02:53,950
I can then mimic the same
definition that we gave for
56
00:02:53,950 --> 00:02:56,890
instantaneous rate of change
and differential calculus.
57
00:02:56,890 --> 00:03:00,060
Namely, I can say, why can't
we define the instantaneous
58
00:03:00,060 --> 00:03:03,510
rate of change to be the average
rate of change with
59
00:03:03,510 --> 00:03:04,520
respect to 'x'?
60
00:03:04,520 --> 00:03:06,350
And then we'll take the
limit as 'delta x'
61
00:03:06,350 --> 00:03:08,130
approaches 0, et cetera.
62
00:03:08,130 --> 00:03:10,420
And if I do that, watch what
happens over here.
63
00:03:10,420 --> 00:03:13,600
You see what I do in terms of
this picture, is I say, OK,
64
00:03:13,600 --> 00:03:15,670
here's the area under
the curve when
65
00:03:15,670 --> 00:03:17,300
'x' is equal to 'x1'.
66
00:03:17,300 --> 00:03:20,720
Now I'll let 'x1' change by some
amount 'delta x', meaning
67
00:03:20,720 --> 00:03:23,120
this'll be the point
'x1 + delta x'.
68
00:03:23,120 --> 00:03:25,770
And this brings about a change
in my area, which
69
00:03:25,770 --> 00:03:27,330
I'll call 'delta A'.
70
00:03:27,330 --> 00:03:29,790
See the change in area
as 'x' goes from 'x1'
71
00:03:29,790 --> 00:03:31,840
to 'x1 + delta x'.
72
00:03:31,840 --> 00:03:35,830
Now, what I would like to find
is 'delta A' divided by 'delta
73
00:03:35,830 --> 00:03:39,390
x', taking the limit as 'delta
x' approaches 0.
74
00:03:39,390 --> 00:03:41,720
Remember, last time we
showed our three
75
00:03:41,720 --> 00:03:43,360
basic axioms for area.
76
00:03:43,360 --> 00:03:45,710
We're obliged to
still use them.
77
00:03:45,710 --> 00:03:46,880
Let's do that here.
78
00:03:46,880 --> 00:03:49,960
For example, the way I've drawn
this curve by rising
79
00:03:49,960 --> 00:03:53,540
this way, notice that the lowest
point in this interval
80
00:03:53,540 --> 00:03:55,360
occurs when 'x' equals 'x1'.
81
00:03:55,360 --> 00:03:59,420
The highest point occurs when
'x' equals 'x1 + delta x'.
82
00:03:59,420 --> 00:04:03,380
Consequently, the rectangle
whose base is the closed
83
00:04:03,380 --> 00:04:07,400
interval from 'x1' to 'x1 +
delta x', the rectangle whose
84
00:04:07,400 --> 00:04:11,680
height corresponds to 'x' equals
'x1' is too small in
85
00:04:11,680 --> 00:04:13,670
area to be 'delta A'.
86
00:04:13,670 --> 00:04:17,160
In other words, that rectangle
is inscribed in 'delta A'.
87
00:04:17,160 --> 00:04:20,760
And in the same way, the
rectangle, with base 'x1' to
88
00:04:20,760 --> 00:04:25,720
'x1 + delta x', whose height is
given by the x-coordinate,
89
00:04:25,720 --> 00:04:30,010
'x1 + delta x', that rectangle
has an area which is too large
90
00:04:30,010 --> 00:04:31,610
to be the exact area.
91
00:04:31,610 --> 00:04:34,830
In other words, notice that
the area of the rectangle,
92
00:04:34,830 --> 00:04:38,900
which is too small to be the
exact area, is 'f of x1'
93
00:04:38,900 --> 00:04:40,690
times 'delta x'.
94
00:04:40,690 --> 00:04:44,310
The area of the rectangle which
is too big to be the
95
00:04:44,310 --> 00:04:50,100
correct 'delta A' is 'f of 'x1
+ delta x'' times 'delta x'.
96
00:04:50,100 --> 00:04:54,530
And the area, 'delta A', is
caught between these two.
97
00:04:54,530 --> 00:04:58,850
In other words, we now have the
inequality that 'f of x1'
98
00:04:58,850 --> 00:05:02,100
times 'delta x' is less than
'delta A', which in turn is
99
00:05:02,100 --> 00:05:06,550
less than 'f of 'x1 + delta
x'' times 'delta x'.
100
00:05:06,550 --> 00:05:09,700
Now, we'll assume that 'delta x'
is positive, the way we've
101
00:05:09,700 --> 00:05:11,940
drawn it in this diagram.
102
00:05:11,940 --> 00:05:15,320
Adjustments have to be made if
'delta x' is negative, in
103
00:05:15,320 --> 00:05:17,920
other words, if we let
'delta x' approach 0
104
00:05:17,920 --> 00:05:19,390
through negative values.
105
00:05:19,390 --> 00:05:22,300
This is taken care of in the
textbook, as well as mentioned
106
00:05:22,300 --> 00:05:22,690
in our notes.
107
00:05:22,690 --> 00:05:25,550
But for the sake of just getting
the main idea here,
108
00:05:25,550 --> 00:05:28,060
we'll assume that 'delta
x' is positive.
109
00:05:28,060 --> 00:05:30,310
We divide through
by 'delta x'.
110
00:05:30,310 --> 00:05:33,310
That gives us 'f of x1' is
less than 'delta a' over
111
00:05:33,310 --> 00:05:35,380
'delta x', which in turn
is less than 'f
112
00:05:35,380 --> 00:05:37,420
of 'x1 + delta x''.
113
00:05:37,420 --> 00:05:39,690
In other words, had 'delta x'
been negative, we would have
114
00:05:39,690 --> 00:05:43,150
had to reverse the inequality
signs, et cetera, but we're
115
00:05:43,150 --> 00:05:45,090
not going to worry about
that right now.
116
00:05:45,090 --> 00:05:46,510
Remember, what is 'delta x'?
117
00:05:46,510 --> 00:05:50,610
It's a non-zero number which is
going to be made to become
118
00:05:50,610 --> 00:05:52,380
arbitrarily close to 0.
119
00:05:52,380 --> 00:05:53,470
Well, here's the idea.
120
00:05:53,470 --> 00:05:56,500
We now have 'delta a' over
'delta x' squeezed between
121
00:05:56,500 --> 00:05:58,180
these two values.
122
00:05:58,180 --> 00:06:02,190
And we say, OK, let's let
'delta x' approach 0.
123
00:06:02,190 --> 00:06:03,490
Now, here's the key point.
124
00:06:03,490 --> 00:06:07,730
As 'delta x' approaches 0,
certainly, 'x1 + delta x'
125
00:06:07,730 --> 00:06:10,110
approaches 'x1'.
