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PROFESSOR: Hi.
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Today we begin our study of
integral calculus, which had
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its roots back in ancient
Greece, roughly 600 BC, and
13
00:00:42,240 --> 00:00:45,560
began with an investigation
of the study of area.
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00:00:45,560 --> 00:00:47,520
In a certain manner of speaking,
today's lecture
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could be called 'Calculus
Revisited Revisited', in the
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00:00:51,000 --> 00:00:54,580
sense that integral calculus
can be studied quite apart
17
00:00:54,580 --> 00:00:56,140
from the fact that differential
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00:00:56,140 --> 00:00:57,900
calculus was ever invented.
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00:00:57,900 --> 00:01:01,260
We'll see later in this block
of material that there is a
20
00:01:01,260 --> 00:01:04,129
wondrous relationship between
integral and differential
21
00:01:04,129 --> 00:01:07,480
calculus, but perhaps the best
proof that integral calculus
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00:01:07,480 --> 00:01:11,160
can be studied independently of
differential calculus lies
23
00:01:11,160 --> 00:01:14,700
in the fact that the ancient
Greek was doing integral
24
00:01:14,700 --> 00:01:19,010
calculus in 600 BC, whereas
differential calculus did not
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00:01:19,010 --> 00:01:23,690
begin until about 1680 AD
with Sir Isaac Newton.
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00:01:23,690 --> 00:01:26,980
Well, at any rate then, what
we'll call today's lecture is
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00:01:26,980 --> 00:01:29,420
simply 'Two-dimensional Area'.
28
00:01:29,420 --> 00:01:31,900
In other words, this is how
the subject began, with
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00:01:31,900 --> 00:01:36,240
studying the amount of space
in plane regions,
30
00:01:36,240 --> 00:01:39,370
two-dimensional area, as opposed
to, say, as we'll talk
31
00:01:39,370 --> 00:01:43,070
about later, three-dimensional
area, which is really a fancy
32
00:01:43,070 --> 00:01:44,700
word for volume, et cetera.
33
00:01:44,700 --> 00:01:46,650
But enough about that
for the time being.
34
00:01:46,650 --> 00:01:50,350
Two-dimensional area, or
Calculus Revisited Revisited.
35
00:01:50,350 --> 00:01:53,780
And to see what's happening over
here, the study of area
36
00:01:53,780 --> 00:01:57,790
was the forerunner of integral
calculus as we now know it.
37
00:01:57,790 --> 00:02:01,630
It began in ancient Greece,
roughly 600 BC.
38
00:02:01,630 --> 00:02:04,840
The branch of calculus that
we've been studying up until
39
00:02:04,840 --> 00:02:08,229
now in our course, differential
calculus, did not
40
00:02:08,229 --> 00:02:10,979
begin until 1680 AD.
41
00:02:10,979 --> 00:02:14,730
Notice, then, the interesting
juxtaposition in time.
42
00:02:14,730 --> 00:02:17,510
In other words, pedagogically,
one seems to study
43
00:02:17,510 --> 00:02:20,790
differential calculus before
integral calculus.
44
00:02:20,790 --> 00:02:24,450
Chronologically, integral
calculus preceded differential
45
00:02:24,450 --> 00:02:27,670
calculus by more than
2,000 years.
46
00:02:27,670 --> 00:02:30,110
Now this is a rather
beautiful study.
47
00:02:30,110 --> 00:02:33,260
It's one of the most aesthetic
parts of elementary
48
00:02:33,260 --> 00:02:36,800
mathematics, namely the
simplicity with which the
49
00:02:36,800 --> 00:02:41,220
Greek was able to tackle the
sophisticated problem of
50
00:02:41,220 --> 00:02:43,160
finding areas in general.
51
00:02:43,160 --> 00:02:46,500
He started with three basically
simple properties of
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00:02:46,500 --> 00:02:50,430
area, so simple that most of us
are willing to accept them
53
00:02:50,430 --> 00:02:52,070
almost intuitively.
54
00:02:52,070 --> 00:02:54,740
We'll call these the
'Axioms for Area'.
55
00:02:54,740 --> 00:02:56,830
And simply stated, they
were the following.
56
00:02:56,830 --> 00:03:00,250
One, that the area of a
rectangle is the base times
57
00:03:00,250 --> 00:03:01,150
the height.
58
00:03:01,150 --> 00:03:06,060
Two, if one region is contained
within another, the
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00:03:06,060 --> 00:03:09,830
area of the contained region is
less than or equal to-- in
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00:03:09,830 --> 00:03:11,740
other words, can be no greater
than-- that of
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00:03:11,740 --> 00:03:12,990
the containing region.
62
00:03:12,990 --> 00:03:15,390
Roughly speaking, the
smaller the region,
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00:03:15,390 --> 00:03:17,160
the smaller its area.
64
00:03:17,160 --> 00:03:21,220
And finally, written slightly
more formally here, the old
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00:03:21,220 --> 00:03:24,670
high-school axiom for plane
geometry, that the area of the
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00:03:24,670 --> 00:03:27,700
whole equals the sum of the
areas of the parts, that if a
67
00:03:27,700 --> 00:03:32,580
region 'R' is subdivided into a
union of mutually exclusive
68
00:03:32,580 --> 00:03:36,670
pieces, then the area of the
entire region is equal to the
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00:03:36,670 --> 00:03:39,640
sum of the areas of the
constituent parts.
70
00:03:39,640 --> 00:03:43,290
And with just these three
axioms, the ancient Greek
71
00:03:43,290 --> 00:03:47,320
invented a technique for finding
areas that today is
72
00:03:47,320 --> 00:03:50,250
still known by its original
name, the 'Method of
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00:03:50,250 --> 00:03:51,980
Exhaustion'.
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00:03:51,980 --> 00:03:55,850
And here, exhaustion does not
refer to the fact that we get
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00:03:55,850 --> 00:03:59,210
tired using the system, even
though, as we shall soon see,
76
00:03:59,210 --> 00:04:00,380
it's quite intricate.
77
00:04:00,380 --> 00:04:03,530
It refers to the fact that we
take the region whose area we
78
00:04:03,530 --> 00:04:09,540
want to find and exhaust the
space by squeezing it between
79
00:04:09,540 --> 00:04:12,690
regions which are made
up of rectangles.
80
00:04:12,690 --> 00:04:16,140
Stated more precisely,
given a region 'R'--
81
00:04:16,140 --> 00:04:19,220
and I'll define this more
precisely later-- we squeeze
82
00:04:19,220 --> 00:04:22,089
it between two networks
of rectangles,
83
00:04:22,089 --> 00:04:23,570
the idea being what?
84
00:04:23,570 --> 00:04:26,550
That we know what the area
of a rectangle is, and
85
00:04:26,550 --> 00:04:30,090
consequently, by the fact that
the area of the whole equals
86
00:04:30,090 --> 00:04:32,630
the sum of the areas of the
parts, if we have a
87
00:04:32,630 --> 00:04:35,940
rectangular network, knowing how
to find the area of each
88
00:04:35,940 --> 00:04:38,460
rectangle, we know how
to find the area of
89
00:04:38,460 --> 00:04:40,020
the rectangular network.
90
00:04:40,020 --> 00:04:43,470
Now, rather than to wax on like
this philosophically,
91
00:04:43,470 --> 00:04:45,550
let's tackle a specific
problem.
92
00:04:45,550 --> 00:04:49,540
In fact, give or take a little
bit, this is specifically the
93
00:04:49,540 --> 00:04:53,470
problem that Archimedes dealt
with in finding the area of
94
00:04:53,470 --> 00:04:56,800
parabolic segments way back,
as I say, in the
95
00:04:56,800 --> 00:05:00,050
300 to 600 BC era.
96
00:05:00,050 --> 00:05:01,930
As an example, consider
the following.
97
00:05:01,930 --> 00:05:06,160
We wish to determine the area of
the region 'R', 'A sub R',
98
00:05:06,160 --> 00:05:10,330
where the region 'R' is that
region which is bounded above
99
00:05:10,330 --> 00:05:15,930
by the curve 'y' equals 'x
squared', below by the x-axis,
100
00:05:15,930 --> 00:05:19,300
on the left by the y-axis,
and on the right by the
101
00:05:19,300 --> 00:05:20,860
line 'x' equals 1.
102
00:05:20,860 --> 00:05:23,960
In other words, this
region in here.
