WEBVTT
00:00:00.040 --> 00:00:01.940
FEMALE SPEAKER: The following
content is provided under a
00:00:01.940 --> 00:00:03.690
Creative 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.
00:00:31.286 --> 00:00:32.290
PROFESSOR: Hi.
00:00:32.290 --> 00:00:35.520
Technically speaking, I suppose,
if time permitted, we
00:00:35.520 --> 00:00:39.030
should spend, or shall we say,
give equal time to integral
00:00:39.030 --> 00:00:42.180
calculus, the same time that
we gave to differential
00:00:42.180 --> 00:00:46.080
calculus, starting from scratch,
building ideas slowly
00:00:46.080 --> 00:00:46.750
and the like.
00:00:46.750 --> 00:00:49.710
But there are two logistic
reasons for not doing this,
00:00:49.710 --> 00:00:53.390
one of which is that the basic
definitions would be the same
00:00:53.390 --> 00:00:55.620
regardless of which branch
we started with.
00:00:55.620 --> 00:00:58.550
The second one is, of course,
that time becomes preciously
00:00:58.550 --> 00:01:02.010
tight, and we have to make the
best of what we can here.
00:01:02.010 --> 00:01:06.750
So what we're going to do now
is try to show, in terms of
00:01:06.750 --> 00:01:10.700
hindsight being better than
foresight, a motivation as to
00:01:10.700 --> 00:01:14.970
how one would have invented
differential calculus had it
00:01:14.970 --> 00:01:18.550
been motivated by the existing
integral calculus.
00:01:18.550 --> 00:01:22.070
In other words, what we want
to do today is to show the
00:01:22.070 --> 00:01:24.580
beautiful interplay between
differential
00:01:24.580 --> 00:01:26.290
and integral calculus.
00:01:26.290 --> 00:01:29.350
Since we have said in our last
lecture that we would like to
00:01:29.350 --> 00:01:32.830
begin with a study of integral
calculus, assuming that
00:01:32.830 --> 00:01:35.990
differential calculus had never
been invented, I think
00:01:35.990 --> 00:01:39.720
that we're obligated to try to
show how, then, differential
00:01:39.720 --> 00:01:42.860
calculus could have been
invented in the environment of
00:01:42.860 --> 00:01:44.510
the studying of areas.
00:01:44.510 --> 00:01:48.030
For this reason, I have called
today's lecture "The Marriage
00:01:48.030 --> 00:01:51.910
of Differential and Integral
Calculus." And essentially,
00:01:51.910 --> 00:01:54.560
the idea goes something
like this.
00:01:54.560 --> 00:01:57.370
Last time, we were talking
about finding
00:01:57.370 --> 00:01:59.240
areas under a curve.
00:01:59.240 --> 00:02:02.170
And to find the area under the
curve, we saw that we could do
00:02:02.170 --> 00:02:04.450
this as the limit of
a certain sum.
00:02:04.450 --> 00:02:07.630
In other words, an infinite sum
having a particular limit.
00:02:07.630 --> 00:02:11.450
Now, to motivate the concept of
rate of change, one could
00:02:11.450 --> 00:02:13.500
ask the following question.
00:02:13.500 --> 00:02:16.710
A curve like 'y' equals 'f of
x', and a starting point, 'x'
00:02:16.710 --> 00:02:17.830
equals 'a'.
00:02:17.830 --> 00:02:21.980
As I move out along the x-axis
from 'a', I could study the
00:02:21.980 --> 00:02:26.670
area under the curve, say, of
the curve, 'y' equals 'f of
00:02:26.670 --> 00:02:30.600
x', from 'x' equals 'a'
to some value 'x1'.
00:02:30.600 --> 00:02:34.280
And the question I could then
ask is, I wonder how fast this
00:02:34.280 --> 00:02:38.040
area is changing at the instant
that 'x' equals 'x1'.
00:02:38.040 --> 00:02:41.860
In other words, how fast is the
area changing under the
00:02:41.860 --> 00:02:47.150
curve as 'x' moves out along the
positive x-axis this way?
00:02:47.150 --> 00:02:48.160
You see?
00:02:48.160 --> 00:02:50.960
Now, the idea is this.
00:02:50.960 --> 00:02:53.950
I can then mimic the same
definition that we gave for
00:02:53.950 --> 00:02:56.890
instantaneous rate of change
and differential calculus.
