WEBVTT
00:00:00.040 --> 00:00:02.400
The following content is
provided under a Creative
00:00:02.400 --> 00:00:03.690
Commons license.
00:00:03.690 --> 00:00:06.630
Your support will help MIT
OpenCourseWare continue to
00:00:06.630 --> 00:00:09.990
offer high-quality educational
resources for free.
00:00:09.990 --> 00:00:12.830
To make a donation, or to view
additional materials from
00:00:12.830 --> 00:00:16.760
hundreds of MIT courses, visit
MIT OpenCourseWare at
00:00:16.760 --> 00:00:18.010
ocw.mit.edu.
00:00:31.070 --> 00:00:32.040
PROFESSOR: Hi.
00:00:32.040 --> 00:00:35.950
Today we begin our study of
integral calculus, which had
00:00:35.950 --> 00:00:42.240
its roots back in ancient
Greece, roughly 600 BC, and
00:00:42.240 --> 00:00:45.560
began with an investigation
of the study of area.
00:00:45.560 --> 00:00:47.520
In a certain manner of speaking,
today's lecture
00:00:47.520 --> 00:00:51.000
could be called 'Calculus
Revisited Revisited', in the
00:00:51.000 --> 00:00:54.580
sense that integral calculus
can be studied quite apart
00:00:54.580 --> 00:00:56.140
from the fact that differential
00:00:56.140 --> 00:00:57.900
calculus was ever invented.
00:00:57.900 --> 00:01:01.260
We'll see later in this block
of material that there is a
00:01:01.260 --> 00:01:04.129
wondrous relationship between
integral and differential
00:01:04.129 --> 00:01:07.480
calculus, but perhaps the best
proof that integral calculus
00:01:07.480 --> 00:01:11.160
can be studied independently of
differential calculus lies
00:01:11.160 --> 00:01:14.700
in the fact that the ancient
Greek was doing integral
00:01:14.700 --> 00:01:19.010
calculus in 600 BC, whereas
differential calculus did not
00:01:19.010 --> 00:01:23.690
begin until about 1680 AD
with Sir Isaac Newton.
00:01:23.690 --> 00:01:26.980
Well, at any rate then, what
we'll call today's lecture is
00:01:26.980 --> 00:01:29.420
simply 'Two-dimensional Area'.
00:01:29.420 --> 00:01:31.900
In other words, this is how
the subject began, with
00:01:31.900 --> 00:01:36.240
studying the amount of space
in plane regions,
00:01:36.240 --> 00:01:39.370
two-dimensional area, as opposed
to, say, as we'll talk
00:01:39.370 --> 00:01:43.070
about later, three-dimensional
area, which is really a fancy
00:01:43.070 --> 00:01:44.700
word for volume, et cetera.
00:01:44.700 --> 00:01:46.650
But enough about that
for the time being.
00:01:46.650 --> 00:01:50.350
Two-dimensional area, or
Calculus Revisited Revisited.
00:01:50.350 --> 00:01:53.780
And to see what's happening over
here, the study of area
00:01:53.780 --> 00:01:57.790
was the forerunner of integral
calculus as we now know it.
00:01:57.790 --> 00:02:01.630
It began in ancient Greece,
roughly 600 BC.
00:02:01.630 --> 00:02:04.840
The branch of calculus that
we've been studying up until
00:02:04.840 --> 00:02:08.229
now in our course, differential
calculus, did not
00:02:08.229 --> 00:02:10.979
begin until 1680 AD.
00:02:10.979 --> 00:02:14.730
Notice, then, the interesting
juxtaposition in time.
00:02:14.730 --> 00:02:17.510
In other words, pedagogically,
one seems to study
00:02:17.510 --> 00:02:20.790
differential calculus before
integral calculus.
00:02:20.790 --> 00:02:24.450
Chronologically, integral
calculus preceded differential
00:02:24.450 --> 00:02:27.670
calculus by more than
2,000 years.
00:02:27.670 --> 00:02:30.110
Now this is a rather
beautiful study.
00:02:30.110 --> 00:02:33.260
It's one of the most aesthetic
parts of elementary
00:02:33.260 --> 00:02:36.800
mathematics, namely the
simplicity with which the
00:02:36.800 --> 00:02:41.220
Greek was able to tackle the
sophisticated problem of
00:02:41.220 --> 00:02:43.160
finding areas in general.
00:02:43.160 --> 00:02:46.500
He started with three basically
simple properties of
00:02:46.500 --> 00:02:50.430
area, so simple that most of us
are willing to accept them
00:02:50.430 --> 00:02:52.070
almost intuitively.
00:02:52.070 --> 00:02:54.740
We'll call these the
'Axioms for Area'.
00:02:54.740 --> 00:02:56.830
And simply stated, they
were the following.
00:02:56.830 --> 00:03:00.250
One, that the area of a
rectangle is the base times
00:03:00.250 --> 00:03:01.150
the height.
00:03:01.150 --> 00:03:06.060
Two, if one region is contained
within another, the
00:03:06.060 --> 00:03:09.830
area of the contained region is
less than or equal to-- in
00:03:09.830 --> 00:03:11.740
other words, can be no greater
than-- that of
00:03:11.740 --> 00:03:12.990
the containing region.
00:03:12.990 --> 00:03:15.390
Roughly speaking, the
smaller the region,
00:03:15.390 --> 00:03:17.160
the smaller its area.
00:03:17.160 --> 00:03:21.220
And finally, written slightly
more formally here, the old
00:03:21.220 --> 00:03:24.670
high-school axiom for plane
geometry, that the area of the
00:03:24.670 --> 00:03:27.700
whole equals the sum of the
areas of the parts, that if a
00:03:27.700 --> 00:03:32.580
region 'R' is subdivided into a
union of mutually exclusive
00:03:32.580 --> 00:03:36.670
pieces, then the area of the
entire region is equal to the
00:03:36.670 --> 00:03:39.640
sum of the areas of the
constituent parts.
00:03:39.640 --> 00:03:43.290
And with just these three
axioms, the ancient Greek
00:03:43.290 --> 00:03:47.320
invented a technique for finding
areas that today is
00:03:47.320 --> 00:03:50.250
still known by its original
name, the 'Method of
00:03:50.250 --> 00:03:51.980
Exhaustion'.
00:03:51.980 --> 00:03:55.850
And here, exhaustion does not
refer to the fact that we get
00:03:55.850 --> 00:03:59.210
tired using the system, even
though, as we shall soon see,
00:03:59.210 --> 00:04:00.380
it's quite intricate.
00:04:00.380 --> 00:04:03.530
It refers to the fact that we
take the region whose area we
00:04:03.530 --> 00:04:09.540
want to find and exhaust the
space by squeezing it between
00:04:09.540 --> 00:04:12.690
regions which are made
up of rectangles.
00:04:12.690 --> 00:04:16.140
Stated more precisely,
given a region 'R'--
00:04:16.140 --> 00:04:19.220
and I'll define this more
precisely later-- we squeeze
00:04:19.220 --> 00:04:22.089
it between two networks
of rectangles,
00:04:22.089 --> 00:04:23.570
the idea being what?
00:04:23.570 --> 00:04:26.550
That we know what the area
of a rectangle is, and
00:04:26.550 --> 00:04:30.090
consequently, by the fact that
the area of the whole equals
00:04:30.090 --> 00:04:32.630
the sum of the areas of the
parts, if we have a
00:04:32.630 --> 00:04:35.940
rectangular network, knowing how
to find the area of each
00:04:35.940 --> 00:04:38.460
rectangle, we know how
to find the area of
00:04:38.460 --> 00:04:40.020
the rectangular network.
00:04:40.020 --> 00:04:43.470
Now, rather than to wax on like
this philosophically,
00:04:43.470 --> 00:04:45.550
let's tackle a specific
problem.
00:04:45.550 --> 00:04:49.540
In fact, give or take a little
bit, this is specifically the
00:04:49.540 --> 00:04:53.470
problem that Archimedes dealt
with in finding the area of
00:04:53.470 --> 00:04:56.800
parabolic segments way back,
as I say, in the
00:04:56.800 --> 00:05:00.050
300 to 600 BC era.