103
00:05:23,960 --> 00:05:26,950
Now here's the way we put the
squeeze on it, and we'll start
104
00:05:26,950 --> 00:05:29,410
this quite gradually.
105
00:05:29,410 --> 00:05:32,900
The first thing that we observe
is that if we draw a
106
00:05:32,900 --> 00:05:37,170
line parallel to the x-axis
through the point (1 , 1), the
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00:05:37,170 --> 00:05:42,000
highest point of our region
here, we construct a rectangle
108
00:05:42,000 --> 00:05:44,390
which contains our region 'R'.
109
00:05:44,390 --> 00:05:48,000
And since 'R' is contained in
the rectangle, the area of the
110
00:05:48,000 --> 00:05:51,410
region 'R' must be less than
the area of the rectangle.
111
00:05:51,410 --> 00:05:53,650
But notice that this particular
rectangle that
112
00:05:53,650 --> 00:05:57,020
we've just constructed has its
base equal to 1, its height
113
00:05:57,020 --> 00:05:58,000
equal to 1.
114
00:05:58,000 --> 00:06:03,160
Therefore by our first axiom,
its area is 1 times 1, or 1.
115
00:06:03,160 --> 00:06:07,140
Also, since intuitively, area
is a positive thing, we see
116
00:06:07,140 --> 00:06:09,950
just from this quick diagram
that whatever the area of the
117
00:06:09,950 --> 00:06:14,370
region is, we now have it
bounded between 0 and 1.
118
00:06:14,370 --> 00:06:15,410
We have an upper bound.
119
00:06:15,410 --> 00:06:17,180
We have a lower bound.
120
00:06:17,180 --> 00:06:19,430
We have this thing caught, OK?
121
00:06:19,430 --> 00:06:21,760
Now, there's still a lot of
space between 0 and 1.
122
00:06:21,760 --> 00:06:25,110
The method of exhaustion
refines this idea.
123
00:06:25,110 --> 00:06:27,460
Namely, what we do next
is we say, look at--
124
00:06:27,460 --> 00:06:31,020
instead of just drawing one big
rectangle like this, why
125
00:06:31,020 --> 00:06:34,010
don't we partition the base
of our region 'R'
126
00:06:34,010 --> 00:06:35,230
into two equal parts?
127
00:06:35,230 --> 00:06:38,510
In other words, let's locate the
point 1/2, which of course
128
00:06:38,510 --> 00:06:41,090
is midway between 0 and 1.
129
00:06:41,090 --> 00:06:46,230
Now what I can do is I can
circumscribe two rectangles,
130
00:06:46,230 --> 00:06:50,400
one of which is the rectangle
whose height corresponds to
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00:06:50,400 --> 00:06:52,890
the x-coordinate
equaling 1/2--
132
00:06:52,890 --> 00:06:55,450
that means the y-coordinate
is 1/4--
133
00:06:55,450 --> 00:06:59,160
and the second rectangle, the
one whose x-coordinate--
134
00:06:59,160 --> 00:07:02,350
the x-coordinate of the height
is 1, and since 'y' equals 'x
135
00:07:02,350 --> 00:07:04,600
squared', the height
is also 1.
136
00:07:04,600 --> 00:07:07,960
You see, notice that by picking
a region which is a
137
00:07:07,960 --> 00:07:11,000
curve which is always rising,
the lowest point of each
138
00:07:11,000 --> 00:07:13,780
region is the point that's
furthest to the left.
139
00:07:13,780 --> 00:07:16,380
And the highest point of each
region is the point that's
140
00:07:16,380 --> 00:07:17,510
furthest to the right.
141
00:07:17,510 --> 00:07:20,980
And keeping this in mind, you
see by making rectangles
142
00:07:20,980 --> 00:07:24,110
corresponding to the points
furthest to the right in each
143
00:07:24,110 --> 00:07:28,170
interval, I get a rectangle
which contains the portion of
144
00:07:28,170 --> 00:07:30,020
the region 'R' that
I'm talking about.
145
00:07:30,020 --> 00:07:33,270
In other words, if I look at
this rectangular network, my
146
00:07:33,270 --> 00:07:35,770
region 'R' is contained
inside that.
147
00:07:35,770 --> 00:07:40,550
Consequently, its area is less
than the area of the
148
00:07:40,550 --> 00:07:41,860
rectangular network.
149
00:07:41,860 --> 00:07:44,350
Now, what is the area of this
rectangular network?
150
00:07:44,350 --> 00:07:48,740
Well, the big rectangle has its
base equal to 1/2 and its
151
00:07:48,740 --> 00:07:50,030
height equal to 1.
152
00:07:50,030 --> 00:07:51,970
So its area is 1/2.
153
00:07:51,970 --> 00:07:55,370
The small rectangle has its
base equal to 1/2 and its
154
00:07:55,370 --> 00:07:56,720
height equal to 1/4.
155
00:07:56,720 --> 00:07:58,390
Since its area is the
base times the
156
00:07:58,390 --> 00:08:00,390
height, its area is 1/8.
157
00:08:00,390 --> 00:08:02,960
1/8 plus 1/2 is 5/8.
158
00:08:02,960 --> 00:08:06,420
In other words, the area of the
rectangular network which
159
00:08:06,420 --> 00:08:08,260
contains 'R' is 5/8.
160
00:08:08,260 --> 00:08:10,450
Consequently, the area of
the region 'R' must
161
00:08:10,450 --> 00:08:12,050
be less than 5/8.
162
00:08:12,050 --> 00:08:16,040
On the other hand, notice that
the lowest point in the second
163
00:08:16,040 --> 00:08:18,900
interval corresponds to
'x' equaling 1/2.
164
00:08:18,900 --> 00:08:22,310
In other words, notice that
this particular rectangle,
165
00:08:22,310 --> 00:08:25,710
whose base is 1/2 and whose
height is 1/4, that this
166
00:08:25,710 --> 00:08:29,440
rectangle here is inscribed
in the region 'R'.
167
00:08:29,440 --> 00:08:31,170
Its area is 1/8.
168
00:08:31,170 --> 00:08:34,380
And since it's contained within
the region 'R', its
169
00:08:34,380 --> 00:08:36,760
area must be less than the
area of the region 'R'.
170
00:08:36,760 --> 00:08:38,880
And so we're now certain that
whatever the area of the
171
00:08:38,880 --> 00:08:42,549
region 'R' is, it's between
1/8 and 5/8.
172
00:08:42,549 --> 00:08:45,510
And by the way, notice in terms
of these shaded regions,
173
00:08:45,510 --> 00:08:48,680
that even though this is an
approximation which is too
174
00:08:48,680 --> 00:08:51,700
large to be the right answer,
it is closer to being the
175
00:08:51,700 --> 00:08:54,420
right answer than this
approximation here.
176
00:08:54,420 --> 00:08:56,210
In other words, notice that
the difference between the
177
00:08:56,210 --> 00:08:59,480
overestimate here and the
overestimate here is this
178
00:08:59,480 --> 00:09:03,220
amount in here.
179
00:09:03,220 --> 00:09:05,120
You see, we've chopped off
part of the error.
180
00:09:05,120 --> 00:09:07,370
But we'll talk about that,
as I say, in more
181
00:09:07,370 --> 00:09:08,470
detail in the notes.
182
00:09:08,470 --> 00:09:11,220
What I'd like to do is to give
you an overall view of what's
183
00:09:11,220 --> 00:09:12,440
happening over here.
184
00:09:12,440 --> 00:09:15,280
You see, to generalize the
method of exhaustion, we do
185
00:09:15,280 --> 00:09:17,010
what the mathematician
usually does.
186
00:09:17,010 --> 00:09:19,980
Instead of saying, let's divide
the base of the region
187
00:09:19,980 --> 00:09:23,250
into two equal parts, or three
equal parts, or four equal
188
00:09:23,250 --> 00:09:27,440
parts, we say, let's divide
it into 'n' equal parts.
189
00:09:27,440 --> 00:09:30,870
Now if we divide this base into
'n' equal parts, since
190
00:09:30,870 --> 00:09:32,370
the whole region--
191
00:09:32,370 --> 00:09:35,720
the whole length is 0 to 1,
dividing into 'n' equal parts
192
00:09:35,720 --> 00:09:39,550
means that my points of division
will be '1/n', '2/n',
193
00:09:39,550 --> 00:09:44,520
'3/n', et cetera, right down
to 'n/n', which is 1.