126
00:06:10,110 --> 00:06:13,310
And here's where we use the fact
that 'f' is continuous.
127
00:06:13,310 --> 00:06:16,460
In other words, notice that the
mere fact that 'x1 + delta
128
00:06:16,460 --> 00:06:20,360
x' approaches 'x1' is not enough
to say, therefore, that
129
00:06:20,360 --> 00:06:23,540
'f of 'x1 + delta x'' approaches
'f of x1'.
130
00:06:23,540 --> 00:06:27,610
As we saw before, that's only
true if 'f' is continuous.
131
00:06:27,610 --> 00:06:31,660
In other words, by continuity,
if 'x1 + delta x' approaches
132
00:06:31,660 --> 00:06:36,830
'x1', then 'f of 'x1 + delta
x'' approaches 'f of x1'.
133
00:06:36,830 --> 00:06:40,210
So assuming, then, that 'f' is
continuous, what do we get we
134
00:06:40,210 --> 00:06:41,360
put the squeeze on?
135
00:06:41,360 --> 00:06:43,910
'x1' is what's picked
to be a fixed point.
136
00:06:43,910 --> 00:06:46,620
Consequently, as 'delta
x' approaches 0, 'f of
137
00:06:46,620 --> 00:06:48,940
x1' stays 'f of x1'.
138
00:06:48,940 --> 00:06:52,480
'f of 'x1 + delta x'' approaches
'f of x1'.
139
00:06:52,480 --> 00:06:57,110
In other words, 'delta A'
divided by 'delta x' is caught
140
00:06:57,110 --> 00:07:02,170
between two numbers, two
sequences, two sets of bounds,
141
00:07:02,170 --> 00:07:05,510
whichever way you want to say
this, both of which converge
142
00:07:05,510 --> 00:07:07,960
to the common limit,
'f of x1'.
143
00:07:07,960 --> 00:07:11,520
In other words, since 'delta A'
over 'delta x' is squeezed
144
00:07:11,520 --> 00:07:15,320
between these two, the limit of
'delta a' over 'delta x',
145
00:07:15,320 --> 00:07:18,670
as 'delta x' approaches 0, must
equal this common limit.
146
00:07:18,670 --> 00:07:23,140
In other words, 'dA/dx', the
limit of 'delta A' over 'delta
147
00:07:23,140 --> 00:07:27,480
x', as 'delta x' approaches 0,
evaluated at 'x' equals 'x1',
148
00:07:27,480 --> 00:07:29,130
is 'f of x1'.
149
00:07:29,130 --> 00:07:34,570
Or another way of saying this,
'dA/dx' is 'f of x'.
150
00:07:34,570 --> 00:07:39,180
Now, that result is partly
intuitive and partly
151
00:07:39,180 --> 00:07:39,950
remarkable.
152
00:07:39,950 --> 00:07:44,170
What it says is, roughly
speaking, is that the area
153
00:07:44,170 --> 00:07:49,480
under the curve is changing at
a rate equal instantaneously
154
00:07:49,480 --> 00:07:52,090
to the height corresponding
to the
155
00:07:52,090 --> 00:07:53,370
x-coordinate at that point.
156
00:07:53,370 --> 00:07:56,970
In other words, to define how
fast the area is changing as
157
00:07:56,970 --> 00:08:01,140
'x' moves, all you have to do
is measure the height of the
158
00:08:01,140 --> 00:08:06,510
curve corresponding to that
particular value of 'x'.
159
00:08:06,510 --> 00:08:09,220
Now, what does this have to
do, then, with relating
160
00:08:09,220 --> 00:08:11,420
integral and differential
calculus?
161
00:08:11,420 --> 00:08:14,310
Notice already we begin to
get some sort of a hint.
162
00:08:14,310 --> 00:08:18,730
What we've shown is now that
the area is going to be
163
00:08:18,730 --> 00:08:22,620
related somehow to the inverse
derivative of 'f of x'.
164
00:08:22,620 --> 00:08:27,310
In fact, this is precisely
what is meant by a rather
165
00:08:27,310 --> 00:08:28,840
important result.
166
00:08:28,840 --> 00:08:31,580
In fact, it's important enough
so it's called the first
167
00:08:31,580 --> 00:08:34,890
fundamental theorem of
integral calculus.
168
00:08:34,890 --> 00:08:36,390
And it simply says this.
169
00:08:36,390 --> 00:08:41,090
Suppose we know explicitly a
function, capital 'G', such
170
00:08:41,090 --> 00:08:44,910
that the derivative of capital
'G' is 'f', where the 'f' now
171
00:08:44,910 --> 00:08:48,940
refers to the 'f' that we're
talking about in our problem.
172
00:08:48,940 --> 00:08:52,640
Now, look it, what we already
know is that 'A', the area
173
00:08:52,640 --> 00:08:55,170
function, has its derivative
equal to 'f'.
174
00:08:55,170 --> 00:08:58,610
Consequently, since the area
function and G have the same
175
00:08:58,610 --> 00:09:01,630
derivative, they must differ
by, at most, a constant.
176
00:09:01,630 --> 00:09:05,180
In other words, the area as
a function of 'x' is this
177
00:09:05,180 --> 00:09:07,310
function, 'G of x' plus 'c'.
178
00:09:07,310 --> 00:09:08,590
And what is this 'G of x'?
179
00:09:08,590 --> 00:09:10,030
It's not any old function.
180
00:09:10,030 --> 00:09:10,710
It's what?
181
00:09:10,710 --> 00:09:14,120
An inverse derivative
of 'f of x'.
182
00:09:14,120 --> 00:09:21,820
Since the area evaluated at
'x' equals 'a' is 0, see,
183
00:09:21,820 --> 00:09:24,920
namely, the area under the
curve, as 'x' goes from 'A' to
184
00:09:24,920 --> 00:09:26,910
'a', see, that's just a line.
185
00:09:26,910 --> 00:09:28,880
A line has no thickness,
has no area.
186
00:09:28,880 --> 00:09:29,680
This is 0.
187
00:09:29,680 --> 00:09:32,690
We have, by plugging back
in to here, that 'A' of
188
00:09:32,690 --> 00:09:34,450
little 'a' is 0.
189
00:09:34,450 --> 00:09:38,100
That in turn is 'G
of a' plus 'c'.
190
00:09:38,100 --> 00:09:41,990
And solving this for 'c', we
find that 'c' is equal to
191
00:09:41,990 --> 00:09:43,340
minus 'G of a'.
192
00:09:43,340 --> 00:09:46,820
In other words, the area under
the curve 'y' equals 'f of x',
193
00:09:46,820 --> 00:09:49,670
as a function of 'x',
is simply what?