00:05:00.050 --> 00:05:01.930
As an example, consider
the following.
00:05:01.930 --> 00:05:06.160
We wish to determine the area of
the region 'R', 'A sub R',
00:05:06.160 --> 00:05:10.330
where the region 'R' is that
region which is bounded above
00:05:10.330 --> 00:05:15.930
by the curve 'y' equals 'x
squared', below by the x-axis,
00:05:15.930 --> 00:05:19.300
on the left by the y-axis,
and on the right by the
00:05:19.300 --> 00:05:20.860
line 'x' equals 1.
00:05:20.860 --> 00:05:23.960
In other words, this
region in here.
00:05:23.960 --> 00:05:26.950
Now here's the way we put the
squeeze on it, and we'll start
00:05:26.950 --> 00:05:29.410
this quite gradually.
00:05:29.410 --> 00:05:32.900
The first thing that we observe
is that if we draw a
00:05:32.900 --> 00:05:37.170
line parallel to the x-axis
through the point (1 , 1), the
00:05:37.170 --> 00:05:42.000
highest point of our region
here, we construct a rectangle
00:05:42.000 --> 00:05:44.390
which contains our region 'R'.
00:05:44.390 --> 00:05:48.000
And since 'R' is contained in
the rectangle, the area of the
00:05:48.000 --> 00:05:51.410
region 'R' must be less than
the area of the rectangle.
00:05:51.410 --> 00:05:53.650
But notice that this particular
rectangle that
00:05:53.650 --> 00:05:57.020
we've just constructed has its
base equal to 1, its height
00:05:57.020 --> 00:05:58.000
equal to 1.
00:05:58.000 --> 00:06:03.160
Therefore by our first axiom,
its area is 1 times 1, or 1.
00:06:03.160 --> 00:06:07.140
Also, since intuitively, area
is a positive thing, we see
00:06:07.140 --> 00:06:09.950
just from this quick diagram
that whatever the area of the
00:06:09.950 --> 00:06:14.370
region is, we now have it
bounded between 0 and 1.
00:06:14.370 --> 00:06:15.410
We have an upper bound.
00:06:15.410 --> 00:06:17.180
We have a lower bound.
00:06:17.180 --> 00:06:19.430
We have this thing caught, OK?
00:06:19.430 --> 00:06:21.760
Now, there's still a lot of
space between 0 and 1.
00:06:21.760 --> 00:06:25.110
The method of exhaustion
refines this idea.
00:06:25.110 --> 00:06:27.460
Namely, what we do next
is we say, look at--
00:06:27.460 --> 00:06:31.020
instead of just drawing one big
rectangle like this, why
00:06:31.020 --> 00:06:34.010
don't we partition the base
of our region 'R'
00:06:34.010 --> 00:06:35.230
into two equal parts?
00:06:35.230 --> 00:06:38.510
In other words, let's locate the
point 1/2, which of course
00:06:38.510 --> 00:06:41.090
is midway between 0 and 1.
00:06:41.090 --> 00:06:46.230
Now what I can do is I can
circumscribe two rectangles,
00:06:46.230 --> 00:06:50.400
one of which is the rectangle
whose height corresponds to
00:06:50.400 --> 00:06:52.890
the x-coordinate
equaling 1/2--
00:06:52.890 --> 00:06:55.450
that means the y-coordinate
is 1/4--
00:06:55.450 --> 00:06:59.160
and the second rectangle, the
one whose x-coordinate--
00:06:59.160 --> 00:07:02.350
the x-coordinate of the height
is 1, and since 'y' equals 'x
00:07:02.350 --> 00:07:04.600
squared', the height
is also 1.
00:07:04.600 --> 00:07:07.960
You see, notice that by picking
a region which is a
00:07:07.960 --> 00:07:11.000
curve which is always rising,
the lowest point of each
00:07:11.000 --> 00:07:13.780
region is the point that's
furthest to the left.
00:07:13.780 --> 00:07:16.380
And the highest point of each
region is the point that's
00:07:16.380 --> 00:07:17.510
furthest to the right.
00:07:17.510 --> 00:07:20.980
And keeping this in mind, you
see by making rectangles
00:07:20.980 --> 00:07:24.110
corresponding to the points
furthest to the right in each
00:07:24.110 --> 00:07:28.170
interval, I get a rectangle
which contains the portion of
00:07:28.170 --> 00:07:30.020
the region 'R' that
I'm talking about.
00:07:30.020 --> 00:07:33.270
In other words, if I look at
this rectangular network, my
00:07:33.270 --> 00:07:35.770
region 'R' is contained
inside that.
00:07:35.770 --> 00:07:40.550
Consequently, its area is less
than the area of the
00:07:40.550 --> 00:07:41.860
rectangular network.
00:07:41.860 --> 00:07:44.350
Now, what is the area of this
rectangular network?
00:07:44.350 --> 00:07:48.740
Well, the big rectangle has its
base equal to 1/2 and its
00:07:48.740 --> 00:07:50.030
height equal to 1.
00:07:50.030 --> 00:07:51.970
So its area is 1/2.
00:07:51.970 --> 00:07:55.370
The small rectangle has its
base equal to 1/2 and its
00:07:55.370 --> 00:07:56.720
height equal to 1/4.
00:07:56.720 --> 00:07:58.390
Since its area is the
base times the
00:07:58.390 --> 00:08:00.390
height, its area is 1/8.
00:08:00.390 --> 00:08:02.960
1/8 plus 1/2 is 5/8.
00:08:02.960 --> 00:08:06.420
In other words, the area of the
rectangular network which
00:08:06.420 --> 00:08:08.260
contains 'R' is 5/8.
00:08:08.260 --> 00:08:10.450
Consequently, the area of
the region 'R' must
00:08:10.450 --> 00:08:12.050
be less than 5/8.
00:08:12.050 --> 00:08:16.040
On the other hand, notice that
the lowest point in the second
00:08:16.040 --> 00:08:18.900
interval corresponds to
'x' equaling 1/2.
00:08:18.900 --> 00:08:22.310
In other words, notice that
this particular rectangle,
00:08:22.310 --> 00:08:25.710
whose base is 1/2 and whose
height is 1/4, that this
00:08:25.710 --> 00:08:29.440
rectangle here is inscribed
in the region 'R'.
00:08:29.440 --> 00:08:31.170
Its area is 1/8.
00:08:31.170 --> 00:08:34.380
And since it's contained within
the region 'R', its
00:08:34.380 --> 00:08:36.760
area must be less than the
area of the region 'R'.
00:08:36.760 --> 00:08:38.880
And so we're now certain that
whatever the area of the
00:08:38.880 --> 00:08:42.549
region 'R' is, it's between
1/8 and 5/8.
00:08:42.549 --> 00:08:45.510
And by the way, notice in terms
of these shaded regions,
00:08:45.510 --> 00:08:48.680
that even though this is an
approximation which is too
00:08:48.680 --> 00:08:51.700
large to be the right answer,
it is closer to being the
00:08:51.700 --> 00:08:54.420
right answer than this
approximation here.
00:08:54.420 --> 00:08:56.210
In other words, notice that
the difference between the
00:08:56.210 --> 00:08:59.480
overestimate here and the
overestimate here is this
00:08:59.480 --> 00:09:03.220
amount in here.
00:09:03.220 --> 00:09:05.120
You see, we've chopped off
part of the error.
00:09:05.120 --> 00:09:07.370
But we'll talk about that,
as I say, in more
00:09:07.370 --> 00:09:08.470
detail in the notes.
00:09:08.470 --> 00:09:11.220
What I'd like to do is to give
you an overall view of what's
00:09:11.220 --> 00:09:12.440
happening over here.
00:09:12.440 --> 00:09:15.280
You see, to generalize the
method of exhaustion, we do
00:09:15.280 --> 00:09:17.010
what the mathematician
usually does.