194
00:09:44,520 --> 00:09:50,460
What I do now is I pick the
right endpoint of each point
195
00:09:50,460 --> 00:09:53,720
in that partition to
draw my rectangle.
196
00:09:53,720 --> 00:09:56,370
You see, each of these
rectangles that I draw this
197
00:09:56,370 --> 00:10:00,130
way contains the corresponding
part of my region 'R'.
198
00:10:00,130 --> 00:10:05,530
Consequently, the area of this
rectangular network must be an
199
00:10:05,530 --> 00:10:08,660
upper bound for the area
of my region 'R'.
200
00:10:08,660 --> 00:10:11,370
Now, because it's going to be an
upper bound, and because it
201
00:10:11,370 --> 00:10:14,830
depends on what the value of
'n' is, I will denote that
202
00:10:14,830 --> 00:10:17,670
upper bound by 'U sub n'.
203
00:10:17,670 --> 00:10:19,350
And what will 'U sub n' be?
204
00:10:19,350 --> 00:10:21,980
It's the sum of the areas
of these circumscribed
205
00:10:21,980 --> 00:10:23,010
rectangles.
206
00:10:23,010 --> 00:10:25,320
And it's not difficult to
see from this picture.
207
00:10:25,320 --> 00:10:26,820
Let's take a look.
208
00:10:26,820 --> 00:10:30,280
Since this rectangle has its
height corresponding to the
209
00:10:30,280 --> 00:10:33,980
x-coordinate equaling '1/n', and
since the y-coordinate is
210
00:10:33,980 --> 00:10:36,090
the square of the x-coordinate,
the height of
211
00:10:36,090 --> 00:10:38,780
this rectangle will be
''1/n' squared'.
212
00:10:38,780 --> 00:10:40,320
The base is '1/n'.
213
00:10:40,320 --> 00:10:43,260
So the area of that
circumscribed rectangle is
214
00:10:43,260 --> 00:10:46,180
''1/n' squared' times '1/n'.
215
00:10:46,180 --> 00:10:49,600
Similarly, the next rectangle
here has its height
216
00:10:49,600 --> 00:10:52,690
corresponding to the
x-coordinate '2/n'.
217
00:10:52,690 --> 00:10:55,550
So its height is ''2/n'
squared'.
218
00:10:55,550 --> 00:10:57,530
Keep in mind the y-coordinate
is the square of the
219
00:10:57,530 --> 00:10:58,880
x-coordinate here.
220
00:10:58,880 --> 00:11:01,730
The base of all of these
rectangles is '1/n'.
221
00:11:01,730 --> 00:11:04,930
And proceeding in this way, I
finally come down to the last
222
00:11:04,930 --> 00:11:07,520
rectangle, whose
height is what?
223
00:11:07,520 --> 00:11:11,080
Well, the x-coordinate is 1,
so the y-coordinate is 1.
224
00:11:11,080 --> 00:11:14,400
To keep my form here, I'll
write that as 'n/n'.
225
00:11:14,400 --> 00:11:18,320
Its height is therefore
''n/n' squared'.
226
00:11:18,320 --> 00:11:21,810
The base of that rectangle is
'1/n', so the area of that
227
00:11:21,810 --> 00:11:26,760
last rectangle is ''n/n'
squared' times '1/n'.
228
00:11:26,760 --> 00:11:28,330
Now you see that's what?
229
00:11:28,330 --> 00:11:32,170
That's an area which is too
large to be 'A sub R'.
230
00:11:32,170 --> 00:11:34,740
Well, let's take a look now and
see what's happening here.
231
00:11:34,740 --> 00:11:38,510
Notice that each of these terms
has an 'n cubed' in the
232
00:11:38,510 --> 00:11:39,450
denominator.
233
00:11:39,450 --> 00:11:42,060
So I can factor out the
'n cubed' term.
234
00:11:42,060 --> 00:11:44,050
And what I'm left
with is what?
235
00:11:44,050 --> 00:11:46,370
The sum of the first
'n' squares.
236
00:11:46,370 --> 00:11:49,995
1 squared plus 2 squared plus 3
squared, et cetera, plus et
237
00:11:49,995 --> 00:11:51,180
cetera, 'n squared'.
238
00:11:51,180 --> 00:11:54,340
In other words, this tells me
how to find 'U sub n' for any
239
00:11:54,340 --> 00:11:55,370
value of 'n'.
240
00:11:55,370 --> 00:11:58,040
And we'll talk about that some
more in a little while.
241
00:11:58,040 --> 00:11:59,690
That's an upper bound.
242
00:11:59,690 --> 00:12:03,810
In a corresponding way, I can
also find a lower bound.
243
00:12:03,810 --> 00:12:08,350
Namely, what I'll do now is pick
the smallest rectangle in
244
00:12:08,350 --> 00:12:08,910
each region.
245
00:12:08,910 --> 00:12:11,860
In other words, I'll now
inscribe rectangles inside my
246
00:12:11,860 --> 00:12:16,140
region 'R', and therefore find
a rectangular region whose
247
00:12:16,140 --> 00:12:19,150
area is less than that
of the region 'R'.
248
00:12:19,150 --> 00:12:22,080
And to do that, without going
through the details, notice
249
00:12:22,080 --> 00:12:24,970
that essentially all I have to
do is shift each of these
250
00:12:24,970 --> 00:12:28,010
rectangles over by
one partition.
251
00:12:28,010 --> 00:12:33,180
Namely, notice, for example,
that the smallest height in
252
00:12:33,180 --> 00:12:36,280
the second partition here is the
one which corresponds to
253
00:12:36,280 --> 00:12:37,530
the height '1/n'.
254
00:12:37,530 --> 00:12:40,010
In other words, what I do
now is, to inscribe the
255
00:12:40,010 --> 00:12:43,430
rectangles, I just shift
everything over like this.
256
00:12:43,430 --> 00:12:46,660
And leaving the details out, and
letting you verify these
257
00:12:46,660 --> 00:12:48,510
for yourselves, I
can again mimic
258
00:12:48,510 --> 00:12:50,160
exactly what I did before.
259
00:12:50,160 --> 00:12:54,030
The only difference being now,
that instead of the height of
260
00:12:54,030 --> 00:12:59,690
the last rectangle being 'n/n',
notice that it's 'n -
261
00:12:59,690 --> 00:13:01,160
1' over 'n', squared.
262
00:13:01,160 --> 00:13:02,650
In other words, the
x-coordinate is
263
00:13:02,650 --> 00:13:04,140
'n - 1' over 'n'.
264
00:13:04,140 --> 00:13:06,620
The y-coordinate is the square
of the x-coordinate.
265
00:13:06,620 --> 00:13:09,340
In other words, notice that this
point here gives rise to
266
00:13:09,340 --> 00:13:12,620
the height of the lowest
rectangle that can be
267
00:13:12,620 --> 00:13:14,490
inscribed in my last
portion here.
268
00:13:14,490 --> 00:13:17,970
So without further ado, it turns
out that 'L sub n', the
269
00:13:17,970 --> 00:13:22,060
lower estimate, is '1 over 'n
cubed'' times the sum of the
270
00:13:22,060 --> 00:13:24,820
first 'n - 1' squares.
271
00:13:24,820 --> 00:13:25,710
OK.
272
00:13:25,710 --> 00:13:28,540
Now, let's keep track of just
what it is that we've done
273
00:13:28,540 --> 00:13:30,380
over here so far.
274
00:13:30,380 --> 00:13:31,410
What we've done is what?
275
00:13:31,410 --> 00:13:37,890
For each 'n', we have squeezed
'A sub R' between two numbers,
276
00:13:37,890 --> 00:13:40,800
one number being an upper
approximation, one number
277
00:13:40,800 --> 00:13:42,720
being a lower approximation.
278
00:13:42,720 --> 00:13:45,370
Now notice that the 'U sub
n' and the 'L sub n' are
279
00:13:45,370 --> 00:13:46,880
functions of 'n'.
280
00:13:46,880 --> 00:13:48,380
'A sub R' is a constant.
281
00:13:48,380 --> 00:13:50,070
That's the area of
the region 'R'.
282
00:13:50,070 --> 00:13:52,670
What the method of exhaustion
means is simply this.
283
00:13:52,670 --> 00:13:53,710
We say, look.
284
00:13:53,710 --> 00:13:56,860
Let's take the limit of the 'L
sub n's as 'n' approaches
285
00:13:56,860 --> 00:14:00,630
infinity, and let's take the
limit of the 'U sub n's as 'n'
286
00:14:00,630 --> 00:14:04,095
approaches infinity, observing
that for each 'n', 'A sub R'
287
00:14:04,095 --> 00:14:06,010
is squeezed between these two.