194
00:09:49,670 --> 00:09:53,640
It's 'G of x' minus 'G of a',
where 'G' is any function
195
00:09:53,640 --> 00:09:55,290
whose derivative is 'f'.
196
00:09:55,290 --> 00:09:59,730
If we now let 'x' equal 'b',
we're back to the situation of
197
00:09:59,730 --> 00:10:03,630
finding the usual region that
we're talking about here,
198
00:10:03,630 --> 00:10:05,490
namely, what is 'A sub R'?
199
00:10:05,490 --> 00:10:08,200
It's the region bounded above--
let's come back here
200
00:10:08,200 --> 00:10:11,380
and take a look at this-- it's
the region bounded above by
201
00:10:11,380 --> 00:10:15,220
the curve, 'y' equals 'f of x',
bounded on the left by 'x'
202
00:10:15,220 --> 00:10:19,910
equals 'a', on the right by 'x'
equals 'b', and below by
203
00:10:19,910 --> 00:10:21,150
the x-axis.
204
00:10:21,150 --> 00:10:24,510
According to this notation, the
area of the region 'R' is
205
00:10:24,510 --> 00:10:29,020
just the area as a function of
'x' when 'x1' is equal to 'b'.
206
00:10:29,020 --> 00:10:32,465
In other words, the area of our
region 'R' is just 'a' of
207
00:10:32,465 --> 00:10:36,260
'b', which is 'G of b' minus
'G of a', where 'G' is any
208
00:10:36,260 --> 00:10:39,880
function whose derivative
is 'f'.
209
00:10:39,880 --> 00:10:40,670
Now remember--
210
00:10:40,670 --> 00:10:42,080
and here's the key point--
211
00:10:42,080 --> 00:10:47,480
remember that the area of the
region 'R' did not necessitate
212
00:10:47,480 --> 00:10:51,080
us having to know a function
whose derivative was 'f'.
213
00:10:51,080 --> 00:10:53,880
In other words, remember, in our
last lecture, how did we
214
00:10:53,880 --> 00:10:56,000
find areas of regions
like 'R'?
215
00:10:56,000 --> 00:10:58,990
What we did was is we formed
these 'U sub n's,
216
00:10:58,990 --> 00:11:00,150
these 'L sub n's.
217
00:11:00,150 --> 00:11:04,170
We inscribed and circumscribed
networks of rectangles, put
218
00:11:04,170 --> 00:11:07,850
the squeeze on by evaluating
this particular limit.
219
00:11:07,850 --> 00:11:10,390
And whatever that limit was,
that was the area of the
220
00:11:10,390 --> 00:11:11,250
region 'R'.
221
00:11:11,250 --> 00:11:14,120
In other words, if I had never
heard of the inverse
222
00:11:14,120 --> 00:11:16,350
derivative, I could still
find the area of the
223
00:11:16,350 --> 00:11:18,290
region 'R' this way.
224
00:11:18,290 --> 00:11:22,510
However, if I just happen to
know a function 'G', whose
225
00:11:22,510 --> 00:11:26,612
derivative is 'f', I have a much
easier way of doing this.
226
00:11:26,612 --> 00:11:29,630
In fact, let me summarize
that.
227
00:11:29,630 --> 00:11:33,340
You see, we can compute
the limit as
228
00:11:33,340 --> 00:11:34,960
'n' approaches infinity.
229
00:11:34,960 --> 00:11:38,700
Summation 'k' goes from 1 to
'n', 'f of 'c sub k'' 'delta
230
00:11:38,700 --> 00:11:41,850
x', by use of inverse
derivatives.
231
00:11:41,850 --> 00:11:44,080
See, again, the highlight
being what?
232
00:11:44,080 --> 00:11:47,590
You can still work with this the
same way as we did in our
233
00:11:47,590 --> 00:11:50,110
last lecture.
234
00:11:50,110 --> 00:11:53,330
There is absolutely no need to
have had to ever heard of a
235
00:11:53,330 --> 00:11:56,030
derivative to solve this
type of problem, even
236
00:11:56,030 --> 00:11:57,450
though it may be messy.
237
00:11:57,450 --> 00:11:59,370
Now, what's the best proof
I have of this?
238
00:11:59,370 --> 00:12:02,680
Well, I guess one of the best
proofs is to specifically
239
00:12:02,680 --> 00:12:07,000
refer to one of the exercises
in the previous unit.
240
00:12:07,000 --> 00:12:08,300
It was a tough exercise.
241
00:12:08,300 --> 00:12:11,050
I did it as a learning exercise
because I felt it was
242
00:12:11,050 --> 00:12:13,870
something that was messy and
that you had to be guided
243
00:12:13,870 --> 00:12:16,880
through in order not to become
hopelessly lost.
244
00:12:16,880 --> 00:12:19,230
By way of review, the
problem was this.
245
00:12:19,230 --> 00:12:23,990
We took as our region 'R' the
region bounded above by 'y'
246
00:12:23,990 --> 00:12:28,850
equals 'sine x', below by the
x-axis, on the right by the
247
00:12:28,850 --> 00:12:31,750
line 'x' equals pi/2.
248
00:12:31,750 --> 00:12:34,620
And the problem was, define the
area of the region 'R'.
249
00:12:34,620 --> 00:12:38,420
And we solved this problem in
the last unit without recourse
250
00:12:38,420 --> 00:12:41,740
to derivatives because, as of
the time that we were in the
251
00:12:41,740 --> 00:12:45,780
last unit, we had no results
relating derivatives
252
00:12:45,780 --> 00:12:48,890
to limits of sums.
253
00:12:48,890 --> 00:12:49,800
How did we do this?
254
00:12:49,800 --> 00:12:54,080
We partitioned this into n
parts, and we formed the sum
255
00:12:54,080 --> 00:12:57,680
limit as 'n' approaches
infinity.
256
00:12:57,680 --> 00:12:59,460
'Sigma k' goes from 1 to 'n'.
257
00:12:59,460 --> 00:13:02,780
We broke this thing up into 'n'
equal parts, so the size
258
00:13:02,780 --> 00:13:04,020
of each piece was what?
259
00:13:04,020 --> 00:13:06,170
pi/2 divided by 'n'.
260
00:13:06,170 --> 00:13:07,440
That's 'pi/2n'.
261
00:13:07,440 --> 00:13:10,240
We computed the sine of each of
these n-points, et cetera.
262
00:13:10,240 --> 00:13:12,880
And whatever that limit was,
that was the area of the
263
00:13:12,880 --> 00:13:13,680
region 'R'.
264
00:13:13,680 --> 00:13:16,400
And as you recall, that homework
problem, we found
265
00:13:16,400 --> 00:13:18,840
that the answer to that problem
was that the area of
266
00:13:18,840 --> 00:13:21,330
the region 'R' was 1.