00:09:17.010 --> 00:09:19.980
Instead of saying, let's divide
the base of the region
00:09:19.980 --> 00:09:23.250
into two equal parts, or three
equal parts, or four equal
00:09:23.250 --> 00:09:27.440
parts, we say, let's divide
it into 'n' equal parts.
00:09:27.440 --> 00:09:30.870
Now if we divide this base into
'n' equal parts, since
00:09:30.870 --> 00:09:32.370
the whole region--
00:09:32.370 --> 00:09:35.720
the whole length is 0 to 1,
dividing into 'n' equal parts
00:09:35.720 --> 00:09:39.550
means that my points of division
will be '1/n', '2/n',
00:09:39.550 --> 00:09:44.520
'3/n', et cetera, right down
to 'n/n', which is 1.
00:09:44.520 --> 00:09:50.460
What I do now is I pick the
right endpoint of each point
00:09:50.460 --> 00:09:53.720
in that partition to
draw my rectangle.
00:09:53.720 --> 00:09:56.370
You see, each of these
rectangles that I draw this
00:09:56.370 --> 00:10:00.130
way contains the corresponding
part of my region 'R'.
00:10:00.130 --> 00:10:05.530
Consequently, the area of this
rectangular network must be an
00:10:05.530 --> 00:10:08.660
upper bound for the area
of my region 'R'.
00:10:08.660 --> 00:10:11.370
Now, because it's going to be an
upper bound, and because it
00:10:11.370 --> 00:10:14.830
depends on what the value of
'n' is, I will denote that
00:10:14.830 --> 00:10:17.670
upper bound by 'U sub n'.
00:10:17.670 --> 00:10:19.350
And what will 'U sub n' be?
00:10:19.350 --> 00:10:21.980
It's the sum of the areas
of these circumscribed
00:10:21.980 --> 00:10:23.010
rectangles.
00:10:23.010 --> 00:10:25.320
And it's not difficult to
see from this picture.
00:10:25.320 --> 00:10:26.820
Let's take a look.
00:10:26.820 --> 00:10:30.280
Since this rectangle has its
height corresponding to the
00:10:30.280 --> 00:10:33.980
x-coordinate equaling '1/n', and
since the y-coordinate is
00:10:33.980 --> 00:10:36.090
the square of the x-coordinate,
the height of
00:10:36.090 --> 00:10:38.780
this rectangle will be
''1/n' squared'.
00:10:38.780 --> 00:10:40.320
The base is '1/n'.
00:10:40.320 --> 00:10:43.260
So the area of that
circumscribed rectangle is
00:10:43.260 --> 00:10:46.180
''1/n' squared' times '1/n'.
00:10:46.180 --> 00:10:49.600
Similarly, the next rectangle
here has its height
00:10:49.600 --> 00:10:52.690
corresponding to the
x-coordinate '2/n'.
00:10:52.690 --> 00:10:55.550
So its height is ''2/n'
squared'.
00:10:55.550 --> 00:10:57.530
Keep in mind the y-coordinate
is the square of the
00:10:57.530 --> 00:10:58.880
x-coordinate here.
00:10:58.880 --> 00:11:01.730
The base of all of these
rectangles is '1/n'.
00:11:01.730 --> 00:11:04.930
And proceeding in this way, I
finally come down to the last
00:11:04.930 --> 00:11:07.520
rectangle, whose
height is what?
00:11:07.520 --> 00:11:11.080
Well, the x-coordinate is 1,
so the y-coordinate is 1.
00:11:11.080 --> 00:11:14.400
To keep my form here, I'll
write that as 'n/n'.
00:11:14.400 --> 00:11:18.320
Its height is therefore
''n/n' squared'.
00:11:18.320 --> 00:11:21.810
The base of that rectangle is
'1/n', so the area of that
00:11:21.810 --> 00:11:26.760
last rectangle is ''n/n'
squared' times '1/n'.
00:11:26.760 --> 00:11:28.330
Now you see that's what?
00:11:28.330 --> 00:11:32.170
That's an area which is too
large to be 'A sub R'.
00:11:32.170 --> 00:11:34.740
Well, let's take a look now and
see what's happening here.
00:11:34.740 --> 00:11:38.510
Notice that each of these terms
has an 'n cubed' in the
00:11:38.510 --> 00:11:39.450
denominator.
00:11:39.450 --> 00:11:42.060
So I can factor out the
'n cubed' term.
00:11:42.060 --> 00:11:44.050
And what I'm left
with is what?
00:11:44.050 --> 00:11:46.370
The sum of the first
'n' squares.
00:11:46.370 --> 00:11:49.995
1 squared plus 2 squared plus 3
squared, et cetera, plus et
00:11:49.995 --> 00:11:51.180
cetera, 'n squared'.
00:11:51.180 --> 00:11:54.340
In other words, this tells me
how to find 'U sub n' for any
00:11:54.340 --> 00:11:55.370
value of 'n'.
00:11:55.370 --> 00:11:58.040
And we'll talk about that some
more in a little while.
00:11:58.040 --> 00:11:59.690
That's an upper bound.
00:11:59.690 --> 00:12:03.810
In a corresponding way, I can
also find a lower bound.
00:12:03.810 --> 00:12:08.350
Namely, what I'll do now is pick
the smallest rectangle in
00:12:08.350 --> 00:12:08.910
each region.
00:12:08.910 --> 00:12:11.860
In other words, I'll now
inscribe rectangles inside my
00:12:11.860 --> 00:12:16.140
region 'R', and therefore find
a rectangular region whose
00:12:16.140 --> 00:12:19.150
area is less than that
of the region 'R'.
00:12:19.150 --> 00:12:22.080
And to do that, without going
through the details, notice
00:12:22.080 --> 00:12:24.970
that essentially all I have to
do is shift each of these
00:12:24.970 --> 00:12:28.010
rectangles over by
one partition.
00:12:28.010 --> 00:12:33.180
Namely, notice, for example,
that the smallest height in
00:12:33.180 --> 00:12:36.280
the second partition here is the
one which corresponds to
00:12:36.280 --> 00:12:37.530
the height '1/n'.
00:12:37.530 --> 00:12:40.010
In other words, what I do
now is, to inscribe the
00:12:40.010 --> 00:12:43.430
rectangles, I just shift
everything over like this.
00:12:43.430 --> 00:12:46.660
And leaving the details out, and
letting you verify these
00:12:46.660 --> 00:12:48.510
for yourselves, I
can again mimic
00:12:48.510 --> 00:12:50.160
exactly what I did before.
00:12:50.160 --> 00:12:54.030
The only difference being now,
that instead of the height of
00:12:54.030 --> 00:12:59.690
the last rectangle being 'n/n',
notice that it's 'n -
00:12:59.690 --> 00:13:01.160
1' over 'n', squared.
00:13:01.160 --> 00:13:02.650
In other words, the
x-coordinate is
00:13:02.650 --> 00:13:04.140
'n - 1' over 'n'.
00:13:04.140 --> 00:13:06.620
The y-coordinate is the square
of the x-coordinate.
00:13:06.620 --> 00:13:09.340
In other words, notice that this
point here gives rise to
00:13:09.340 --> 00:13:12.620
the height of the lowest
rectangle that can be
00:13:12.620 --> 00:13:14.490
inscribed in my last
portion here.
00:13:14.490 --> 00:13:17.970
So without further ado, it turns
out that 'L sub n', the
00:13:17.970 --> 00:13:22.060
lower estimate, is '1 over 'n
cubed'' times the sum of the
00:13:22.060 --> 00:13:24.820
first 'n - 1' squares.
00:13:24.820 --> 00:13:25.710
OK.
00:13:25.710 --> 00:13:28.540
Now, let's keep track of just
what it is that we've done
00:13:28.540 --> 00:13:30.380
over here so far.
00:13:30.380 --> 00:13:31.410
What we've done is what?
00:13:31.410 --> 00:13:37.890
For each 'n', we have squeezed
'A sub R' between two numbers,
00:13:37.890 --> 00:13:40.800
one number being an upper
approximation, one number
00:13:40.800 --> 00:13:42.720
being a lower approximation.
00:13:42.720 --> 00:13:45.370
Now notice that the 'U sub
n' and the 'L sub n' are
00:13:45.370 --> 00:13:46.880
functions of 'n'.