00:02:56.890 --> 00:03:00.060
Namely, I can say, why can't
we define the instantaneous
00:03:00.060 --> 00:03:03.510
rate of change to be the average
rate of change with
00:03:03.510 --> 00:03:04.520
respect to 'x'?
00:03:04.520 --> 00:03:06.350
And then we'll take the
limit as 'delta x'
00:03:06.350 --> 00:03:08.130
approaches 0, et cetera.
00:03:08.130 --> 00:03:10.420
And if I do that, watch what
happens over here.
00:03:10.420 --> 00:03:13.600
You see what I do in terms of
this picture, is I say, OK,
00:03:13.600 --> 00:03:15.670
here's the area under
the curve when
00:03:15.670 --> 00:03:17.300
'x' is equal to 'x1'.
00:03:17.300 --> 00:03:20.720
Now I'll let 'x1' change by some
amount 'delta x', meaning
00:03:20.720 --> 00:03:23.120
this'll be the point
'x1 + delta x'.
00:03:23.120 --> 00:03:25.770
And this brings about a change
in my area, which
00:03:25.770 --> 00:03:27.330
I'll call 'delta A'.
00:03:27.330 --> 00:03:29.790
See the change in area
as 'x' goes from 'x1'
00:03:29.790 --> 00:03:31.840
to 'x1 + delta x'.
00:03:31.840 --> 00:03:35.830
Now, what I would like to find
is 'delta A' divided by 'delta
00:03:35.830 --> 00:03:39.390
x', taking the limit as 'delta
x' approaches 0.
00:03:39.390 --> 00:03:41.720
Remember, last time we
showed our three
00:03:41.720 --> 00:03:43.360
basic axioms for area.
00:03:43.360 --> 00:03:45.710
We're obliged to
still use them.
00:03:45.710 --> 00:03:46.880
Let's do that here.
00:03:46.880 --> 00:03:49.960
For example, the way I've drawn
this curve by rising
00:03:49.960 --> 00:03:53.540
this way, notice that the lowest
point in this interval
00:03:53.540 --> 00:03:55.360
occurs when 'x' equals 'x1'.
00:03:55.360 --> 00:03:59.420
The highest point occurs when
'x' equals 'x1 + delta x'.
00:03:59.420 --> 00:04:03.380
Consequently, the rectangle
whose base is the closed
00:04:03.380 --> 00:04:07.400
interval from 'x1' to 'x1 +
delta x', the rectangle whose
00:04:07.400 --> 00:04:11.680
height corresponds to 'x' equals
'x1' is too small in
00:04:11.680 --> 00:04:13.670
area to be 'delta A'.
00:04:13.670 --> 00:04:17.160
In other words, that rectangle
is inscribed in 'delta A'.
00:04:17.160 --> 00:04:20.760
And in the same way, the
rectangle, with base 'x1' to
00:04:20.760 --> 00:04:25.720
'x1 + delta x', whose height is
given by the x-coordinate,
00:04:25.720 --> 00:04:30.010
'x1 + delta x', that rectangle
has an area which is too large
00:04:30.010 --> 00:04:31.610
to be the exact area.
00:04:31.610 --> 00:04:34.830
In other words, notice that
the area of the rectangle,
00:04:34.830 --> 00:04:38.900
which is too small to be the
exact area, is 'f of x1'
00:04:38.900 --> 00:04:40.690
times 'delta x'.
00:04:40.690 --> 00:04:44.310
The area of the rectangle which
is too big to be the
00:04:44.310 --> 00:04:50.100
correct 'delta A' is 'f of 'x1
+ delta x'' times 'delta x'.
00:04:50.100 --> 00:04:54.530
And the area, 'delta A', is
caught between these two.
00:04:54.530 --> 00:04:58.850
In other words, we now have the
inequality that 'f of x1'
00:04:58.850 --> 00:05:02.100
times 'delta x' is less than
'delta A', which in turn is
00:05:02.100 --> 00:05:06.550
less than 'f of 'x1 + delta
x'' times 'delta x'.
00:05:06.550 --> 00:05:09.700
Now, we'll assume that 'delta x'
is positive, the way we've
00:05:09.700 --> 00:05:11.940
drawn it in this diagram.
00:05:11.940 --> 00:05:15.320
Adjustments have to be made if
'delta x' is negative, in
00:05:15.320 --> 00:05:17.920
other words, if we let
'delta x' approach 0
00:05:17.920 --> 00:05:19.390
through negative values.