288
00:14:06,010 --> 00:14:08,960
Consequently, since 'A sub R' is
squeezed between these two
289
00:14:08,960 --> 00:14:12,640
for each 'n', it must be
squeezed between these two
290
00:14:12,640 --> 00:14:13,770
when we go to the limit.
291
00:14:13,770 --> 00:14:18,100
In other words, whatever 'A sub
R' is, it must be between
292
00:14:18,100 --> 00:14:19,460
these two limits.
293
00:14:19,460 --> 00:14:21,170
Now here's the key step.
294
00:14:21,170 --> 00:14:25,420
It turns out, at least in this
particular problem, that the
295
00:14:25,420 --> 00:14:29,700
limit of 'L sub n' as 'n' goes
to infinity is the same as the
296
00:14:29,700 --> 00:14:32,180
limit of 'U sub n' as 'n'
goes to infinity.
297
00:14:32,180 --> 00:14:34,740
And the best way to see that
without becoming too fancy at
298
00:14:34,740 --> 00:14:38,880
this stage of the game is to
observe that for a given 'n',
299
00:14:38,880 --> 00:14:41,120
the difference between
'U sub n' and 'L
300
00:14:41,120 --> 00:14:43,880
sub n' is just '1/n'.
301
00:14:43,880 --> 00:14:47,560
And again, I'll just indicate
to you from this diagram how
302
00:14:47,560 --> 00:14:48,630
we can see that.
303
00:14:48,630 --> 00:14:51,820
Notice that back here, the way
we went from 'U sub n' to 'L
304
00:14:51,820 --> 00:14:54,580
sub n' was we just pushed
over this whole
305
00:14:54,580 --> 00:14:56,370
network by one unit.
306
00:14:56,370 --> 00:14:59,770
In other words, the rectangle
that we squeezed out in going
307
00:14:59,770 --> 00:15:03,050
from the upper sum to the
lower sum was just this
308
00:15:03,050 --> 00:15:05,940
particular rectangle whose
height was 1 and
309
00:15:05,940 --> 00:15:07,600
whose base was '1/n'.
310
00:15:07,600 --> 00:15:10,720
In other words, the area that
was kicked out in going from
311
00:15:10,720 --> 00:15:13,870
'L sub n' to 'U sub
n' is '1/n'.
312
00:15:13,870 --> 00:15:16,400
This is what this thing
here says here.
313
00:15:16,400 --> 00:15:18,180
This is worked out in
detail in the notes.
314
00:15:18,180 --> 00:15:20,440
But again, I just want to go
through this thing quickly so
315
00:15:20,440 --> 00:15:22,500
we get the idea of what's
happening over here.
316
00:15:22,500 --> 00:15:26,490
In other words, since 'U sub n'
minus 'L sub n' is '1/n',
317
00:15:26,490 --> 00:15:28,170
that's just another way
of saying what?
318
00:15:28,170 --> 00:15:30,610
That the limit of this
difference is just the limit
319
00:15:30,610 --> 00:15:32,720
of '1/n' as 'n' goes
to infinity.
320
00:15:32,720 --> 00:15:34,260
But that's clearly 0.
321
00:15:34,260 --> 00:15:38,140
In other words, these two limits
are the same because
322
00:15:38,140 --> 00:15:40,840
their difference goes
to 0 in the limit.
323
00:15:40,840 --> 00:15:43,960
Consequently, since these two
things are the same, and 'A
324
00:15:43,960 --> 00:15:48,170
sub R' is caught between these
two, it must be that 'A sub R'
325
00:15:48,170 --> 00:15:51,320
is equal to this common limit.
326
00:15:51,320 --> 00:15:54,390
And that's precisely what the
method of exhaustion was.
327
00:15:54,390 --> 00:15:57,880
We squeeze what we were looking
for between two
328
00:15:57,880 --> 00:16:00,070
estimates which converge
towards each
329
00:16:00,070 --> 00:16:01,450
other as 'n' got large.
330
00:16:01,450 --> 00:16:04,130
Now, by the way, I'm going to
point out the fact that it's
331
00:16:04,130 --> 00:16:06,510
very, very difficult
in general to find
332
00:16:06,510 --> 00:16:07,480
what this limit is.
333
00:16:07,480 --> 00:16:09,220
We're going to see
that in working a
334
00:16:09,220 --> 00:16:10,440
few specific problems.
335
00:16:10,440 --> 00:16:12,590
What I thought might
be informative--
336
00:16:12,590 --> 00:16:15,000
without going through the
details right now, I have
337
00:16:15,000 --> 00:16:18,800
taken the liberty of computing
'L sub n' and 'U sub n' in
338
00:16:18,800 --> 00:16:21,970
this problem for 'n'
equals 1,000.
339
00:16:21,970 --> 00:16:24,980
In other words, if we divided
this region into 1,000 equal
340
00:16:24,980 --> 00:16:28,350
parts and took the network of
circumscribed rectangles and
341
00:16:28,350 --> 00:16:32,710
inscribed rectangles, it would
turn out that the area of the
342
00:16:32,710 --> 00:16:40,690
inscribed rectangles would
be 0.3328335, and for the
343
00:16:40,690 --> 00:16:46,950
circumscribed rectangles,
0.3338335.
344
00:16:46,950 --> 00:16:49,300
And by the way, notice how
this is borne out.
345
00:16:49,300 --> 00:16:52,440
Notice that the difference
between these two is precisely
346
00:16:52,440 --> 00:16:57,270
1/1000, and that is '1/n'
with 'n' equal to 1,000.
347
00:16:57,270 --> 00:16:59,310
But that's not the important
point here.
348
00:16:59,310 --> 00:17:01,980
The important point is that just
looking at this decimal
349
00:17:01,980 --> 00:17:05,359
expansion, if I didn't know
anything else, I know that 'A
350
00:17:05,359 --> 00:17:08,720
sub R' is caught between
these two.
351
00:17:08,720 --> 00:17:12,200
Consequently, to two decimal
places, I can be sure that 'A
352
00:17:12,200 --> 00:17:15,020
sub R' is 0.33.
353
00:17:15,020 --> 00:17:17,470
And notice that if I want more
and more decimal place
354
00:17:17,470 --> 00:17:21,579
accuracy, especially if I have
access to a desk calculator, I
355
00:17:21,579 --> 00:17:24,040
don't really have to compute
the limit exactly.
356
00:17:24,040 --> 00:17:27,490
I can feed the formula into
the machine and put the
357
00:17:27,490 --> 00:17:29,740
squeeze on and get as
many decimal place
358
00:17:29,740 --> 00:17:31,940
accuracy as I want.
359
00:17:31,940 --> 00:17:34,720
By the way, as an aside, this is
exactly what we did in high
360
00:17:34,720 --> 00:17:39,470
school, when we said things
like let pi equal 22/7.
361
00:17:39,470 --> 00:17:42,090
Pi is not equal to 22/7.
362
00:17:42,090 --> 00:17:45,450
Among other things, 22/7
is a rational number.
363
00:17:45,450 --> 00:17:47,030
It's the ratio of two
whole numbers.
364
00:17:47,030 --> 00:17:48,610
Pi is an irrational number.
365
00:17:48,610 --> 00:17:51,260
What people really meant was
that you can't tell the
366
00:17:51,260 --> 00:17:55,470
difference between pi and 22/7
to two decimal places.
367
00:17:55,470 --> 00:17:58,050
They both begin 3.14.
368
00:17:58,050 --> 00:18:01,200
And consequently, if all you
wanted to measure was to two
369
00:18:01,200 --> 00:18:04,200
decimal digits, you could
use 22/7 as the
370
00:18:04,200 --> 00:18:05,610
approximation for pi.
371
00:18:05,610 --> 00:18:08,150
But if you wanted to squeeze out
more places, you'd have to
372
00:18:08,150 --> 00:18:10,210
use a more refined approach.
373
00:18:10,210 --> 00:18:13,200
By the way, let me point out
that one of the reasons that I
374
00:18:13,200 --> 00:18:17,130
chose the curve 'y' equals 'x
squared' to work with was--
375
00:18:17,130 --> 00:18:22,010
you may recall that back in our
discussion of mathematical
376
00:18:22,010 --> 00:18:25,890
induction, one of the problems
that we worked on was to show
377
00:18:25,890 --> 00:18:29,830
how we find by induction the
recipe for the sum of the
378
00:18:29,830 --> 00:18:31,740
first 'n' squares.