00:13:46.880 --> 00:13:48.380
'A sub R' is a constant.
00:13:48.380 --> 00:13:50.070
That's the area of
the region 'R'.
00:13:50.070 --> 00:13:52.670
What the method of exhaustion
means is simply this.
00:13:52.670 --> 00:13:53.710
We say, look.
00:13:53.710 --> 00:13:56.860
Let's take the limit of the 'L
sub n's as 'n' approaches
00:13:56.860 --> 00:14:00.630
infinity, and let's take the
limit of the 'U sub n's as 'n'
00:14:00.630 --> 00:14:04.095
approaches infinity, observing
that for each 'n', 'A sub R'
00:14:04.095 --> 00:14:06.010
is squeezed between these two.
00:14:06.010 --> 00:14:08.960
Consequently, since 'A sub R' is
squeezed between these two
00:14:08.960 --> 00:14:12.640
for each 'n', it must be
squeezed between these two
00:14:12.640 --> 00:14:13.770
when we go to the limit.
00:14:13.770 --> 00:14:18.100
In other words, whatever 'A sub
R' is, it must be between
00:14:18.100 --> 00:14:19.460
these two limits.
00:14:19.460 --> 00:14:21.170
Now here's the key step.
00:14:21.170 --> 00:14:25.420
It turns out, at least in this
particular problem, that the
00:14:25.420 --> 00:14:29.700
limit of 'L sub n' as 'n' goes
to infinity is the same as the
00:14:29.700 --> 00:14:32.180
limit of 'U sub n' as 'n'
goes to infinity.
00:14:32.180 --> 00:14:34.740
And the best way to see that
without becoming too fancy at
00:14:34.740 --> 00:14:38.880
this stage of the game is to
observe that for a given 'n',
00:14:38.880 --> 00:14:41.120
the difference between
'U sub n' and 'L
00:14:41.120 --> 00:14:43.880
sub n' is just '1/n'.
00:14:43.880 --> 00:14:47.560
And again, I'll just indicate
to you from this diagram how
00:14:47.560 --> 00:14:48.630
we can see that.
00:14:48.630 --> 00:14:51.820
Notice that back here, the way
we went from 'U sub n' to 'L
00:14:51.820 --> 00:14:54.580
sub n' was we just pushed
over this whole
00:14:54.580 --> 00:14:56.370
network by one unit.
00:14:56.370 --> 00:14:59.770
In other words, the rectangle
that we squeezed out in going
00:14:59.770 --> 00:15:03.050
from the upper sum to the
lower sum was just this
00:15:03.050 --> 00:15:05.940
particular rectangle whose
height was 1 and
00:15:05.940 --> 00:15:07.600
whose base was '1/n'.
00:15:07.600 --> 00:15:10.720
In other words, the area that
was kicked out in going from
00:15:10.720 --> 00:15:13.870
'L sub n' to 'U sub
n' is '1/n'.
00:15:13.870 --> 00:15:16.400
This is what this thing
here says here.
00:15:16.400 --> 00:15:18.180
This is worked out in
detail in the notes.
00:15:18.180 --> 00:15:20.440
But again, I just want to go
through this thing quickly so
00:15:20.440 --> 00:15:22.500
we get the idea of what's
happening over here.
00:15:22.500 --> 00:15:26.490
In other words, since 'U sub n'
minus 'L sub n' is '1/n',
00:15:26.490 --> 00:15:28.170
that's just another way
of saying what?
00:15:28.170 --> 00:15:30.610
That the limit of this
difference is just the limit
00:15:30.610 --> 00:15:32.720
of '1/n' as 'n' goes
to infinity.
00:15:32.720 --> 00:15:34.260
But that's clearly 0.
00:15:34.260 --> 00:15:38.140
In other words, these two limits
are the same because
00:15:38.140 --> 00:15:40.840
their difference goes
to 0 in the limit.
00:15:40.840 --> 00:15:43.960
Consequently, since these two
things are the same, and 'A
00:15:43.960 --> 00:15:48.170
sub R' is caught between these
two, it must be that 'A sub R'
00:15:48.170 --> 00:15:51.320
is equal to this common limit.
00:15:51.320 --> 00:15:54.390
And that's precisely what the
method of exhaustion was.
00:15:54.390 --> 00:15:57.880
We squeeze what we were looking
for between two
00:15:57.880 --> 00:16:00.070
estimates which converge
towards each
00:16:00.070 --> 00:16:01.450
other as 'n' got large.
00:16:01.450 --> 00:16:04.130
Now, by the way, I'm going to
point out the fact that it's
00:16:04.130 --> 00:16:06.510
very, very difficult
in general to find
00:16:06.510 --> 00:16:07.480
what this limit is.
00:16:07.480 --> 00:16:09.220
We're going to see
that in working a
00:16:09.220 --> 00:16:10.440
few specific problems.
00:16:10.440 --> 00:16:12.590
What I thought might
be informative--
00:16:12.590 --> 00:16:15.000
without going through the
details right now, I have
00:16:15.000 --> 00:16:18.800
taken the liberty of computing
'L sub n' and 'U sub n' in
00:16:18.800 --> 00:16:21.970
this problem for 'n'
equals 1,000.
00:16:21.970 --> 00:16:24.980
In other words, if we divided
this region into 1,000 equal
00:16:24.980 --> 00:16:28.350
parts and took the network of
circumscribed rectangles and
00:16:28.350 --> 00:16:32.710
inscribed rectangles, it would
turn out that the area of the
00:16:32.710 --> 00:16:40.690
inscribed rectangles would
be 0.3328335, and for the
00:16:40.690 --> 00:16:46.950
circumscribed rectangles,
0.3338335.
00:16:46.950 --> 00:16:49.300
And by the way, notice how
this is borne out.
00:16:49.300 --> 00:16:52.440
Notice that the difference
between these two is precisely
00:16:52.440 --> 00:16:57.270
1/1000, and that is '1/n'
with 'n' equal to 1,000.
00:16:57.270 --> 00:16:59.310
But that's not the important
point here.
00:16:59.310 --> 00:17:01.980
The important point is that just
looking at this decimal
00:17:01.980 --> 00:17:05.359
expansion, if I didn't know
anything else, I know that 'A
00:17:05.359 --> 00:17:08.720
sub R' is caught between
these two.
00:17:08.720 --> 00:17:12.200
Consequently, to two decimal
places, I can be sure that 'A
00:17:12.200 --> 00:17:15.020
sub R' is 0.33.
00:17:15.020 --> 00:17:17.470
And notice that if I want more
and more decimal place
00:17:17.470 --> 00:17:21.579
accuracy, especially if I have
access to a desk calculator, I
00:17:21.579 --> 00:17:24.040
don't really have to compute
the limit exactly.
00:17:24.040 --> 00:17:27.490
I can feed the formula into
the machine and put the
00:17:27.490 --> 00:17:29.740
squeeze on and get as
many decimal place
00:17:29.740 --> 00:17:31.940
accuracy as I want.
00:17:31.940 --> 00:17:34.720
By the way, as an aside, this is
exactly what we did in high
00:17:34.720 --> 00:17:39.470
school, when we said things
like let pi equal 22/7.
00:17:39.470 --> 00:17:42.090
Pi is not equal to 22/7.
00:17:42.090 --> 00:17:45.450
Among other things, 22/7
is a rational number.
00:17:45.450 --> 00:17:47.030
It's the ratio of two
whole numbers.
00:17:47.030 --> 00:17:48.610
Pi is an irrational number.
00:17:48.610 --> 00:17:51.260
What people really meant was
that you can't tell the
00:17:51.260 --> 00:17:55.470
difference between pi and 22/7
to two decimal places.
00:17:55.470 --> 00:17:58.050
They both begin 3.14.
00:17:58.050 --> 00:18:01.200
And consequently, if all you
wanted to measure was to two
00:18:01.200 --> 00:18:04.200
decimal digits, you could
use 22/7 as the
00:18:04.200 --> 00:18:05.610
approximation for pi.