267
00:13:21,330 --> 00:13:24,670
Now again, as messy as that was,
it proved that we could
268
00:13:24,670 --> 00:13:27,330
at least solve the problem
with no knowledge of
269
00:13:27,330 --> 00:13:28,200
derivatives.
270
00:13:28,200 --> 00:13:31,230
How does the first fundamental
theorem apply here?
271
00:13:31,230 --> 00:13:32,390
The way the first fundamental
theorem
272
00:13:32,390 --> 00:13:34,020
applies is the following.
273
00:13:34,020 --> 00:13:35,890
The first fundamental
theorem says this.
274
00:13:35,890 --> 00:13:40,060
Look it, to evaluate this
particular sum, all we have to
275
00:13:40,060 --> 00:13:44,750
do is find a function whose
derivative is 'sine x'.
276
00:13:44,750 --> 00:13:48,290
See, notice in this problem, the
general 'a' and 'b' of the
277
00:13:48,290 --> 00:13:53,120
above is now played by 'a'
equals 0 and 'b' equals pi/2.
278
00:13:53,120 --> 00:13:54,500
All we have to do is what?
279
00:13:54,500 --> 00:13:58,260
Find the function 'G' whose
derivative is 'sine x', and
280
00:13:58,260 --> 00:14:02,670
the answer should then be 'G
of pi/2' minus 'G of 0'.
281
00:14:02,670 --> 00:14:05,800
Well, you see this happens
to be one that, with our
282
00:14:05,800 --> 00:14:08,970
knowledge of differential
calculus, we
283
00:14:08,970 --> 00:14:09,970
can do rather easily.
284
00:14:09,970 --> 00:14:12,570
In other words, do we, at the
tip of our tongues, have a
285
00:14:12,570 --> 00:14:15,990
function whose derivative
is 'sine x'?
286
00:14:15,990 --> 00:14:17,030
And the answer is yes.
287
00:14:17,030 --> 00:14:20,130
Since the derivative of 'cosine
x' is 'minus sine x',
288
00:14:20,130 --> 00:14:23,600
the derivative of 'minus
cosine x' is 'sine x'.
289
00:14:23,600 --> 00:14:26,530
In other words, we can choose
for our 'G of x' here
290
00:14:26,530 --> 00:14:29,020
'minus cosine x'.
291
00:14:29,020 --> 00:14:30,530
In other words, that the
area of the region
292
00:14:30,530 --> 00:14:32,050
'R' should be what?
293
00:14:32,050 --> 00:14:36,460
'G of pi/2' minus 'G of
0', where 'G of x' is
294
00:14:36,460 --> 00:14:38,180
'minus cosine x'.
295
00:14:38,180 --> 00:14:41,940
Well, you see, cosine
pi/2 is 0.
296
00:14:41,940 --> 00:14:49,120
Cosine of 0 is 1, so 0 minus
minus 1 is equal to 1.
297
00:14:49,120 --> 00:14:52,440
And notice that, first of all,
we get the same answer.
298
00:14:52,440 --> 00:14:55,510
And secondly, notice this
two-line job over here.
299
00:14:55,510 --> 00:14:58,550
We not only get the correct
answer, but we get the answer
300
00:14:58,550 --> 00:15:01,450
very much more rapidly
than we did by the
301
00:15:01,450 --> 00:15:03,620
so-called limit process.
302
00:15:03,620 --> 00:15:06,640
Now, I'll come back to this in a
minute, because you may be a
303
00:15:06,640 --> 00:15:08,190
little bit angry at me now.
304
00:15:08,190 --> 00:15:10,810
Namely, you may be asking,
after having to do this
305
00:15:10,810 --> 00:15:14,220
problem the hard way, why did
I have to do the usual
306
00:15:14,220 --> 00:15:17,510
teacher's trick here and show
you the hard way of doing this
307
00:15:17,510 --> 00:15:20,000
and then waiting until the next
unit before I show you
308
00:15:20,000 --> 00:15:21,120
the easy way?
309
00:15:21,120 --> 00:15:23,780
Well, there happens to be a
catch here that I'll come back
310
00:15:23,780 --> 00:15:25,490
to in just a moment.
311
00:15:25,490 --> 00:15:29,340
But before I do that, I'd like
to make an aside, an aside
312
00:15:29,340 --> 00:15:30,410
that's rather important.
313
00:15:30,410 --> 00:15:34,220
Namely, you may recall, as I
started to work over here,
314
00:15:34,220 --> 00:15:38,290
that this reminded you of a
notation that we were using
315
00:15:38,290 --> 00:15:42,280
back when we introduced the
concept of the inverse of
316
00:15:42,280 --> 00:15:43,700
differentiation.
317
00:15:43,700 --> 00:15:48,070
Namely, we earlier used a
notation, integral from 'a' to
318
00:15:48,070 --> 00:15:53,160
'b', 'f of x' 'dx' to denote 'G
of b' minus 'G of a', where
319
00:15:53,160 --> 00:15:55,600
'G prime' equals 'f'.
320
00:15:55,600 --> 00:15:58,300
What we have shown by the first
fundamental theorem,
321
00:15:58,300 --> 00:16:02,770
then, using this notation, is
that the area of the region
322
00:16:02,770 --> 00:16:06,990
'R' is integral from 'a'
to 'b', 'f of x' 'dx'.
323
00:16:06,990 --> 00:16:08,480
In other words, this
means what?
324
00:16:08,480 --> 00:16:14,910
It means 'G of b' minus
'G of a', where 'G
325
00:16:14,910 --> 00:16:16,880
prime' equals 'f'.
326
00:16:16,880 --> 00:16:19,220
OK, let's pause here
for a moment.
327
00:16:19,220 --> 00:16:22,010
See, as we developed
our course, this is
328
00:16:22,010 --> 00:16:23,930
how this would evolve.
329
00:16:23,930 --> 00:16:28,470
The interesting thing is that,
historically, this notation
330
00:16:28,470 --> 00:16:33,400
was not used to define 'G of
b' minus 'G of a', where 'G
331
00:16:33,400 --> 00:16:34,750
prime' equaled 'f'.
332
00:16:34,750 --> 00:16:38,400
Historically, what happened was
that this notation, called
333
00:16:38,400 --> 00:16:40,970
the definite integral, the
integral from 'a' to 'b', 'f
334
00:16:40,970 --> 00:16:42,910
of x' 'dx', was--
335
00:16:42,910 --> 00:16:45,620
I don't know if it's proper to
say invented, but let me just
336
00:16:45,620 --> 00:16:47,650
say it in quotation marks
to play it safe--
337
00:16:47,650 --> 00:16:51,490
it was "invented" to denote
the limit of this sum.