00:05:19.390 --> 00:05:22.300
This is taken care of in the
textbook, as well as mentioned
00:05:22.300 --> 00:05:22.690
in our notes.
00:05:22.690 --> 00:05:25.550
But for the sake of just getting
the main idea here,
00:05:25.550 --> 00:05:28.060
we'll assume that 'delta
x' is positive.
00:05:28.060 --> 00:05:30.310
We divide through
by 'delta x'.
00:05:30.310 --> 00:05:33.310
That gives us 'f of x1' is
less than 'delta a' over
00:05:33.310 --> 00:05:35.380
'delta x', which in turn
is less than 'f
00:05:35.380 --> 00:05:37.420
of 'x1 + delta x''.
00:05:37.420 --> 00:05:39.690
In other words, had 'delta x'
been negative, we would have
00:05:39.690 --> 00:05:43.150
had to reverse the inequality
signs, et cetera, but we're
00:05:43.150 --> 00:05:45.090
not going to worry about
that right now.
00:05:45.090 --> 00:05:46.510
Remember, what is 'delta x'?
00:05:46.510 --> 00:05:50.610
It's a non-zero number which is
going to be made to become
00:05:50.610 --> 00:05:52.380
arbitrarily close to 0.
00:05:52.380 --> 00:05:53.470
Well, here's the idea.
00:05:53.470 --> 00:05:56.500
We now have 'delta a' over
'delta x' squeezed between
00:05:56.500 --> 00:05:58.180
these two values.
00:05:58.180 --> 00:06:02.190
And we say, OK, let's let
'delta x' approach 0.
00:06:02.190 --> 00:06:03.490
Now, here's the key point.
00:06:03.490 --> 00:06:07.730
As 'delta x' approaches 0,
certainly, 'x1 + delta x'
00:06:07.730 --> 00:06:10.110
approaches 'x1'.
00:06:10.110 --> 00:06:13.310
And here's where we use the fact
that 'f' is continuous.
00:06:13.310 --> 00:06:16.460
In other words, notice that the
mere fact that 'x1 + delta
00:06:16.460 --> 00:06:20.360
x' approaches 'x1' is not enough
to say, therefore, that
00:06:20.360 --> 00:06:23.540
'f of 'x1 + delta x'' approaches
'f of x1'.
00:06:23.540 --> 00:06:27.610
As we saw before, that's only
true if 'f' is continuous.
00:06:27.610 --> 00:06:31.660
In other words, by continuity,
if 'x1 + delta x' approaches
00:06:31.660 --> 00:06:36.830
'x1', then 'f of 'x1 + delta
x'' approaches 'f of x1'.
00:06:36.830 --> 00:06:40.210
So assuming, then, that 'f' is
continuous, what do we get we
00:06:40.210 --> 00:06:41.360
put the squeeze on?
00:06:41.360 --> 00:06:43.910
'x1' is what's picked
to be a fixed point.
00:06:43.910 --> 00:06:46.620
Consequently, as 'delta
x' approaches 0, 'f of
00:06:46.620 --> 00:06:48.940
x1' stays 'f of x1'.
00:06:48.940 --> 00:06:52.480
'f of 'x1 + delta x'' approaches
'f of x1'.
00:06:52.480 --> 00:06:57.110
In other words, 'delta A'
divided by 'delta x' is caught
00:06:57.110 --> 00:07:02.170
between two numbers, two
sequences, two sets of bounds,
00:07:02.170 --> 00:07:05.510
whichever way you want to say
this, both of which converge
00:07:05.510 --> 00:07:07.960
to the common limit,
'f of x1'.
00:07:07.960 --> 00:07:11.520
In other words, since 'delta A'
over 'delta x' is squeezed
00:07:11.520 --> 00:07:15.320
between these two, the limit of
'delta a' over 'delta x',
00:07:15.320 --> 00:07:18.670
as 'delta x' approaches 0, must
equal this common limit.
00:07:18.670 --> 00:07:23.140
In other words, 'dA/dx', the
limit of 'delta A' over 'delta
00:07:23.140 --> 00:07:27.480
x', as 'delta x' approaches 0,
evaluated at 'x' equals 'x1',
00:07:27.480 --> 00:07:29.130
is 'f of x1'.
00:07:29.130 --> 00:07:34.570
Or another way of saying this,
'dA/dx' is 'f of x'.