00:18:05.610 --> 00:18:08.150
But if you wanted to squeeze out
more places, you'd have to
00:18:08.150 --> 00:18:10.210
use a more refined approach.
00:18:10.210 --> 00:18:13.200
By the way, let me point out
that one of the reasons that I
00:18:13.200 --> 00:18:17.130
chose the curve 'y' equals 'x
squared' to work with was--
00:18:17.130 --> 00:18:22.010
you may recall that back in our
discussion of mathematical
00:18:22.010 --> 00:18:25.890
induction, one of the problems
that we worked on was to show
00:18:25.890 --> 00:18:29.830
how we find by induction the
recipe for the sum of the
00:18:29.830 --> 00:18:31.740
first 'n' squares.
00:18:31.740 --> 00:18:36.260
And by way of review, let me
recall for you the fact that
00:18:36.260 --> 00:18:39.830
the sum of the first 'n' squares
was given by ''n'
00:18:39.830 --> 00:18:45.220
times 'n + 1' times
'2n + 1'' over 6.
00:18:45.220 --> 00:18:45.860
OK?
00:18:45.860 --> 00:18:49.250
In other words, going back to
our recipe for 'U sub n', I
00:18:49.250 --> 00:18:52.570
can now replace 1 squared plus
2 squared plus et cetera 'n
00:18:52.570 --> 00:18:57.580
squared' by ''n' times 'n +
1' times '2n + 1'' over 6.
00:18:57.580 --> 00:19:01.570
And if I now divide through by
'n cubed' judiciously, namely
00:19:01.570 --> 00:19:05.330
canceling out one 'n' with this
'n', dividing 'n + 1' by
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'
00:19:10.800 --> 00:19:13.440
over 'n'' times ''2n
+ 1' over 'n''.
00:19:13.440 --> 00:19:17.510
And that can be written even
more suggestively as 1/6 times
00:19:17.510 --> 00:19:20.680
'1 + '1/n'' times '2 + '1/n''.
00:19:20.680 --> 00:19:23.100
And I can now put a very
nice mathematical
00:19:23.100 --> 00:19:24.610
interpretation on this.
00:19:24.610 --> 00:19:29.680
Mainly, notice that no matter
how big 'n' is, '1 + '1/n'' is
00:19:29.680 --> 00:19:32.880
bigger than 1, you see, because
'1/n' is positive.
00:19:32.880 --> 00:19:35.350
And '2 + '1/n'' is
bigger than 2.
00:19:35.350 --> 00:19:40.450
Consequently, whatever this is,
it's bigger than 1/6 times
00:19:40.450 --> 00:19:43.930
1 times 2, which is 1/3.
00:19:43.930 --> 00:19:48.790
In other words, for each 'n', 'U
sub n' is greater than 1/3.
00:19:48.790 --> 00:19:52.340
It also happens that
as 'n' gets larger,
00:19:52.340 --> 00:19:54.300
'U sub n' gets smaller.
00:19:54.300 --> 00:19:58.060
Again, looking at this recipe,
the bigger 'n' is, the bigger
00:19:58.060 --> 00:20:00.890
is our denominator, and the
bigger the denominator, the
00:20:00.890 --> 00:20:02.950
smaller the fraction.
00:20:02.950 --> 00:20:07.560
And finally, notice that as 'n'
goes to infinity, 'U sub
00:20:07.560 --> 00:20:12.140
n' gets arbitrarily close to
1/3 in value, because '1/n'
00:20:12.140 --> 00:20:13.850
approaches 0 in the limit.
00:20:13.850 --> 00:20:15.580
In fact, it is 0 in the limit.
00:20:15.580 --> 00:20:19.380
So summarizing then, each 'U
sub n' is bigger than 1/3.
00:20:19.380 --> 00:20:23.280
As 'n' increases, 'U sub n'
decreases, and the limit of 'U
00:20:23.280 --> 00:20:26.490
sub n' as 'n' approaches
infinity is 1/3.
00:20:26.490 --> 00:20:29.790
Pictorially, what this means
is that if we locate 1/3 on
00:20:29.790 --> 00:20:35.580
the number line, the 'U sub n's
converge uniformly in the
00:20:35.580 --> 00:20:40.130
sense of moving steadily towards
the left, towards 1/3
00:20:40.130 --> 00:20:41.950
as the limit.
00:20:41.950 --> 00:20:43.770
Well, that's an upper squeeze.
00:20:43.770 --> 00:20:46.960
The lower squeeze comes from
the fact that in a similar
00:20:46.960 --> 00:20:51.000
way, we can show that
'L sub n' is 1/6--
00:20:51.000 --> 00:20:53.590
and look at how close this
comes to parallelling the
00:20:53.590 --> 00:20:59.460
structure of 'U sub n'-- '1 -
'1/n'' times '2 - '1/n''.
00:20:59.460 --> 00:21:02.310
Minuses here instead of pluses
as we had above.
00:21:02.310 --> 00:21:05.000
Now, mimicking what we did
before, notice now
00:21:05.000 --> 00:21:06.390
that for each 'n'--
00:21:06.390 --> 00:21:08.325
since we're subtracting
over here--
00:21:08.325 --> 00:21:12.220
for each 'n', 'L sub n'
is less than 1/3.
00:21:12.220 --> 00:21:14.940
But now since the fractions
get smaller as 'n' gets
00:21:14.940 --> 00:21:17.130
bigger, and you're subtracting
them, the
00:21:17.130 --> 00:21:19.120
difference becomes larger.
00:21:19.120 --> 00:21:21.770
In other words, now notice
that each 'L sub n'
00:21:21.770 --> 00:21:23.170
is less than 1/3.
00:21:23.170 --> 00:21:26.470
As 'n' increases, 'L
sub n' increases.
00:21:26.470 --> 00:21:28.540
And the limit of 'L sub
n' as 'n' approaches
00:21:28.540 --> 00:21:30.750
infinity is also 1/3.
00:21:30.750 --> 00:21:33.920
In other words, if we draw this
in back here, look what
00:21:33.920 --> 00:21:35.290
the 'L sub n's are doing.
00:21:35.290 --> 00:21:38.750
They're all less than 1/3, but
they move steadily towards the
00:21:38.750 --> 00:21:42.020
right as 'n' increases,
pushing in on 1/3.
00:21:42.020 --> 00:21:46.040
And since 'A sub R' is always
caught between these two, and
00:21:46.040 --> 00:21:49.460
these two converge relentlessly
upon 1/3, it must
00:21:49.460 --> 00:21:52.690
be, by definition, if there is
an area at all, that the area
00:21:52.690 --> 00:21:55.290
of the region 'R' must
be exactly 1/3.
00:21:55.290 --> 00:21:58.250
And by the way, just as a
quick aside, notice how
00:21:58.250 --> 00:22:01.840
important it is that we not
only have upper and lower
00:22:01.840 --> 00:22:03.790
bounds which converge.
00:22:03.790 --> 00:22:06.570
They must converge to
the same value.
00:22:06.570 --> 00:22:09.640
In other words, what I'm saying
is suppose all the 'U
00:22:09.640 --> 00:22:13.290
sub n's get arbitrarily close
to what I call script 'L sub
00:22:13.290 --> 00:22:16.120
1', and all the 'L sub n's
get arbitrarily close
00:22:16.120 --> 00:22:17.750
to 'L2' over here.
00:22:17.750 --> 00:22:22.160
Then all I'm saying is, all we
would know is that the area
00:22:22.160 --> 00:22:24.380
was someplace between
'L2' and 'L1'.
00:22:24.380 --> 00:22:27.470
We couldn't conclude that it
was exactly equal to 1/3.
00:22:27.470 --> 00:22:30.460
You see, the method of
exhaustion implies that all
00:22:30.460 --> 00:22:34.560
the error, all the room for
doubt, is squeezed out.
00:22:34.560 --> 00:22:37.100
By the way, there was nothing
sacred about choosing 'y'
00:22:37.100 --> 00:22:38.020
equals 'x squared'.
00:22:38.020 --> 00:22:39.630
We could generalize this.