379
00:18:31,740 --> 00:18:36,260
And by way of review, let me
recall for you the fact that
380
00:18:36,260 --> 00:18:39,830
the sum of the first 'n' squares
was given by ''n'
381
00:18:39,830 --> 00:18:45,220
times 'n + 1' times
'2n + 1'' over 6.
382
00:18:45,220 --> 00:18:45,860
OK?
383
00:18:45,860 --> 00:18:49,250
In other words, going back to
our recipe for 'U sub n', I
384
00:18:49,250 --> 00:18:52,570
can now replace 1 squared plus
2 squared plus et cetera 'n
385
00:18:52,570 --> 00:18:57,580
squared' by ''n' times 'n +
1' times '2n + 1'' over 6.
386
00:18:57,580 --> 00:19:01,570
And if I now divide through by
'n cubed' judiciously, namely
387
00:19:01,570 --> 00:19:05,330
canceling out one 'n' with this
'n', dividing 'n + 1' by
388
00:19:05,330 --> 00:19:10,800
'n' and '2n + 1' by 'n', I get
that 'U sub n' is '1/6 'n + 1'
389
00:19:10,800 --> 00:19:13,440
over 'n'' times ''2n
+ 1' over 'n''.
390
00:19:13,440 --> 00:19:17,510
And that can be written even
more suggestively as 1/6 times
391
00:19:17,510 --> 00:19:20,680
'1 + '1/n'' times '2 + '1/n''.
392
00:19:20,680 --> 00:19:23,100
And I can now put a very
nice mathematical
393
00:19:23,100 --> 00:19:24,610
interpretation on this.
394
00:19:24,610 --> 00:19:29,680
Mainly, notice that no matter
how big 'n' is, '1 + '1/n'' is
395
00:19:29,680 --> 00:19:32,880
bigger than 1, you see, because
'1/n' is positive.
396
00:19:32,880 --> 00:19:35,350
And '2 + '1/n'' is
bigger than 2.
397
00:19:35,350 --> 00:19:40,450
Consequently, whatever this is,
it's bigger than 1/6 times
398
00:19:40,450 --> 00:19:43,930
1 times 2, which is 1/3.
399
00:19:43,930 --> 00:19:48,790
In other words, for each 'n', 'U
sub n' is greater than 1/3.
400
00:19:48,790 --> 00:19:52,340
It also happens that
as 'n' gets larger,
401
00:19:52,340 --> 00:19:54,300
'U sub n' gets smaller.
402
00:19:54,300 --> 00:19:58,060
Again, looking at this recipe,
the bigger 'n' is, the bigger
403
00:19:58,060 --> 00:20:00,890
is our denominator, and the
bigger the denominator, the
404
00:20:00,890 --> 00:20:02,950
smaller the fraction.
405
00:20:02,950 --> 00:20:07,560
And finally, notice that as 'n'
goes to infinity, 'U sub
406
00:20:07,560 --> 00:20:12,140
n' gets arbitrarily close to
1/3 in value, because '1/n'
407
00:20:12,140 --> 00:20:13,850
approaches 0 in the limit.
408
00:20:13,850 --> 00:20:15,580
In fact, it is 0 in the limit.
409
00:20:15,580 --> 00:20:19,380
So summarizing then, each 'U
sub n' is bigger than 1/3.
410
00:20:19,380 --> 00:20:23,280
As 'n' increases, 'U sub n'
decreases, and the limit of 'U
411
00:20:23,280 --> 00:20:26,490
sub n' as 'n' approaches
infinity is 1/3.
412
00:20:26,490 --> 00:20:29,790
Pictorially, what this means
is that if we locate 1/3 on
413
00:20:29,790 --> 00:20:35,580
the number line, the 'U sub n's
converge uniformly in the
414
00:20:35,580 --> 00:20:40,130
sense of moving steadily towards
the left, towards 1/3
415
00:20:40,130 --> 00:20:41,950
as the limit.
416
00:20:41,950 --> 00:20:43,770
Well, that's an upper squeeze.
417
00:20:43,770 --> 00:20:46,960
The lower squeeze comes from
the fact that in a similar
418
00:20:46,960 --> 00:20:51,000
way, we can show that
'L sub n' is 1/6--
419
00:20:51,000 --> 00:20:53,590
and look at how close this
comes to parallelling the
420
00:20:53,590 --> 00:20:59,460
structure of 'U sub n'-- '1 -
'1/n'' times '2 - '1/n''.
421
00:20:59,460 --> 00:21:02,310
Minuses here instead of pluses
as we had above.
422
00:21:02,310 --> 00:21:05,000
Now, mimicking what we did
before, notice now
423
00:21:05,000 --> 00:21:06,390
that for each 'n'--
424
00:21:06,390 --> 00:21:08,325
since we're subtracting
over here--
425
00:21:08,325 --> 00:21:12,220
for each 'n', 'L sub n'
is less than 1/3.
426
00:21:12,220 --> 00:21:14,940
But now since the fractions
get smaller as 'n' gets
427
00:21:14,940 --> 00:21:17,130
bigger, and you're subtracting
them, the
428
00:21:17,130 --> 00:21:19,120
difference becomes larger.
429
00:21:19,120 --> 00:21:21,770
In other words, now notice
that each 'L sub n'
430
00:21:21,770 --> 00:21:23,170
is less than 1/3.
431
00:21:23,170 --> 00:21:26,470
As 'n' increases, 'L
sub n' increases.
432
00:21:26,470 --> 00:21:28,540
And the limit of 'L sub
n' as 'n' approaches
433
00:21:28,540 --> 00:21:30,750
infinity is also 1/3.
434
00:21:30,750 --> 00:21:33,920
In other words, if we draw this
in back here, look what
435
00:21:33,920 --> 00:21:35,290
the 'L sub n's are doing.
436
00:21:35,290 --> 00:21:38,750
They're all less than 1/3, but
they move steadily towards the
437
00:21:38,750 --> 00:21:42,020
right as 'n' increases,
pushing in on 1/3.
438
00:21:42,020 --> 00:21:46,040
And since 'A sub R' is always
caught between these two, and
439
00:21:46,040 --> 00:21:49,460
these two converge relentlessly
upon 1/3, it must
440
00:21:49,460 --> 00:21:52,690
be, by definition, if there is
an area at all, that the area
441
00:21:52,690 --> 00:21:55,290
of the region 'R' must
be exactly 1/3.
442
00:21:55,290 --> 00:21:58,250
And by the way, just as a
quick aside, notice how
443
00:21:58,250 --> 00:22:01,840
important it is that we not
only have upper and lower
444
00:22:01,840 --> 00:22:03,790
bounds which converge.
445
00:22:03,790 --> 00:22:06,570
They must converge to
the same value.
446
00:22:06,570 --> 00:22:09,640
In other words, what I'm saying
is suppose all the 'U
447
00:22:09,640 --> 00:22:13,290
sub n's get arbitrarily close
to what I call script 'L sub
448
00:22:13,290 --> 00:22:16,120
1', and all the 'L sub n's
get arbitrarily close
449
00:22:16,120 --> 00:22:17,750
to 'L2' over here.
450
00:22:17,750 --> 00:22:22,160
Then all I'm saying is, all we
would know is that the area
451
00:22:22,160 --> 00:22:24,380
was someplace between
'L2' and 'L1'.
452
00:22:24,380 --> 00:22:27,470
We couldn't conclude that it
was exactly equal to 1/3.
453
00:22:27,470 --> 00:22:30,460
You see, the method of
exhaustion implies that all
454
00:22:30,460 --> 00:22:34,560
the error, all the room for
doubt, is squeezed out.
455
00:22:34,560 --> 00:22:37,100
By the way, there was nothing
sacred about choosing 'y'
456
00:22:37,100 --> 00:22:38,020
equals 'x squared'.
457
00:22:38,020 --> 00:22:39,630
We could generalize this.
458
00:22:39,630 --> 00:22:42,190
And we'll do this kind of
rapidly because I just want
459
00:22:42,190 --> 00:22:43,880
you to hear what I'm saying.
460
00:22:43,880 --> 00:22:46,280
This is all written out
in great detail in our
461
00:22:46,280 --> 00:22:47,280
supplementary notes.
462
00:22:47,280 --> 00:22:48,530
The idea is this.