00:07:34.570 --> 00:07:39.180
Now, that result is partly
intuitive and partly
00:07:39.180 --> 00:07:39.950
remarkable.
00:07:39.950 --> 00:07:44.170
What it says is, roughly
speaking, is that the area
00:07:44.170 --> 00:07:49.480
under the curve is changing at
a rate equal instantaneously
00:07:49.480 --> 00:07:52.090
to the height corresponding
to the
00:07:52.090 --> 00:07:53.370
x-coordinate at that point.
00:07:53.370 --> 00:07:56.970
In other words, to define how
fast the area is changing as
00:07:56.970 --> 00:08:01.140
'x' moves, all you have to do
is measure the height of the
00:08:01.140 --> 00:08:06.510
curve corresponding to that
particular value of 'x'.
00:08:06.510 --> 00:08:09.220
Now, what does this have to
do, then, with relating
00:08:09.220 --> 00:08:11.420
integral and differential
calculus?
00:08:11.420 --> 00:08:14.310
Notice already we begin to
get some sort of a hint.
00:08:14.310 --> 00:08:18.730
What we've shown is now that
the area is going to be
00:08:18.730 --> 00:08:22.620
related somehow to the inverse
derivative of 'f of x'.
00:08:22.620 --> 00:08:27.310
In fact, this is precisely
what is meant by a rather
00:08:27.310 --> 00:08:28.840
important result.
00:08:28.840 --> 00:08:31.580
In fact, it's important enough
so it's called the first
00:08:31.580 --> 00:08:34.890
fundamental theorem of
integral calculus.
00:08:34.890 --> 00:08:36.390
And it simply says this.
00:08:36.390 --> 00:08:41.090
Suppose we know explicitly a
function, capital 'G', such
00:08:41.090 --> 00:08:44.910
that the derivative of capital
'G' is 'f', where the 'f' now
00:08:44.910 --> 00:08:48.940
refers to the 'f' that we're
talking about in our problem.
00:08:48.940 --> 00:08:52.640
Now, look it, what we already
know is that 'A', the area
00:08:52.640 --> 00:08:55.170
function, has its derivative
equal to 'f'.
00:08:55.170 --> 00:08:58.610
Consequently, since the area
function and G have the same
00:08:58.610 --> 00:09:01.630
derivative, they must differ
by, at most, a constant.
00:09:01.630 --> 00:09:05.180
In other words, the area as
a function of 'x' is this
00:09:05.180 --> 00:09:07.310
function, 'G of x' plus 'c'.
00:09:07.310 --> 00:09:08.590
And what is this 'G of x'?
00:09:08.590 --> 00:09:10.030
It's not any old function.
00:09:10.030 --> 00:09:10.710
It's what?
00:09:10.710 --> 00:09:14.120
An inverse derivative
of 'f of x'.
00:09:14.120 --> 00:09:21.820
Since the area evaluated at
'x' equals 'a' is 0, see,
00:09:21.820 --> 00:09:24.920
namely, the area under the
curve, as 'x' goes from 'A' to
00:09:24.920 --> 00:09:26.910
'a', see, that's just a line.
00:09:26.910 --> 00:09:28.880
A line has no thickness,
has no area.
00:09:28.880 --> 00:09:29.680
This is 0.
00:09:29.680 --> 00:09:32.690
We have, by plugging back
in to here, that 'A' of
00:09:32.690 --> 00:09:34.450
little 'a' is 0.
00:09:34.450 --> 00:09:38.100
That in turn is 'G
of a' plus 'c'.
00:09:38.100 --> 00:09:41.990
And solving this for 'c', we
find that 'c' is equal to
00:09:41.990 --> 00:09:43.340
minus 'G of a'.
00:09:43.340 --> 00:09:46.820
In other words, the area under
the curve 'y' equals 'f of x',
00:09:46.820 --> 00:09:49.670
as a function of 'x',
is simply what?
00:09:49.670 --> 00:09:53.640
It's 'G of x' minus 'G of a',
where 'G' is any function
00:09:53.640 --> 00:09:55.290
whose derivative is 'f'.
00:09:55.290 --> 00:09:59.730
If we now let 'x' equal 'b',
we're back to the situation of
00:09:59.730 --> 00:10:03.630
finding the usual region that
we're talking about here,
00:10:03.630 --> 00:10:05.490
namely, what is 'A sub R'?