338
00:16:51,490 --> 00:16:55,840
In fact, notice how much more
meaningful the symbol looks in
339
00:16:55,840 --> 00:16:56,930
this connotation.
340
00:16:56,930 --> 00:16:59,320
In other words, this is
the limit of a sum.
341
00:16:59,320 --> 00:17:02,100
You can think of sum as
beginning with 's', and the
342
00:17:02,100 --> 00:17:05,180
integral of sine as
an elongated 's'.
343
00:17:05,180 --> 00:17:08,690
In other words, when one first
wrote this symbol, the
344
00:17:08,690 --> 00:17:13,040
definite integral, it was meant
to denote this limit.
345
00:17:13,040 --> 00:17:16,160
The important point is that,
by the first fundamental
346
00:17:16,160 --> 00:17:19,355
theorem, the definite integral,
whether it's a limit
347
00:17:19,355 --> 00:17:23,230
or not, turns out to be 'G of
b' minus 'G of a', where 'G
348
00:17:23,230 --> 00:17:24,589
prime' equals 'f'.
349
00:17:24,589 --> 00:17:28,089
In other words, it really makes
no difference in terms
350
00:17:28,089 --> 00:17:30,540
of what answer you get, whether
you think of this
351
00:17:30,540 --> 00:17:34,010
definite integral as meaning
this, or whether you think of
352
00:17:34,010 --> 00:17:36,940
the definite integral
as meaning this.
353
00:17:36,940 --> 00:17:38,800
See, numerically, they'll
be the same.
354
00:17:38,800 --> 00:17:41,820
But conceptually, they're
quite different.
355
00:17:41,820 --> 00:17:46,260
And that's precisely the point
that leads to the most, I
356
00:17:46,260 --> 00:17:49,720
think, probably the most
difficult part all the course,
357
00:17:49,720 --> 00:17:51,090
at least up until now.
358
00:17:51,090 --> 00:17:53,670
It's something that more
people cause and have
359
00:17:53,670 --> 00:17:55,210
misinterpretation over.
360
00:17:55,210 --> 00:17:57,090
And so I'd like to present
this in a very,
361
00:17:57,090 --> 00:17:58,580
very gradual way.
362
00:17:58,580 --> 00:18:01,540
See, the result will ultimately
be known as the
363
00:18:01,540 --> 00:18:04,490
second fundamental theorem
of integral calculus.
364
00:18:04,490 --> 00:18:07,510
I've taken the liberty of
writing that over here.
365
00:18:07,510 --> 00:18:10,620
But it's going to be quite a
while before I actually say
366
00:18:10,620 --> 00:18:12,760
what the theorem
is explicitly.
367
00:18:12,760 --> 00:18:16,280
What I'd like to do is to pick
up where I left off when I
368
00:18:16,280 --> 00:18:18,120
made my aside about
the notation
369
00:18:18,120 --> 00:18:19,770
of a definite integral.
370
00:18:19,770 --> 00:18:21,820
And what I'd like to mention
now is, for the person who
371
00:18:21,820 --> 00:18:23,760
says, look it, why did
you make us find the
372
00:18:23,760 --> 00:18:24,900
areas the hard way?
373
00:18:24,900 --> 00:18:29,250
Why didn't you just say let us
compute 'G of b' minus 'G of
374
00:18:29,250 --> 00:18:32,200
a', where 'G prime' equals 'f',
and the heck with all
375
00:18:32,200 --> 00:18:34,090
this summation process?
376
00:18:34,090 --> 00:18:36,690
Now, I think the best way to
answer that question--
377
00:18:36,690 --> 00:18:38,510
and this is probably the
ultimate in teaching
378
00:18:38,510 --> 00:18:41,270
technique, when somebody thinks
that he has an easier
379
00:18:41,270 --> 00:18:44,410
way than the way that
he was taught--
380
00:18:44,410 --> 00:18:47,980
give them a problem that can't
be done by that easier way.
381
00:18:47,980 --> 00:18:51,040
In fact, as a sarcastic aside,
if you can't invent that kind
382
00:18:51,040 --> 00:18:53,990
of a problem, maybe the
so-called easier way is the
383
00:18:53,990 --> 00:18:55,330
better way of doing a thing.
384
00:18:55,330 --> 00:18:57,640
But let me just show you what
I'm driving at over here.
385
00:18:57,640 --> 00:19:00,620
Suppose I say to you, let's find
the area of the region
386
00:19:00,620 --> 00:19:03,220
'R' where 'R' is now
the following.
387
00:19:03,220 --> 00:19:07,130
It's bounded above by the
curve, 'y' equals '1/x'.
388
00:19:07,130 --> 00:19:09,950
It's bounded below
by the x-axis.
389
00:19:09,950 --> 00:19:13,220
It's bounded on the left by the
line, 'x' equals 1, and on
390
00:19:13,220 --> 00:19:15,630
the right by the line,
'x' equals 2.
391
00:19:15,630 --> 00:19:18,870
I would like to find the area
of this region 'R'.
392
00:19:18,870 --> 00:19:24,520
Now look it, from a purely
conceptual point of view, what
393
00:19:24,520 --> 00:19:27,810
difference is there between this
problem and the problem
394
00:19:27,810 --> 00:19:31,620
where the upper curve was
'y' equals 'x squared'?
395
00:19:31,620 --> 00:19:33,840
Conceptually, what's going
on is the same thing.
396
00:19:33,840 --> 00:19:37,280
We have a bounded region and we
want to compute the area.
397
00:19:37,280 --> 00:19:41,490
To this end, notice that, if I
use the precise definition of
398
00:19:41,490 --> 00:19:44,890
a definite integral, the area
of the region 'R' is what?
399
00:19:44,890 --> 00:19:49,370
The definite integral
from 1 to 2, 'dx/x'.
400
00:19:49,370 --> 00:19:50,970
What does that mean?
401
00:19:50,970 --> 00:19:52,900
It means a particular limit.
402
00:19:52,900 --> 00:19:53,890
What limit?
403
00:19:53,890 --> 00:19:56,990
Well, you partition this
integral from 1 to 2 into n
404
00:19:56,990 --> 00:20:02,200
equal parts, pick a point, 'c
sub k' in each partition, then
405
00:20:02,200 --> 00:20:05,880
'f of 'c sub k'' in this problem
is just '1/'c sub k''.
406
00:20:05,880 --> 00:20:08,230
In other words, what we want
to do is to compute this
407
00:20:08,230 --> 00:20:12,870
limit. 'Sigma k' goes from
1 to 'n', '1/'c sub k'',
408
00:20:12,870 --> 00:20:14,430
times 'delta x'.