463
00:22:48,530 --> 00:22:51,775
Let 'f' be any positive,
continuous function on 'a',
464
00:22:51,775 --> 00:22:54,100
'b' and non-decreasing.
465
00:22:54,100 --> 00:22:57,000
The non-decreasing part is just
for the convenience of
466
00:22:57,000 --> 00:23:00,210
being able to locate the high
and low points of each
467
00:23:00,210 --> 00:23:02,930
partition point conveniently.
468
00:23:02,930 --> 00:23:06,850
Partition 'a', 'b' into 'n'
equal parts, calling the first
469
00:23:06,850 --> 00:23:11,460
partition point 'x sub 0', the
last partition point 'x sub
470
00:23:11,460 --> 00:23:17,470
n', and the points in between
'x1' up through ''x 'n - 1''.
471
00:23:17,470 --> 00:23:18,760
In other words, we've
partitioned this closed
472
00:23:18,760 --> 00:23:21,230
interval into 'n' equal parts.
473
00:23:21,230 --> 00:23:22,980
And what we can now
do is what?
474
00:23:22,980 --> 00:23:26,630
Pick the lowest point in each
region, the highest point in
475
00:23:26,630 --> 00:23:28,360
each region, to form
a rectangle.
476
00:23:28,360 --> 00:23:32,280
We can form 'U sub
n' and 'L sub n'.
477
00:23:32,280 --> 00:23:35,490
And you see it's just what?
478
00:23:35,490 --> 00:23:39,830
For 'L sub n', you just pick
this height, which is 'f of 'x
479
00:23:39,830 --> 00:23:42,310
sub 0'', 'f of a'.
480
00:23:42,310 --> 00:23:44,270
The base is 'delta x'.
481
00:23:44,270 --> 00:23:47,620
You just add these all up until
you get to your last
482
00:23:47,620 --> 00:23:49,340
inscribed partition point.
483
00:23:49,340 --> 00:23:51,810
That's 'x sub 'n - 1''.
484
00:23:51,810 --> 00:23:52,570
OK.
485
00:23:52,570 --> 00:23:56,950
'U sub n' is the same thing,
only now your first rectangle
486
00:23:56,950 --> 00:24:00,220
has its height corresponding
to 'x' equals 'x1'.
487
00:24:00,220 --> 00:24:02,670
Consequently, the height
is 'f of x1'.
488
00:24:02,670 --> 00:24:04,460
The base is 'delta x'.
489
00:24:04,460 --> 00:24:06,940
Where in each of these, 'delta
x' is just what?
490
00:24:06,940 --> 00:24:09,730
This total length, which
is 'b - a', divided
491
00:24:09,730 --> 00:24:11,680
into 'n' equal parts.
492
00:24:11,680 --> 00:24:16,630
And to review the so-called
sigma notation that we have in
493
00:24:16,630 --> 00:24:17,760
our course--
494
00:24:17,760 --> 00:24:20,110
in this section in the textbook,
and we'll have
495
00:24:20,110 --> 00:24:23,130
exercises on this to make sure
that you are familiar with
496
00:24:23,130 --> 00:24:27,580
this notation, the shortcut
notation for writing this sum
497
00:24:27,580 --> 00:24:28,890
is given by this.
498
00:24:28,890 --> 00:24:32,260
We add up 'f of 'x sub
k'' times 'delta x'.
499
00:24:32,260 --> 00:24:37,680
As a subscript, 'k' is allowed
to vary from 1 to 'n'.
500
00:24:37,680 --> 00:24:42,220
And here we form the same sum,
only now the subscript 'k'
501
00:24:42,220 --> 00:24:45,960
varies from 0 to 'n - 1'.
502
00:24:45,960 --> 00:24:49,510
At any rate, the observations
are this.
503
00:24:49,510 --> 00:24:53,230
That 'U sub n' and 'L sub n'
converge to the same limit.
504
00:24:53,230 --> 00:24:55,150
In fact, this isn't
too hard to show.
505
00:24:55,150 --> 00:24:58,780
If you just algebraically
subtract 'L sub n' from 'U sub
506
00:24:58,780 --> 00:25:03,280
n', notice that the only terms
that won't cancel are
507
00:25:03,280 --> 00:25:04,890
the last term here.
508
00:25:04,890 --> 00:25:08,400
See, 'f of 'x sub n'' times
'delta x' has no counterpart
509
00:25:08,400 --> 00:25:12,310
here, because this ends with
the subscript 'n - 1'.
510
00:25:12,310 --> 00:25:16,590
Similarly, there is no 0
subscript in 'U sub n', so
511
00:25:16,590 --> 00:25:18,350
this term won't cancel.
512
00:25:18,350 --> 00:25:21,630
To make a long story short, if
we just subtract 'L sub n'
513
00:25:21,630 --> 00:25:25,420
from 'U sub n', that difference
will be 'f of 'x
514
00:25:25,420 --> 00:25:29,430
sub n'' times 'delta x'
minus 'f of 'x sub 0''
515
00:25:29,430 --> 00:25:30,910
times 'delta x'.
516
00:25:30,910 --> 00:25:33,740
See, that's just this difference
over here.
517
00:25:33,740 --> 00:25:36,610
'Delta x' is 'b - a' over 'n'.
518
00:25:36,610 --> 00:25:38,780
'x sub n' was called 'b'.
519
00:25:38,780 --> 00:25:41,220
'x sub 0' was called 'a'.
520
00:25:41,220 --> 00:25:44,480
In other words, the difference
between 'U sub n' and 'L sub
521
00:25:44,480 --> 00:25:48,020
n' is just 'f of b' minus
'f of a' times
522
00:25:48,020 --> 00:25:49,760
''b - a' over 'n''.
523
00:25:49,760 --> 00:25:54,000
And the important thing to
observe is that 'a', 'b', 'f
524
00:25:54,000 --> 00:25:56,960
of a', and 'f of b' are
fixed constants.
525
00:25:56,960 --> 00:26:00,460
The only thing that depends on
'n' is this denominator.
526
00:26:00,460 --> 00:26:03,620
Consequently, as 'n' goes to
infinity, since the rest of
527
00:26:03,620 --> 00:26:07,990
this thing is a constant, the
whole term goes to 0.
528
00:26:07,990 --> 00:26:11,050
In other words, the difference
between 'Un' and 'Ln' goes to
529
00:26:11,050 --> 00:26:12,200
0 in the limit.
530
00:26:12,200 --> 00:26:15,060
That means that the limit of 'U
sub n', as 'n' approaches
531
00:26:15,060 --> 00:26:18,190
infinity, equals the limit
of 'L sub n' as
532
00:26:18,190 --> 00:26:19,740
'n' approaches infinity.
533
00:26:19,740 --> 00:26:23,900
'A sub R' is always caught
between 'L sub n' and 'U sub
534
00:26:23,900 --> 00:26:26,480
n' by this very construction.
535
00:26:26,480 --> 00:26:30,170
Consequently, what this means is
that the area of the region
536
00:26:30,170 --> 00:26:34,830
'R' must equal this
common limit.
537
00:26:34,830 --> 00:26:37,360
Now by the way, this
looks very hard.
538
00:26:37,360 --> 00:26:38,360
And it is difficult.
539
00:26:38,360 --> 00:26:41,510
In fact, in general, these
limits are very hard to find.
540
00:26:41,510 --> 00:26:43,780
The point that I thought would
be interesting to mention at
541
00:26:43,780 --> 00:26:47,210
this stage of the game is that
if all we want is an estimate
542
00:26:47,210 --> 00:26:50,120
for the area under the curve,
there are faster and better
543
00:26:50,120 --> 00:26:52,020
ways of doing this.
544
00:26:52,020 --> 00:26:54,790
You see, the beauty of this
technique here is that it's a
545
00:26:54,790 --> 00:26:57,240
technique for exhausting
the space completely.
546
00:26:57,240 --> 00:26:59,850
We find the exact
areas this way.
547
00:26:59,850 --> 00:27:02,690
Let me give you a for
instance here.
548
00:27:02,690 --> 00:27:05,650
Namely, let me explain to you
what we mean by trapezoidal
549
00:27:05,650 --> 00:27:06,760
approximations.
550
00:27:06,760 --> 00:27:10,510
Let's take the same region,
'y' equals 'x squared'.
551
00:27:10,510 --> 00:27:13,370
And now, we'll divide it
into two parts here.