00:10:05.490 --> 00:10:08.200
It's the region bounded above--
let's come back here
00:10:08.200 --> 00:10:11.380
and take a look at this-- it's
the region bounded above by
00:10:11.380 --> 00:10:15.220
the curve, 'y' equals 'f of x',
bounded on the left by 'x'
00:10:15.220 --> 00:10:19.910
equals 'a', on the right by 'x'
equals 'b', and below by
00:10:19.910 --> 00:10:21.150
the x-axis.
00:10:21.150 --> 00:10:24.510
According to this notation, the
area of the region 'R' is
00:10:24.510 --> 00:10:29.020
just the area as a function of
'x' when 'x1' is equal to 'b'.
00:10:29.020 --> 00:10:32.465
In other words, the area of our
region 'R' is just 'a' of
00:10:32.465 --> 00:10:36.260
'b', which is 'G of b' minus
'G of a', where 'G' is any
00:10:36.260 --> 00:10:39.880
function whose derivative
is 'f'.
00:10:39.880 --> 00:10:40.670
Now remember--
00:10:40.670 --> 00:10:42.080
and here's the key point--
00:10:42.080 --> 00:10:47.480
remember that the area of the
region 'R' did not necessitate
00:10:47.480 --> 00:10:51.080
us having to know a function
whose derivative was 'f'.
00:10:51.080 --> 00:10:53.880
In other words, remember, in our
last lecture, how did we
00:10:53.880 --> 00:10:56.000
find areas of regions
like 'R'?
00:10:56.000 --> 00:10:58.990
What we did was is we formed
these 'U sub n's,
00:10:58.990 --> 00:11:00.150
these 'L sub n's.
00:11:00.150 --> 00:11:04.170
We inscribed and circumscribed
networks of rectangles, put
00:11:04.170 --> 00:11:07.850
the squeeze on by evaluating
this particular limit.
00:11:07.850 --> 00:11:10.390
And whatever that limit was,
that was the area of the
00:11:10.390 --> 00:11:11.250
region 'R'.
00:11:11.250 --> 00:11:14.120
In other words, if I had never
heard of the inverse
00:11:14.120 --> 00:11:16.350
derivative, I could still
find the area of the
00:11:16.350 --> 00:11:18.290
region 'R' this way.
00:11:18.290 --> 00:11:22.510
However, if I just happen to
know a function 'G', whose
00:11:22.510 --> 00:11:26.612
derivative is 'f', I have a much
easier way of doing this.
00:11:26.612 --> 00:11:29.630
In fact, let me summarize
that.
00:11:29.630 --> 00:11:33.340
You see, we can compute
the limit as
00:11:33.340 --> 00:11:34.960
'n' approaches infinity.
00:11:34.960 --> 00:11:38.700
Summation 'k' goes from 1 to
'n', 'f of 'c sub k'' 'delta
00:11:38.700 --> 00:11:41.850
x', by use of inverse
derivatives.
00:11:41.850 --> 00:11:44.080
See, again, the highlight
being what?
00:11:44.080 --> 00:11:47.590
You can still work with this the
same way as we did in our
00:11:47.590 --> 00:11:50.110
last lecture.
00:11:50.110 --> 00:11:53.330
There is absolutely no need to
have had to ever heard of a
00:11:53.330 --> 00:11:56.030
derivative to solve this
type of problem, even
00:11:56.030 --> 00:11:57.450
though it may be messy.
00:11:57.450 --> 00:11:59.370
Now, what's the best proof
I have of this?
00:11:59.370 --> 00:12:02.680
Well, I guess one of the best
proofs is to specifically
00:12:02.680 --> 00:12:07.000
refer to one of the exercises
in the previous unit.
00:12:07.000 --> 00:12:08.300
It was a tough exercise.
00:12:08.300 --> 00:12:11.050
I did it as a learning exercise
because I felt it was
00:12:11.050 --> 00:12:13.870
something that was messy and
that you had to be guided
00:12:13.870 --> 00:12:16.880
through in order not to become
hopelessly lost.
00:12:16.880 --> 00:12:19.230
By way of review, the
problem was this.
00:12:19.230 --> 00:12:23.990
We took as our region 'R' the
region bounded above by 'y'
00:12:23.990 --> 00:12:28.850
equals 'sine x', below by the
x-axis, on the right by the
00:12:28.850 --> 00:12:31.750
line 'x' equals pi/2.