409
00:20:14,430 --> 00:20:17,570
Now, the thing to notice is,
this thing may be a mess but
410
00:20:17,570 --> 00:20:18,990
it's computable.
411
00:20:18,990 --> 00:20:24,320
We could use specially designed
graph paper or
412
00:20:24,320 --> 00:20:27,650
measuring devices to count the
units of area under here.
413
00:20:27,650 --> 00:20:31,720
We can find all sorts of ways of
getting estimates, even the
414
00:20:31,720 --> 00:20:35,520
long hard way of the 'U sub
n's and the 'L sub n's to
415
00:20:35,520 --> 00:20:39,080
pinpoint the area of the region
'R' to as close a
416
00:20:39,080 --> 00:20:41,340
degree of accuracy as
we want, et cetera.
417
00:20:41,340 --> 00:20:44,650
The point is that, if we have
never heard of a derivative,
418
00:20:44,650 --> 00:20:50,070
the area of the region 'R' is
given precisely by this sum.
419
00:20:50,070 --> 00:20:53,970
And admittedly, the
sum is a mess.
420
00:20:53,970 --> 00:20:57,200
So let's try to do it the
so-called easier way.
421
00:20:57,200 --> 00:20:58,530
The skeptic looks at
this thing and
422
00:20:58,530 --> 00:21:00,630
says, who needs this?
423
00:21:00,630 --> 00:21:03,460
The area of the region 'R',
we just saw by the first
424
00:21:03,460 --> 00:21:08,370
fundamental theorem, is 'G of
2', 'G of 2', minus 'G of 1',
425
00:21:08,370 --> 00:21:11,360
where 'G' is any function of
'x' whose derivative with
426
00:21:11,360 --> 00:21:14,010
respect to 'x' is '1/x'.
427
00:21:14,010 --> 00:21:15,860
And the answer to that is yes.
428
00:21:15,860 --> 00:21:17,770
So far, so good.
429
00:21:17,770 --> 00:21:21,840
But do we know a function 'G'
whose derivative with respect
430
00:21:21,840 --> 00:21:23,610
to 'x' is '1/x'?
431
00:21:23,610 --> 00:21:25,870
Remember, this is calculus
revisited.
432
00:21:25,870 --> 00:21:27,790
For those of you who remember
calculus from
433
00:21:27,790 --> 00:21:29,100
the first time around--
434
00:21:29,100 --> 00:21:30,770
and I'll talk about
this later--
435
00:21:30,770 --> 00:21:33,570
it's going to turn out that
the required 'G' is a
436
00:21:33,570 --> 00:21:35,110
logarithmic function.
437
00:21:35,110 --> 00:21:37,570
For those of you don't remember
that, there's no harm
438
00:21:37,570 --> 00:21:38,770
in not remembering that.
439
00:21:38,770 --> 00:21:40,850
All I'm trying to bring out is
that, whether you remember it
440
00:21:40,850 --> 00:21:44,000
or you don't, as far as we're
concerned so far in this
441
00:21:44,000 --> 00:21:47,090
course, as far as what we've
developed in this course, we
442
00:21:47,090 --> 00:21:50,720
do not know a function 'G' whose
derivative with respect
443
00:21:50,720 --> 00:21:52,710
to 'x' is '1/x'.
444
00:21:52,710 --> 00:21:56,450
That's what we mean by saying
we can't exhibit 'G'
445
00:21:56,450 --> 00:21:58,340
explicitly.
446
00:21:58,340 --> 00:21:59,490
What do I mean by that?
447
00:21:59,490 --> 00:22:01,150
Somebody says, I am thinking
of a function whose
448
00:22:01,150 --> 00:22:02,660
derivative is '1/x'.
449
00:22:02,660 --> 00:22:04,310
I say, well, that's simple.
450
00:22:04,310 --> 00:22:05,150
It's 'G of x'.
451
00:22:05,150 --> 00:22:06,470
He says, well, what's
'G of x'?
452
00:22:06,470 --> 00:22:08,600
I say, that's the function whose
derivative with respect
453
00:22:08,600 --> 00:22:10,340
to 'x' is '1/x'.
454
00:22:10,340 --> 00:22:13,210
Well, you see, that implicitly
tells me what 'G' is like.
455
00:22:13,210 --> 00:22:15,680
But in terms of concrete
measurements, I don't know
456
00:22:15,680 --> 00:22:16,630
anything about 'G'.
457
00:22:16,630 --> 00:22:18,390
I can't express it in terms of
458
00:22:18,390 --> 00:22:20,060
well-known, familiar functions.
459
00:22:20,060 --> 00:22:21,610
You see, I'm hung up now.
460
00:22:21,610 --> 00:22:23,620
Namely, this is precise.
461
00:22:23,620 --> 00:22:26,970
The answer to this problem will
be 'G of 2' minus 'G of
462
00:22:26,970 --> 00:22:29,790
1', where 'G prime' is '1/x'.
463
00:22:29,790 --> 00:22:33,850
But I don't know explicitly--
464
00:22:33,850 --> 00:22:35,020
yikes--
465
00:22:35,020 --> 00:22:37,470
I don't know, explicitly
such a 'G'.
466
00:22:37,470 --> 00:22:40,290
And I put the exclamation point
out here to emphasize
467
00:22:40,290 --> 00:22:42,690
that particular fact.
468
00:22:42,690 --> 00:22:43,980
That's the hang-up.
469
00:22:43,980 --> 00:22:47,130
The statement that says that the
area is 'G of b' minus 'G
470
00:22:47,130 --> 00:22:50,960
of a', where 'G prime' equals
'f' hinges on the fact that
471
00:22:50,960 --> 00:22:54,120
you can explicitly exhibit
such as 'G'.
472
00:22:54,120 --> 00:22:56,710
Certainly, if you can't exhibit
that 'G', you can
473
00:22:56,710 --> 00:22:58,910
still compute this
area as a limit.
474
00:22:58,910 --> 00:23:03,650
It would be tragic to say, oh,
the area doesn't exist because
475
00:23:03,650 --> 00:23:06,560
I don't know a 'G' whose
derivative is '1/x'.
476
00:23:06,560 --> 00:23:09,320
Certainly, this region
'R' has an area.
477
00:23:09,320 --> 00:23:13,550
In fact, because we can pinpoint
the area of the
478
00:23:13,550 --> 00:23:17,920
region 'R', we can actually
construct the 'G' such that 'G
479
00:23:17,920 --> 00:23:20,220
prime of x' equals '1/x'.
480
00:23:20,220 --> 00:23:22,580
And by the way, there's plenty
of drill on this.
481
00:23:22,580 --> 00:23:24,210
This is a hard concept.