00:22:39.630 --> 00:22:42.190
And we'll do this kind of
rapidly because I just want
00:22:42.190 --> 00:22:43.880
you to hear what I'm saying.
00:22:43.880 --> 00:22:46.280
This is all written out
in great detail in our
00:22:46.280 --> 00:22:47.280
supplementary notes.
00:22:47.280 --> 00:22:48.530
The idea is this.
00:22:48.530 --> 00:22:51.775
Let 'f' be any positive,
continuous function on 'a',
00:22:51.775 --> 00:22:54.100
'b' and non-decreasing.
00:22:54.100 --> 00:22:57.000
The non-decreasing part is just
for the convenience of
00:22:57.000 --> 00:23:00.210
being able to locate the high
and low points of each
00:23:00.210 --> 00:23:02.930
partition point conveniently.
00:23:02.930 --> 00:23:06.850
Partition 'a', 'b' into 'n'
equal parts, calling the first
00:23:06.850 --> 00:23:11.460
partition point 'x sub 0', the
last partition point 'x sub
00:23:11.460 --> 00:23:17.470
n', and the points in between
'x1' up through ''x 'n - 1''.
00:23:17.470 --> 00:23:18.760
In other words, we've
partitioned this closed
00:23:18.760 --> 00:23:21.230
interval into 'n' equal parts.
00:23:21.230 --> 00:23:22.980
And what we can now
do is what?
00:23:22.980 --> 00:23:26.630
Pick the lowest point in each
region, the highest point in
00:23:26.630 --> 00:23:28.360
each region, to form
a rectangle.
00:23:28.360 --> 00:23:32.280
We can form 'U sub
n' and 'L sub n'.
00:23:32.280 --> 00:23:35.490
And you see it's just what?
00:23:35.490 --> 00:23:39.830
For 'L sub n', you just pick
this height, which is 'f of 'x
00:23:39.830 --> 00:23:42.310
sub 0'', 'f of a'.
00:23:42.310 --> 00:23:44.270
The base is 'delta x'.
00:23:44.270 --> 00:23:47.620
You just add these all up until
you get to your last
00:23:47.620 --> 00:23:49.340
inscribed partition point.
00:23:49.340 --> 00:23:51.810
That's 'x sub 'n - 1''.
00:23:51.810 --> 00:23:52.570
OK.
00:23:52.570 --> 00:23:56.950
'U sub n' is the same thing,
only now your first rectangle
00:23:56.950 --> 00:24:00.220
has its height corresponding
to 'x' equals 'x1'.
00:24:00.220 --> 00:24:02.670
Consequently, the height
is 'f of x1'.
00:24:02.670 --> 00:24:04.460
The base is 'delta x'.
00:24:04.460 --> 00:24:06.940
Where in each of these, 'delta
x' is just what?
00:24:06.940 --> 00:24:09.730
This total length, which
is 'b - a', divided
00:24:09.730 --> 00:24:11.680
into 'n' equal parts.
00:24:11.680 --> 00:24:16.630
And to review the so-called
sigma notation that we have in
00:24:16.630 --> 00:24:17.760
our course--
00:24:17.760 --> 00:24:20.110
in this section in the textbook,
and we'll have
00:24:20.110 --> 00:24:23.130
exercises on this to make sure
that you are familiar with
00:24:23.130 --> 00:24:27.580
this notation, the shortcut
notation for writing this sum
00:24:27.580 --> 00:24:28.890
is given by this.
00:24:28.890 --> 00:24:32.260
We add up 'f of 'x sub
k'' times 'delta x'.
00:24:32.260 --> 00:24:37.680
As a subscript, 'k' is allowed
to vary from 1 to 'n'.
00:24:37.680 --> 00:24:42.220
And here we form the same sum,
only now the subscript 'k'
00:24:42.220 --> 00:24:45.960
varies from 0 to 'n - 1'.
00:24:45.960 --> 00:24:49.510
At any rate, the observations
are this.
00:24:49.510 --> 00:24:53.230
That 'U sub n' and 'L sub n'
converge to the same limit.
00:24:53.230 --> 00:24:55.150
In fact, this isn't
too hard to show.
00:24:55.150 --> 00:24:58.780
If you just algebraically
subtract 'L sub n' from 'U sub
00:24:58.780 --> 00:25:03.280
n', notice that the only terms
that won't cancel are
00:25:03.280 --> 00:25:04.890
the last term here.
00:25:04.890 --> 00:25:08.400
See, 'f of 'x sub n'' times
'delta x' has no counterpart
00:25:08.400 --> 00:25:12.310
here, because this ends with
the subscript 'n - 1'.
00:25:12.310 --> 00:25:16.590
Similarly, there is no 0
subscript in 'U sub n', so
00:25:16.590 --> 00:25:18.350
this term won't cancel.
00:25:18.350 --> 00:25:21.630
To make a long story short, if
we just subtract 'L sub n'
00:25:21.630 --> 00:25:25.420
from 'U sub n', that difference
will be 'f of 'x
00:25:25.420 --> 00:25:29.430
sub n'' times 'delta x'
minus 'f of 'x sub 0''
00:25:29.430 --> 00:25:30.910
times 'delta x'.
00:25:30.910 --> 00:25:33.740
See, that's just this difference
over here.
00:25:33.740 --> 00:25:36.610
'Delta x' is 'b - a' over 'n'.
00:25:36.610 --> 00:25:38.780
'x sub n' was called 'b'.
00:25:38.780 --> 00:25:41.220
'x sub 0' was called 'a'.
00:25:41.220 --> 00:25:44.480
In other words, the difference
between 'U sub n' and 'L sub
00:25:44.480 --> 00:25:48.020
n' is just 'f of b' minus
'f of a' times
00:25:48.020 --> 00:25:49.760
''b - a' over 'n''.
00:25:49.760 --> 00:25:54.000
And the important thing to
observe is that 'a', 'b', 'f
00:25:54.000 --> 00:25:56.960
of a', and 'f of b' are
fixed constants.
00:25:56.960 --> 00:26:00.460
The only thing that depends on
'n' is this denominator.
00:26:00.460 --> 00:26:03.620
Consequently, as 'n' goes to
infinity, since the rest of
00:26:03.620 --> 00:26:07.990
this thing is a constant, the
whole term goes to 0.
00:26:07.990 --> 00:26:11.050
In other words, the difference
between 'Un' and 'Ln' goes to
00:26:11.050 --> 00:26:12.200
0 in the limit.
00:26:12.200 --> 00:26:15.060
That means that the limit of 'U
sub n', as 'n' approaches
00:26:15.060 --> 00:26:18.190
infinity, equals the limit
of 'L sub n' as
00:26:18.190 --> 00:26:19.740
'n' approaches infinity.
00:26:19.740 --> 00:26:23.900
'A sub R' is always caught
between 'L sub n' and 'U sub
00:26:23.900 --> 00:26:26.480
n' by this very construction.
00:26:26.480 --> 00:26:30.170
Consequently, what this means is
that the area of the region
00:26:30.170 --> 00:26:34.830
'R' must equal this
common limit.
00:26:34.830 --> 00:26:37.360
Now by the way, this
looks very hard.
00:26:37.360 --> 00:26:38.360
And it is difficult.
00:26:38.360 --> 00:26:41.510
In fact, in general, these
limits are very hard to find.
00:26:41.510 --> 00:26:43.780
The point that I thought would
be interesting to mention at
00:26:43.780 --> 00:26:47.210
this stage of the game is that
if all we want is an estimate
00:26:47.210 --> 00:26:50.120
for the area under the curve,
there are faster and better
00:26:50.120 --> 00:26:52.020
ways of doing this.
00:26:52.020 --> 00:26:54.790
You see, the beauty of this
technique here is that it's a
00:26:54.790 --> 00:26:57.240
technique for exhausting
the space completely.
00:26:57.240 --> 00:26:59.850
We find the exact
areas this way.
00:26:59.850 --> 00:27:02.690
Let me give you a for
instance here.
00:27:02.690 --> 00:27:05.650
Namely, let me explain to you
what we mean by trapezoidal
00:27:05.650 --> 00:27:06.760
approximations.