552
00:27:13,370 --> 00:27:17,130
And what we'll do is we'll
replace the arc of the curve
553
00:27:17,130 --> 00:27:20,360
by the straight line segment,
the chord, that joins two
554
00:27:20,360 --> 00:27:21,640
points here.
555
00:27:21,640 --> 00:27:24,540
In other words, using the
accented chalk here to
556
00:27:24,540 --> 00:27:25,410
illustrate this.
557
00:27:25,410 --> 00:27:28,310
What I'm going to do is instead
of finding the area of
558
00:27:28,310 --> 00:27:31,400
the region 'R', I'm going to
find the area of the region
559
00:27:31,400 --> 00:27:34,640
which has the top of
'R' replaced by
560
00:27:34,640 --> 00:27:36,520
these two line segments.
561
00:27:36,520 --> 00:27:40,050
All I want you to observe is
that this first region here is
562
00:27:40,050 --> 00:27:45,080
a triangle whose height is 1/4
and whose base is 1/2.
563
00:27:45,080 --> 00:27:47,700
Consequently, its area, being
one half the base times the
564
00:27:47,700 --> 00:27:50,380
height, is 1/16.
565
00:27:50,380 --> 00:27:55,370
The second region here is a
trapezoid whose bases are 1
566
00:27:55,370 --> 00:27:58,280
and 1/4 and whose
height is 1/2.
567
00:27:58,280 --> 00:28:01,150
And since the area of a
trapezoid is half the sum of
568
00:28:01,150 --> 00:28:06,870
the bases times the height, we
get that the area of this
569
00:28:06,870 --> 00:28:09,310
trapezoid is 5/16.
570
00:28:09,310 --> 00:28:12,950
Therefore, the area of the
triangle plus the trapezoid is
571
00:28:12,950 --> 00:28:17,400
6/16 or 3/8, which is 0.375.
572
00:28:17,400 --> 00:28:21,110
Recall that 1/3, we've just
seen, was the exact answer.
573
00:28:21,110 --> 00:28:24,450
And notice how close this
approximation is to the exact
574
00:28:24,450 --> 00:28:28,360
answer, even with just two
subdivisions over here.
575
00:28:28,360 --> 00:28:32,770
In fact, sparing you the
details, if you now divide the
576
00:28:32,770 --> 00:28:36,820
base here into four equal
parts, thus forming your
577
00:28:36,820 --> 00:28:39,790
trapezoidal approximations
corresponding to what?
578
00:28:39,790 --> 00:28:46,430
Altitudes of 1/16, 1/4, 9/16,
and 1, and adding up the areas
579
00:28:46,430 --> 00:28:48,400
of all these trapezoids--
580
00:28:48,400 --> 00:28:50,660
by the way, this is
a bit unfortunate.
581
00:28:50,660 --> 00:28:52,110
This is a triangle.
582
00:28:52,110 --> 00:28:56,110
But we can view a triangle as
being a degenerate trapezoid,
583
00:28:56,110 --> 00:28:58,990
meaning the height here just
happens to be 0 at this
584
00:28:58,990 --> 00:29:00,030
particular point.
585
00:29:00,030 --> 00:29:01,790
But let's not worry
about that.
586
00:29:01,790 --> 00:29:03,560
Let's just get the idea of
what the trapezoidal
587
00:29:03,560 --> 00:29:05,080
approximation idea means.
588
00:29:05,080 --> 00:29:10,280
We find the area of these four
trapezoids, and we use that as
589
00:29:10,280 --> 00:29:13,060
an approximation for the
area under the curve.
590
00:29:13,060 --> 00:29:16,070
By the way, just coming back
here for a moment, notice that
591
00:29:16,070 --> 00:29:19,850
because this curve is always
holding water, the chord will
592
00:29:19,850 --> 00:29:22,620
always lie above the arc.
593
00:29:22,620 --> 00:29:25,390
And consequently, not only do we
get an approximation using
594
00:29:25,390 --> 00:29:28,710
trapezoids this way that's
reasonably close, but we also
595
00:29:28,710 --> 00:29:32,210
know by the geometry here that
our approximation must be too
596
00:29:32,210 --> 00:29:34,170
large to be the right answer.
597
00:29:34,170 --> 00:29:36,730
In other words, we're going to
get an over-approximation.
598
00:29:36,730 --> 00:29:38,380
But watch how close we come.
599
00:29:38,380 --> 00:29:43,350
Leaving these details for you to
verify, all I show here is
600
00:29:43,350 --> 00:29:46,690
that if you add up the areas of
these four regions, we wind
601
00:29:46,690 --> 00:29:53,010
up with 44/128, which
is 11/32.
602
00:29:53,010 --> 00:29:57,590
And 11/32 is mighty close
to 1/3, being what?
603
00:29:57,590 --> 00:29:59,120
11/33.
604
00:29:59,120 --> 00:30:01,420
In other words, look at how
close, with just four
605
00:30:01,420 --> 00:30:04,900
subdivisions, we get to an
approximation that's good for
606
00:30:04,900 --> 00:30:07,360
the area under the curve,
without having to put the
607
00:30:07,360 --> 00:30:08,460
squeeze on.
608
00:30:08,460 --> 00:30:11,410
But to get the exact area,
we need the squeeze.
609
00:30:11,410 --> 00:30:14,740
We have to push the thing from
above, from below, and show
610
00:30:14,740 --> 00:30:17,630
that these two things that we're
squeezing it between, as
611
00:30:17,630 --> 00:30:20,700
gruesome as it sounds, converge
towards each other.
612
00:30:20,700 --> 00:30:23,970
The thing that we're looking
for is caught between them,
613
00:30:23,970 --> 00:30:26,560
and consequently must
be the common limit.
614
00:30:26,560 --> 00:30:29,690
That was the method of
exhaustion as known by the
615
00:30:29,690 --> 00:30:30,370
ancient Greeks.
616
00:30:30,370 --> 00:30:32,530
And if this seems tough to
you, think of it from two
617
00:30:32,530 --> 00:30:33,290
points of view.
618
00:30:33,290 --> 00:30:35,040
One, it is tough.
619
00:30:35,040 --> 00:30:39,510
And secondly, if people of some
2,500 years ago were able
620
00:30:39,510 --> 00:30:42,420
to do this, it should be at
least plausible that with a
621
00:30:42,420 --> 00:30:44,800
little bit of effort, we can get
a good feeling for what's
622
00:30:44,800 --> 00:30:45,840
going on here.
623
00:30:45,840 --> 00:30:48,880
In fact, hopefully, the
exercises to this particular
624
00:30:48,880 --> 00:30:51,440
unit will make this a little
bit easier for
625
00:30:51,440 --> 00:30:53,440
you to see in action.
626
00:30:53,440 --> 00:30:55,100
But at any rate--
627
00:30:55,100 --> 00:30:57,790
let me just make a few
asides over here.
628
00:30:57,790 --> 00:31:01,200
The deeper asides will be made
in greater detail in our
629
00:31:01,200 --> 00:31:02,750
supplementary notes.
630
00:31:02,750 --> 00:31:05,830
Notice that when we dealt with
differential calculus, it was
631
00:31:05,830 --> 00:31:08,960
very, very important in
differential calculus to have
632
00:31:08,960 --> 00:31:10,380
smooth curves.
633
00:31:10,380 --> 00:31:14,750
For finding areas, all you need
are continuous curves.
634
00:31:14,750 --> 00:31:16,490
In other words, for example,
we can find
635
00:31:16,490 --> 00:31:18,280
the area of a square.
636
00:31:18,280 --> 00:31:21,270
But certainly, a square
has sharp corners.
637
00:31:21,270 --> 00:31:23,960
In other words, those corners
are continuous, but they're
638
00:31:23,960 --> 00:31:25,820
not differentiable in terms
of the language of
639
00:31:25,820 --> 00:31:27,240
differential calculus.
640
00:31:27,240 --> 00:31:29,930
The point is that even
continuity can be weakened.
641
00:31:29,930 --> 00:31:31,630
Let me give you a definition.
642
00:31:31,630 --> 00:31:34,820
'f' is called piecewise
continuous on the interval
643
00:31:34,820 --> 00:31:39,910
from 'a' to 'b' if and only if
'f' is continuous except at a
644
00:31:39,910 --> 00:31:43,700
finite number of points where
it has jump discontinuities.