482
00:23:24,210 --> 00:23:26,690
And as a result, you'll notice
that the exercises in this
483
00:23:26,690 --> 00:23:30,250
section hammer home on this
point, because it's a point
484
00:23:30,250 --> 00:23:33,090
that I'm positive that, if
you're having trouble at all
485
00:23:33,090 --> 00:23:35,580
with integral calculus, this
is certainly the most
486
00:23:35,580 --> 00:23:38,910
sophisticated part of what
we're doing right now.
487
00:23:38,910 --> 00:23:41,730
You see, look it, suppose I want
to construct a function
488
00:23:41,730 --> 00:23:45,770
'G' whose derivative is '1/x'.
489
00:23:45,770 --> 00:23:47,470
The idea looks something
like this.
490
00:23:47,470 --> 00:23:48,990
I'm on the interval
from 1 to 2 here.
491
00:23:48,990 --> 00:23:52,970
What I do is, I plot the curve,
'y' equals '1/x'.
492
00:23:52,970 --> 00:23:57,450
And I compute the area of the
region formed by the curve,
493
00:23:57,450 --> 00:23:59,410
'y' equals '1/x'.
494
00:23:59,410 --> 00:24:00,740
'x' equals 1.
495
00:24:00,740 --> 00:24:06,500
The x-axis and the line,
'x' equals 'x1'.
496
00:24:06,500 --> 00:24:09,160
This is my region 'R'.
497
00:24:09,160 --> 00:24:11,210
And what I say is, look
it, what do I
498
00:24:11,210 --> 00:24:13,220
know about this area?
499
00:24:13,220 --> 00:24:16,120
Remember, what I started this
lecture with was the knowledge
500
00:24:16,120 --> 00:24:19,870
that a prime of 'x' is 'f
of x'. 'f of x', in
501
00:24:19,870 --> 00:24:21,360
this case, is '1/x'.
502
00:24:21,360 --> 00:24:24,460
In other words, what property
does this area function have?
503
00:24:24,460 --> 00:24:27,970
It has the property that its
derivative with respect to 'x'
504
00:24:27,970 --> 00:24:29,890
is going to be '1/x'.
505
00:24:29,890 --> 00:24:33,440
In other words, if I now say,
look it, boys, I can compute
506
00:24:33,440 --> 00:24:37,700
this area, if not exactly, at
least to as many decimal place
507
00:24:37,700 --> 00:24:41,500
accuracy as I wish, so that the
area function is something
508
00:24:41,500 --> 00:24:42,640
I can construct.
509
00:24:42,640 --> 00:24:48,100
Let me define 'G of x1' simply
to be the area under this
510
00:24:48,100 --> 00:24:50,150
curve when 'x' is
equal to 'x1'.
511
00:24:50,150 --> 00:24:51,940
In other words, written
now as a limit--
512
00:24:51,940 --> 00:24:53,600
see how this definite integral
comes in here--
513
00:24:53,600 --> 00:24:58,980
written as integral from 1 to 'x
sub 1', 'dx/x', which I now
514
00:24:58,980 --> 00:25:01,460
can compute as closely
as I want as a limit.
515
00:25:01,460 --> 00:25:02,840
The beauty is what?
516
00:25:02,840 --> 00:25:07,270
That 'G prime of x1' is, by
definition, 'A prime of 'x1'.
517
00:25:07,270 --> 00:25:11,120
And we have already seen that
the derivative of 'A' is 'f',
518
00:25:11,120 --> 00:25:13,350
where 'f' is the top curve.
519
00:25:13,350 --> 00:25:16,440
In this case, the derivative
of 'A' is '1/x'.
520
00:25:16,440 --> 00:25:20,090
In other words, the area under
the curve explicitly can be
521
00:25:20,090 --> 00:25:23,020
computed as a function of 'x'.
522
00:25:23,020 --> 00:25:25,930
And that function has the
property, but its derivative
523
00:25:25,930 --> 00:25:28,380
with respect to 'x', in
this case, is '1/x'.
524
00:25:28,380 --> 00:25:30,910
525
00:25:30,910 --> 00:25:34,670
And by the way, let me make just
a quick aside over here.
526
00:25:34,670 --> 00:25:37,010
I happened to throw in the
word "logarithms" before.
527
00:25:37,010 --> 00:25:39,340
It's going to turn out, as we'll
talk about in our next
528
00:25:39,340 --> 00:25:43,030
block of material, that 'G of x'
turns out to be the natural
529
00:25:43,030 --> 00:25:46,480
log of 'x', which means that the
area of the region 'R' is
530
00:25:46,480 --> 00:25:49,760
precisely natural log 2.
531
00:25:49,760 --> 00:25:51,540
Now, the idea is--
532
00:25:51,540 --> 00:25:53,460
and this is exactly what's going
to happen in our last
533
00:25:53,460 --> 00:25:54,890
block in our course--
534
00:25:54,890 --> 00:25:56,790
that this is exactly how
the log tables are
535
00:25:56,790 --> 00:25:58,450
constructed, in fact.
536
00:25:58,450 --> 00:26:00,820
I mean, how do you think they
find the log of 2 to eight
537
00:26:00,820 --> 00:26:01,630
decimal places?
538
00:26:01,630 --> 00:26:04,070
Did they strand somebody on a
desert island and tell them to
539
00:26:04,070 --> 00:26:08,330
compute these things, measure
it with laser beams, what?
540
00:26:08,330 --> 00:26:08,740
No.
541
00:26:08,740 --> 00:26:11,490
The way we do it is, we know
it's an area, and knowing how
542
00:26:11,490 --> 00:26:14,710
to find the area as a limit to
as many decimal place accuracy
543
00:26:14,710 --> 00:26:17,910
as we wish, give or take a few
tricks of the trade, this is
544
00:26:17,910 --> 00:26:21,080
how we find logs, and later
on, trigonometric
545
00:26:21,080 --> 00:26:22,330
tables, and the like.
546
00:26:22,330 --> 00:26:25,050
But I just mention this as
an aside to show you how
547
00:26:25,050 --> 00:26:28,260
important knowing area
under a curve is.
548
00:26:28,260 --> 00:26:31,750
At any rate, to summarize what
we're saying over here, we
549
00:26:31,750 --> 00:26:32,670
have the following.
550
00:26:32,670 --> 00:26:36,520
In general, if 'f' is any
continuous function on the
551
00:26:36,520 --> 00:26:40,080
closed interval from 'a' to 'b',
we define a function 'G'
552
00:26:40,080 --> 00:26:41,320
as follows.
553
00:26:41,320 --> 00:26:46,180
'G of x1' is the definite
integral from 'a' to 'x1', 'f
554
00:26:46,180 --> 00:26:49,510
of x' 'dx', where 'x1' is any
point in the closed interval
555
00:26:49,510 --> 00:26:50,430
for 'a' to 'b'.