00:27:06.760 --> 00:27:10.510
Let's take the same region,
'y' equals 'x squared'.
00:27:10.510 --> 00:27:13.370
And now, we'll divide it
into two parts here.
00:27:13.370 --> 00:27:17.130
And what we'll do is we'll
replace the arc of the curve
00:27:17.130 --> 00:27:20.360
by the straight line segment,
the chord, that joins two
00:27:20.360 --> 00:27:21.640
points here.
00:27:21.640 --> 00:27:24.540
In other words, using the
accented chalk here to
00:27:24.540 --> 00:27:25.410
illustrate this.
00:27:25.410 --> 00:27:28.310
What I'm going to do is instead
of finding the area of
00:27:28.310 --> 00:27:31.400
the region 'R', I'm going to
find the area of the region
00:27:31.400 --> 00:27:34.640
which has the top of
'R' replaced by
00:27:34.640 --> 00:27:36.520
these two line segments.
00:27:36.520 --> 00:27:40.050
All I want you to observe is
that this first region here is
00:27:40.050 --> 00:27:45.080
a triangle whose height is 1/4
and whose base is 1/2.
00:27:45.080 --> 00:27:47.700
Consequently, its area, being
one half the base times the
00:27:47.700 --> 00:27:50.380
height, is 1/16.
00:27:50.380 --> 00:27:55.370
The second region here is a
trapezoid whose bases are 1
00:27:55.370 --> 00:27:58.280
and 1/4 and whose
height is 1/2.
00:27:58.280 --> 00:28:01.150
And since the area of a
trapezoid is half the sum of
00:28:01.150 --> 00:28:06.870
the bases times the height, we
get that the area of this
00:28:06.870 --> 00:28:09.310
trapezoid is 5/16.
00:28:09.310 --> 00:28:12.950
Therefore, the area of the
triangle plus the trapezoid is
00:28:12.950 --> 00:28:17.400
6/16 or 3/8, which is 0.375.
00:28:17.400 --> 00:28:21.110
Recall that 1/3, we've just
seen, was the exact answer.
00:28:21.110 --> 00:28:24.450
And notice how close this
approximation is to the exact
00:28:24.450 --> 00:28:28.360
answer, even with just two
subdivisions over here.
00:28:28.360 --> 00:28:32.770
In fact, sparing you the
details, if you now divide the
00:28:32.770 --> 00:28:36.820
base here into four equal
parts, thus forming your
00:28:36.820 --> 00:28:39.790
trapezoidal approximations
corresponding to what?
00:28:39.790 --> 00:28:46.430
Altitudes of 1/16, 1/4, 9/16,
and 1, and adding up the areas
00:28:46.430 --> 00:28:48.400
of all these trapezoids--
00:28:48.400 --> 00:28:50.660
by the way, this is
a bit unfortunate.
00:28:50.660 --> 00:28:52.110
This is a triangle.
00:28:52.110 --> 00:28:56.110
But we can view a triangle as
being a degenerate trapezoid,
00:28:56.110 --> 00:28:58.990
meaning the height here just
happens to be 0 at this
00:28:58.990 --> 00:29:00.030
particular point.
00:29:00.030 --> 00:29:01.790
But let's not worry
about that.
00:29:01.790 --> 00:29:03.560
Let's just get the idea of
what the trapezoidal
00:29:03.560 --> 00:29:05.080
approximation idea means.
00:29:05.080 --> 00:29:10.280
We find the area of these four
trapezoids, and we use that as
00:29:10.280 --> 00:29:13.060
an approximation for the
area under the curve.
00:29:13.060 --> 00:29:16.070
By the way, just coming back
here for a moment, notice that
00:29:16.070 --> 00:29:19.850
because this curve is always
holding water, the chord will
00:29:19.850 --> 00:29:22.620
always lie above the arc.
00:29:22.620 --> 00:29:25.390
And consequently, not only do we
get an approximation using
00:29:25.390 --> 00:29:28.710
trapezoids this way that's
reasonably close, but we also
00:29:28.710 --> 00:29:32.210
know by the geometry here that
our approximation must be too
00:29:32.210 --> 00:29:34.170
large to be the right answer.
00:29:34.170 --> 00:29:36.730
In other words, we're going to
get an over-approximation.
00:29:36.730 --> 00:29:38.380
But watch how close we come.
00:29:38.380 --> 00:29:43.350
Leaving these details for you to
verify, all I show here is
00:29:43.350 --> 00:29:46.690
that if you add up the areas of
these four regions, we wind
00:29:46.690 --> 00:29:53.010
up with 44/128, which
is 11/32.
00:29:53.010 --> 00:29:57.590
And 11/32 is mighty close
to 1/3, being what?
00:29:57.590 --> 00:29:59.120
11/33.
00:29:59.120 --> 00:30:01.420
In other words, look at how
close, with just four
00:30:01.420 --> 00:30:04.900
subdivisions, we get to an
approximation that's good for
00:30:04.900 --> 00:30:07.360
the area under the curve,
without having to put the
00:30:07.360 --> 00:30:08.460
squeeze on.
00:30:08.460 --> 00:30:11.410
But to get the exact area,
we need the squeeze.
00:30:11.410 --> 00:30:14.740
We have to push the thing from
above, from below, and show
00:30:14.740 --> 00:30:17.630
that these two things that we're
squeezing it between, as
00:30:17.630 --> 00:30:20.700
gruesome as it sounds, converge
towards each other.
00:30:20.700 --> 00:30:23.970
The thing that we're looking
for is caught between them,
00:30:23.970 --> 00:30:26.560
and consequently must
be the common limit.
00:30:26.560 --> 00:30:29.690
That was the method of
exhaustion as known by the
00:30:29.690 --> 00:30:30.370
ancient Greeks.
00:30:30.370 --> 00:30:32.530
And if this seems tough to
you, think of it from two
00:30:32.530 --> 00:30:33.290
points of view.
00:30:33.290 --> 00:30:35.040
One, it is tough.
00:30:35.040 --> 00:30:39.510
And secondly, if people of some
2,500 years ago were able
00:30:39.510 --> 00:30:42.420
to do this, it should be at
least plausible that with a
00:30:42.420 --> 00:30:44.800
little bit of effort, we can get
a good feeling for what's
00:30:44.800 --> 00:30:45.840
going on here.
00:30:45.840 --> 00:30:48.880
In fact, hopefully, the
exercises to this particular
00:30:48.880 --> 00:30:51.440
unit will make this a little
bit easier for
00:30:51.440 --> 00:30:53.440
you to see in action.
00:30:53.440 --> 00:30:55.100
But at any rate--
00:30:55.100 --> 00:30:57.790
let me just make a few
asides over here.
00:30:57.790 --> 00:31:01.200
The deeper asides will be made
in greater detail in our
00:31:01.200 --> 00:31:02.750
supplementary notes.
00:31:02.750 --> 00:31:05.830
Notice that when we dealt with
differential calculus, it was
00:31:05.830 --> 00:31:08.960
very, very important in
differential calculus to have
00:31:08.960 --> 00:31:10.380
smooth curves.
00:31:10.380 --> 00:31:14.750
For finding areas, all you need
are continuous curves.
00:31:14.750 --> 00:31:16.490
In other words, for example,
we can find
00:31:16.490 --> 00:31:18.280
the area of a square.
00:31:18.280 --> 00:31:21.270
But certainly, a square
has sharp corners.
00:31:21.270 --> 00:31:23.960
In other words, those corners
are continuous, but they're
00:31:23.960 --> 00:31:25.820
not differentiable in terms
of the language of
00:31:25.820 --> 00:31:27.240
differential calculus.
00:31:27.240 --> 00:31:29.930
The point is that even
continuity can be weakened.
00:31:29.930 --> 00:31:31.630
Let me give you a definition.
00:31:31.630 --> 00:31:34.820
'f' is called piecewise
continuous on the interval
00:31:34.820 --> 00:31:39.910
from 'a' to 'b' if and only if
'f' is continuous except at a
00:31:39.910 --> 00:31:43.700
finite number of points where
it has jump discontinuities.