645
00:31:43,700 --> 00:31:47,210
For example, in terms of this
particular diagram, notice
646
00:31:47,210 --> 00:31:51,270
that my curve, 'y' equals 'f of
x', is discontinuous at a
647
00:31:51,270 --> 00:31:53,320
finite number of points,
namely at 'c1'
648
00:31:53,320 --> 00:31:55,290
and 'c2', two points.
649
00:31:55,290 --> 00:31:57,810
Notice that what happens at
those two points is you have
650
00:31:57,810 --> 00:32:00,300
just a finite jump
discontinuity.
651
00:32:00,300 --> 00:32:03,520
The point is that whereas this
curve is not continuous,
652
00:32:03,520 --> 00:32:06,550
notice that since a straight
line having no thickness has
653
00:32:06,550 --> 00:32:12,060
no area, if we replace the
given curve by this one--
654
00:32:12,060 --> 00:32:14,010
you see, putting in these
vertical lines--
655
00:32:14,010 --> 00:32:17,200
notice that this does form
a closed region.
656
00:32:17,200 --> 00:32:20,890
And to find the area of this
closed region, I can pretend
657
00:32:20,890 --> 00:32:24,880
it was made up of these three
particular regular regions.
658
00:32:24,880 --> 00:32:28,370
In other words, that even if I
just have jumps, since the
659
00:32:28,370 --> 00:32:31,250
jump does not contribute towards
the area, there is no
660
00:32:31,250 --> 00:32:34,920
harm done when one talks about
piecewise continuous in
661
00:32:34,920 --> 00:32:38,500
finding areas rather
than continuous.
662
00:32:38,500 --> 00:32:39,800
A second aside--
663
00:32:39,800 --> 00:32:42,090
well, I've actually called this
one aside number one,
664
00:32:42,090 --> 00:32:43,870
because this definition I
didn't call an aside--
665
00:32:43,870 --> 00:32:45,230
but the idea is this.
666
00:32:45,230 --> 00:32:49,010
Notice that up until now, we
were assuming that we had to
667
00:32:49,010 --> 00:32:52,680
form our 'U sub n's and our 'L
sub n's by choosing either the
668
00:32:52,680 --> 00:32:56,350
left endpoint of a partition
or the right endpoint.
669
00:32:56,350 --> 00:32:59,910
What I'd like you to see also as
a generalization is that if
670
00:32:59,910 --> 00:33:05,340
I pick any point between these
two extremes and form--
671
00:33:05,340 --> 00:33:07,720
look, I'll call it
'c sub k' between
672
00:33:07,720 --> 00:33:09,000
these two points here--
673
00:33:09,000 --> 00:33:13,510
and form this particular sum,
where 'c sub k' replaces
674
00:33:13,510 --> 00:33:18,530
either 'x sub k' or 'x sub 'k -
1'', that this sum here also
675
00:33:18,530 --> 00:33:20,990
gives me the area of
the region 'R'.
676
00:33:20,990 --> 00:33:23,540
In other words, that whatever
this sum is, it's caught
677
00:33:23,540 --> 00:33:25,970
between 'U sub n'
and 'L sub n'.
678
00:33:25,970 --> 00:33:28,900
And since 'U sub n' and 'L
sub n' have a common
679
00:33:28,900 --> 00:33:30,620
limit of 'A sub R'--
680
00:33:30,620 --> 00:33:32,855
in other words, since both
of these squeeze in
681
00:33:32,855 --> 00:33:34,220
towards 'A sub R'--
682
00:33:34,220 --> 00:33:37,860
this being caught between them
must also be 'A sub R'.
683
00:33:37,860 --> 00:33:39,790
That's what I've
said over here.
684
00:33:39,790 --> 00:33:42,650
And to see this thing in terms
of a picture, all we're saying
685
00:33:42,650 --> 00:33:48,460
is that when you pick the lowest
point in the interval,
686
00:33:48,460 --> 00:33:52,870
you get the rectangle that
contributes to 'L sub n'.
687
00:33:52,870 --> 00:33:55,440
When you pick the highest point
in the interval, you get
688
00:33:55,440 --> 00:33:58,320
the rectangle that contributes
to 'U sub n'.
689
00:33:58,320 --> 00:34:01,990
Consequently, for any point
between these two extremes--
690
00:34:01,990 --> 00:34:04,630
call that--that's what 'c sub k'
is, it's any point between
691
00:34:04,630 --> 00:34:05,525
these two--
692
00:34:05,525 --> 00:34:07,640
pick the height that corresponds
to that.
693
00:34:07,640 --> 00:34:11,409
And whatever rectangle you form
this way, that rectangle
694
00:34:11,409 --> 00:34:15,040
must have a greater area than
the rectangle that went into
695
00:34:15,040 --> 00:34:18,800
forming 'L sub n', but a lesser
area than the rectangle
696
00:34:18,800 --> 00:34:21,100
that went into forming
'U sub n'.
697
00:34:21,100 --> 00:34:23,909
And consequently, that's
where this particular
698
00:34:23,909 --> 00:34:25,260
result comes from.
699
00:34:25,260 --> 00:34:28,389
Again, this is done in more
detail in the notes, but I
700
00:34:28,389 --> 00:34:31,050
think some of these things you
should hear me say out loud
701
00:34:31,050 --> 00:34:33,300
rather than to rely on
your reading it.
702
00:34:33,300 --> 00:34:36,620
And finally, one more
important point.
703
00:34:36,620 --> 00:34:39,429
Up until now, it's been clear
that since we're talking about
704
00:34:39,429 --> 00:34:43,790
area, our region has to
lie above the x-axis.
705
00:34:43,790 --> 00:34:48,570
We can remove the restriction
that 'f' be non-negative if we
706
00:34:48,570 --> 00:34:51,210
replace area by net area.
707
00:34:51,210 --> 00:34:54,159
In other words, all I want you
to see here is that if my
708
00:34:54,159 --> 00:34:57,390
region happens to look something
like this, notice
709
00:34:57,390 --> 00:35:00,800
that between 'c' and 'b',
'f of x' is negative.
710
00:35:00,800 --> 00:35:04,010
Consequently, when I form things
of the form 'f of x'
711
00:35:04,010 --> 00:35:07,990
times 'delta x', 'delta x' being
positive, 'f of x' being
712
00:35:07,990 --> 00:35:10,960
negative, is going to give
me a negative result.
713
00:35:10,960 --> 00:35:15,270
In other words, algebraically,
if I form my summation, this
714
00:35:15,270 --> 00:35:17,840
portion in here will give
me a positive result.
715
00:35:17,840 --> 00:35:20,350
This portion in here will give
me a negative result.
716
00:35:20,350 --> 00:35:23,230
Consequently, working
algebraically, what I will
717
00:35:23,230 --> 00:35:27,280
find is not the true area
but the positive
718
00:35:27,280 --> 00:35:29,670
minus this amount here.
719
00:35:29,670 --> 00:35:32,680
In other words, what I'll
find is the net area.
720
00:35:32,680 --> 00:35:36,860
And again, this will all be
worked out in the notes.
721
00:35:36,860 --> 00:35:39,670
In addition, there'll be other
refinements made in the notes,
722
00:35:39,670 --> 00:35:43,130
such as that these partitions
don't have to be into equal
723
00:35:43,130 --> 00:35:44,950
parts or things like this.
724
00:35:44,950 --> 00:35:48,570
The important point from today's
lecture is this.
725
00:35:48,570 --> 00:35:55,000
Observe that this entire study
of area is done independently
726
00:35:55,000 --> 00:35:56,880
of differential calculus.
727
00:35:56,880 --> 00:36:00,730
In our next lecture, we are
going to show a truly
728
00:36:00,730 --> 00:36:05,020
remarkable, a wonderful
relationship between the
729
00:36:05,020 --> 00:36:08,620
differential calculus of before
and the so-called area
730
00:36:08,620 --> 00:36:10,980
or integral calculus
that we did today.
731
00:36:10,980 --> 00:36:13,030
At any rate, until next
time, goodbye.
732
00:36:13,030 --> 00:36:15,910
733
00:36:15,910 --> 00:36:19,110
Funding for the publication of
this video was provided by the
734
00:36:19,110 --> 00:36:23,160
Gabriella and Paul Rosenbaum
Foundation.
735
00:36:23,160 --> 00:36:27,330
Help OCW continue to provide
free and open access to MIT
736
00:36:27,330 --> 00:36:31,540
courses by making a donation
at ocw.mit.edu/donate.
737
00:36:31,540 --> 00:36:36,266