556
00:26:50,430 --> 00:26:52,670
What does this thing mean,
geometrically?
557
00:26:52,670 --> 00:26:57,200
It's the area under the curve,
'y' equals 'f of x' on top.
558
00:26:57,200 --> 00:27:00,940
'x' equals the x-axis on the
bottom, the line 'x' equals
559
00:27:00,940 --> 00:27:05,190
'a' on the left, the line 'x'
equals 'x1' on the right.
560
00:27:05,190 --> 00:27:06,800
And that area is what?
561
00:27:06,800 --> 00:27:08,750
It's the limit as 'n'
approaches infinity.
562
00:27:08,750 --> 00:27:12,730
'Sigma k' goes from 1 to 'n', 'f
of 'c sub k'' 'delta x', et
563
00:27:12,730 --> 00:27:16,060
cetera, et cetera, et cetera,
meaning we can go through this
564
00:27:16,060 --> 00:27:19,740
the same way as we did in
our previous lecture.
565
00:27:19,740 --> 00:27:21,370
This function, 'G of x1'--
566
00:27:21,370 --> 00:27:25,060
which we can now compute
explicitly as an area, because
567
00:27:25,060 --> 00:27:27,710
we have the curve 'f' drawn,
we can approximate it to as
568
00:27:27,710 --> 00:27:29,510
close a degree of accuracy
as we want--
569
00:27:29,510 --> 00:27:33,420
whatever that function 'G' is,
what property does it have?
570
00:27:33,420 --> 00:27:38,380
It has the property that 'G
prime' is equal to 'f'.
571
00:27:38,380 --> 00:27:41,090
And that's exactly what the
second fundamental theorem
572
00:27:41,090 --> 00:27:42,020
really means.
573
00:27:42,020 --> 00:27:45,430
The second fundamental theorem
says, look it, if you can
574
00:27:45,430 --> 00:27:49,480
compute the area, you can now
reverse this procedure that
575
00:27:49,480 --> 00:27:52,150
we're talking about in the first
fundamental theorem.
576
00:27:52,150 --> 00:27:54,890
See, in the first fundamental
theorem, what did we do?
577
00:27:54,890 --> 00:27:57,510
In the first fundamental
theorem, we found a quick way
578
00:27:57,510 --> 00:28:00,100
of computing the area as
a limit by knowing the
579
00:28:00,100 --> 00:28:01,980
appropriate inverse
derivative.
580
00:28:01,980 --> 00:28:05,060
The second fundamental theorem,
in a sense, is the
581
00:28:05,060 --> 00:28:07,980
inverse of the first fundamental
theorem.
582
00:28:07,980 --> 00:28:11,110
Namely, it switches the roles.
583
00:28:11,110 --> 00:28:12,950
Namely, with the second
fundamental theorem, what we
584
00:28:12,950 --> 00:28:13,800
say is this.
585
00:28:13,800 --> 00:28:16,760
If we know how to find the
area under the curve, 'y'
586
00:28:16,760 --> 00:28:21,640
equals 'f of x', that gives us
a way of computing a function
587
00:28:21,640 --> 00:28:23,910
'G' whose derivative is 'f'.
588
00:28:23,910 --> 00:28:26,100
In fact, to summarize this--
589
00:28:26,100 --> 00:28:27,340
and I think this is
very important.
590
00:28:27,340 --> 00:28:28,860
Let me just summarize
this, then.
591
00:28:28,860 --> 00:28:32,560
First of all, the first
fundamental theorem allows us
592
00:28:32,560 --> 00:28:37,480
to compute this particular
limit, provided we can find a
593
00:28:37,480 --> 00:28:40,320
'G', such that 'G prime'
equals 'f'.
594
00:28:40,320 --> 00:28:43,690
In fact, in this case, the
limit, as 'n' approaches
595
00:28:43,690 --> 00:28:47,990
infinity, summation 'k' goes
from 1 to 'n', 'f of 'c sub
596
00:28:47,990 --> 00:28:52,660
k'' 'delta x', is given very,
very ingeniously, beautifully,
597
00:28:52,660 --> 00:28:57,110
and concisely by 'G of
b' minus 'G of a'.
598
00:28:57,110 --> 00:29:01,400
Secondly, the second fundamental
theorem allows us,
599
00:29:01,400 --> 00:29:05,010
given 'f', to construct
'G', such that 'G
600
00:29:05,010 --> 00:29:06,480
prime' equals 'f'.
601
00:29:06,480 --> 00:29:10,430
Namely, the required 'G' is
just the definite integral
602
00:29:10,430 --> 00:29:16,095
from 'a' to 'x1', 'f of x' 'dx',
which in turn is a limit
603
00:29:16,095 --> 00:29:20,120
of a network of rectangles,
namely the limit as 'n'
604
00:29:20,120 --> 00:29:24,420
approaches infinity, summation
'k' goes from 1 to 'n', 'f of
605
00:29:24,420 --> 00:29:27,040
'c sub k'' times 'delta x'.
606
00:29:27,040 --> 00:29:32,150
This then is perhaps the most
beautiful lecture in calculus,
607
00:29:32,150 --> 00:29:33,280
not necessarily the
way I've given it,
608
00:29:33,280 --> 00:29:34,620
but in terms of what?
609
00:29:34,620 --> 00:29:39,250
Relating two apparently diverse
branches of calculus,
610
00:29:39,250 --> 00:29:41,330
integral and differential
calculus.
611
00:29:41,330 --> 00:29:43,210
This is a difficult subject.
612
00:29:43,210 --> 00:29:45,980
I have taken great pains to try
to write this clearly in
613
00:29:45,980 --> 00:29:47,250
the supplementary notes.
614
00:29:47,250 --> 00:29:50,850
I have tried to pick typical
exercises that bring home
615
00:29:50,850 --> 00:29:52,550
these highlights.
616
00:29:52,550 --> 00:29:57,470
And in addition, our next two
lectures will emphasize the
617
00:29:57,470 --> 00:30:02,150
relationship between derivatives
and integrals as
618
00:30:02,150 --> 00:30:04,540
we study volume and
arc length.
619
00:30:04,540 --> 00:30:06,670
At any rate, until next
time, goodbye.
620
00:30:06,670 --> 00:30:09,710
621
00:30:09,710 --> 00:30:12,240
MALE SPEAKER: Funding for the
publication of this video was
622
00:30:12,240 --> 00:30:16,960
provided by the Gabriella and
Paul Rosenbaum Foundation.
623
00:30:16,960 --> 00:30:21,130
Help OCW continue to provide
free and open access to MIT
624
00:30:21,130 --> 00:30:25,330
courses by making a donation
at ocw.mit.edu/donate.
625
00:30:25,330 --> 00:30:30,079