00:31:43.700 --> 00:31:47.210
For example, in terms of this
particular diagram, notice
00:31:47.210 --> 00:31:51.270
that my curve, 'y' equals 'f of
x', is discontinuous at a
00:31:51.270 --> 00:31:53.320
finite number of points,
namely at 'c1'
00:31:53.320 --> 00:31:55.290
and 'c2', two points.
00:31:55.290 --> 00:31:57.810
Notice that what happens at
those two points is you have
00:31:57.810 --> 00:32:00.300
just a finite jump
discontinuity.
00:32:00.300 --> 00:32:03.520
The point is that whereas this
curve is not continuous,
00:32:03.520 --> 00:32:06.550
notice that since a straight
line having no thickness has
00:32:06.550 --> 00:32:12.060
no area, if we replace the
given curve by this one--
00:32:12.060 --> 00:32:14.010
you see, putting in these
vertical lines--
00:32:14.010 --> 00:32:17.200
notice that this does form
a closed region.
00:32:17.200 --> 00:32:20.890
And to find the area of this
closed region, I can pretend
00:32:20.890 --> 00:32:24.880
it was made up of these three
particular regular regions.
00:32:24.880 --> 00:32:28.370
In other words, that even if I
just have jumps, since the
00:32:28.370 --> 00:32:31.250
jump does not contribute towards
the area, there is no
00:32:31.250 --> 00:32:34.920
harm done when one talks about
piecewise continuous in
00:32:34.920 --> 00:32:38.500
finding areas rather
than continuous.
00:32:38.500 --> 00:32:39.800
A second aside--
00:32:39.800 --> 00:32:42.090
well, I've actually called this
one aside number one,
00:32:42.090 --> 00:32:43.870
because this definition I
didn't call an aside--
00:32:43.870 --> 00:32:45.230
but the idea is this.
00:32:45.230 --> 00:32:49.010
Notice that up until now, we
were assuming that we had to
00:32:49.010 --> 00:32:52.680
form our 'U sub n's and our 'L
sub n's by choosing either the
00:32:52.680 --> 00:32:56.350
left endpoint of a partition
or the right endpoint.
00:32:56.350 --> 00:32:59.910
What I'd like you to see also as
a generalization is that if
00:32:59.910 --> 00:33:05.340
I pick any point between these
two extremes and form--
00:33:05.340 --> 00:33:07.720
look, I'll call it
'c sub k' between
00:33:07.720 --> 00:33:09.000
these two points here--
00:33:09.000 --> 00:33:13.510
and form this particular sum,
where 'c sub k' replaces
00:33:13.510 --> 00:33:18.530
either 'x sub k' or 'x sub 'k -
1'', that this sum here also
00:33:18.530 --> 00:33:20.990
gives me the area of
the region 'R'.
00:33:20.990 --> 00:33:23.540
In other words, that whatever
this sum is, it's caught
00:33:23.540 --> 00:33:25.970
between 'U sub n'
and 'L sub n'.
00:33:25.970 --> 00:33:28.900
And since 'U sub n' and 'L
sub n' have a common
00:33:28.900 --> 00:33:30.620
limit of 'A sub R'--
00:33:30.620 --> 00:33:32.855
in other words, since both
of these squeeze in
00:33:32.855 --> 00:33:34.220
towards 'A sub R'--
00:33:34.220 --> 00:33:37.860
this being caught between them
must also be 'A sub R'.
00:33:37.860 --> 00:33:39.790
That's what I've
said over here.
00:33:39.790 --> 00:33:42.650
And to see this thing in terms
of a picture, all we're saying
00:33:42.650 --> 00:33:48.460
is that when you pick the lowest
point in the interval,
00:33:48.460 --> 00:33:52.870
you get the rectangle that
contributes to 'L sub n'.
00:33:52.870 --> 00:33:55.440
When you pick the highest point
in the interval, you get
00:33:55.440 --> 00:33:58.320
the rectangle that contributes
to 'U sub n'.
00:33:58.320 --> 00:34:01.990
Consequently, for any point
between these two extremes--
00:34:01.990 --> 00:34:04.630
call that--that's what 'c sub k'
is, it's any point between
00:34:04.630 --> 00:34:05.525
these two--
00:34:05.525 --> 00:34:07.640
pick the height that corresponds
to that.
00:34:07.640 --> 00:34:11.409
And whatever rectangle you form
this way, that rectangle
00:34:11.409 --> 00:34:15.040
must have a greater area than
the rectangle that went into
00:34:15.040 --> 00:34:18.800
forming 'L sub n', but a lesser
area than the rectangle
00:34:18.800 --> 00:34:21.100
that went into forming
'U sub n'.
00:34:21.100 --> 00:34:23.909
And consequently, that's
where this particular
00:34:23.909 --> 00:34:25.260
result comes from.
00:34:25.260 --> 00:34:28.389
Again, this is done in more
detail in the notes, but I
00:34:28.389 --> 00:34:31.050
think some of these things you
should hear me say out loud
00:34:31.050 --> 00:34:33.300
rather than to rely on
your reading it.
00:34:33.300 --> 00:34:36.620
And finally, one more
important point.
00:34:36.620 --> 00:34:39.429
Up until now, it's been clear
that since we're talking about
00:34:39.429 --> 00:34:43.790
area, our region has to
lie above the x-axis.
00:34:43.790 --> 00:34:48.570
We can remove the restriction
that 'f' be non-negative if we
00:34:48.570 --> 00:34:51.210
replace area by net area.
00:34:51.210 --> 00:34:54.159
In other words, all I want you
to see here is that if my
00:34:54.159 --> 00:34:57.390
region happens to look something
like this, notice
00:34:57.390 --> 00:35:00.800
that between 'c' and 'b',
'f of x' is negative.
00:35:00.800 --> 00:35:04.010
Consequently, when I form things
of the form 'f of x'
00:35:04.010 --> 00:35:07.990
times 'delta x', 'delta x' being
positive, 'f of x' being
00:35:07.990 --> 00:35:10.960
negative, is going to give
me a negative result.
00:35:10.960 --> 00:35:15.270
In other words, algebraically,
if I form my summation, this
00:35:15.270 --> 00:35:17.840
portion in here will give
me a positive result.
00:35:17.840 --> 00:35:20.350
This portion in here will give
me a negative result.
00:35:20.350 --> 00:35:23.230
Consequently, working
algebraically, what I will
00:35:23.230 --> 00:35:27.280
find is not the true area
but the positive
00:35:27.280 --> 00:35:29.670
minus this amount here.
00:35:29.670 --> 00:35:32.680
In other words, what I'll
find is the net area.
00:35:32.680 --> 00:35:36.860
And again, this will all be
worked out in the notes.
00:35:36.860 --> 00:35:39.670
In addition, there'll be other
refinements made in the notes,
00:35:39.670 --> 00:35:43.130
such as that these partitions
don't have to be into equal
00:35:43.130 --> 00:35:44.950
parts or things like this.
00:35:44.950 --> 00:35:48.570
The important point from today's
lecture is this.
00:35:48.570 --> 00:35:55.000
Observe that this entire study
of area is done independently
00:35:55.000 --> 00:35:56.880
of differential calculus.
00:35:56.880 --> 00:36:00.730
In our next lecture, we are
going to show a truly
00:36:00.730 --> 00:36:05.020
remarkable, a wonderful
relationship between the
00:36:05.020 --> 00:36:08.620
differential calculus of before
and the so-called area
00:36:08.620 --> 00:36:10.980
or integral calculus
that we did today.
00:36:10.980 --> 00:36:13.030
At any rate, until next
time, goodbye.
00:36:15.910 --> 00:36:19.110
Funding for the publication of
this video was provided by the
00:36:19.110 --> 00:36:23.160
Gabriella and Paul Rosenbaum
Foundation.
00:36:23.160 --> 00:36:27.330
Help OCW continue to provide
free and open access to MIT
00:36:27.330 --> 00:36:31.540
courses by making a donation
at ocw.mit.edu/donate.