00:12:31.750 --> 00:12:34.620
And the problem was, define the
area of the region 'R'.
00:12:34.620 --> 00:12:38.420
And we solved this problem in
the last unit without recourse
00:12:38.420 --> 00:12:41.740
to derivatives because, as of
the time that we were in the
00:12:41.740 --> 00:12:45.780
last unit, we had no results
relating derivatives
00:12:45.780 --> 00:12:48.890
to limits of sums.
00:12:48.890 --> 00:12:49.800
How did we do this?
00:12:49.800 --> 00:12:54.080
We partitioned this into n
parts, and we formed the sum
00:12:54.080 --> 00:12:57.680
limit as 'n' approaches
infinity.
00:12:57.680 --> 00:12:59.460
'Sigma k' goes from 1 to 'n'.
00:12:59.460 --> 00:13:02.780
We broke this thing up into 'n'
equal parts, so the size
00:13:02.780 --> 00:13:04.020
of each piece was what?
00:13:04.020 --> 00:13:06.170
pi/2 divided by 'n'.
00:13:06.170 --> 00:13:07.440
That's 'pi/2n'.
00:13:07.440 --> 00:13:10.240
We computed the sine of each of
these n-points, et cetera.
00:13:10.240 --> 00:13:12.880
And whatever that limit was,
that was the area of the
00:13:12.880 --> 00:13:13.680
region 'R'.
00:13:13.680 --> 00:13:16.400
And as you recall, that homework
problem, we found
00:13:16.400 --> 00:13:18.840
that the answer to that problem
was that the area of
00:13:18.840 --> 00:13:21.330
the region 'R' was 1.
00:13:21.330 --> 00:13:24.670
Now again, as messy as that was,
it proved that we could
00:13:24.670 --> 00:13:27.330
at least solve the problem
with no knowledge of
00:13:27.330 --> 00:13:28.200
derivatives.
00:13:28.200 --> 00:13:31.230
How does the first fundamental
theorem apply here?
00:13:31.230 --> 00:13:32.390
The way the first fundamental
theorem
00:13:32.390 --> 00:13:34.020
applies is the following.
00:13:34.020 --> 00:13:35.890
The first fundamental
theorem says this.
00:13:35.890 --> 00:13:40.060
Look it, to evaluate this
particular sum, all we have to
00:13:40.060 --> 00:13:44.750
do is find a function whose
derivative is 'sine x'.
00:13:44.750 --> 00:13:48.290
See, notice in this problem, the
general 'a' and 'b' of the
00:13:48.290 --> 00:13:53.120
above is now played by 'a'
equals 0 and 'b' equals pi/2.
00:13:53.120 --> 00:13:54.500
All we have to do is what?
00:13:54.500 --> 00:13:58.260
Find the function 'G' whose
derivative is 'sine x', and
00:13:58.260 --> 00:14:02.670
the answer should then be 'G
of pi/2' minus 'G of 0'.
00:14:02.670 --> 00:14:05.800
Well, you see this happens
to be one that, with our
00:14:05.800 --> 00:14:08.970
knowledge of differential
calculus, we
00:14:08.970 --> 00:14:09.970
can do rather easily.
00:14:09.970 --> 00:14:12.570
In other words, do we, at the
tip of our tongues, have a
00:14:12.570 --> 00:14:15.990
function whose derivative
is 'sine x'?
00:14:15.990 --> 00:14:17.030
And the answer is yes.
00:14:17.030 --> 00:14:20.130
Since the derivative of 'cosine
x' is 'minus sine x',
00:14:20.130 --> 00:14:23.600
the derivative of 'minus
cosine x' is 'sine x'.
00:14:23.600 --> 00:14:26.530
In other words, we can choose
for our 'G of x' here
00:14:26.530 --> 00:14:29.020
'minus cosine x'.
00:14:29.020 --> 00:14:30.530
In other words, that the
area of the region
00:14:30.530 --> 00:14:32.050
'R' should be what?
00:14:32.050 --> 00:14:36.460
'G of pi/2' minus 'G of
0', where 'G of x' is
00:14:36.460 --> 00:14:38.180
'minus cosine x'.
00:14:38.180 --> 00:14:41.940
Well, you see, cosine
pi/2 is 0.
00:14:41.940 --> 00:14:49.120
Cosine of 0 is 1, so 0 minus
minus 1 is equal to 1.
00:14:49.120 --> 00:14:52.440
And notice that, first of all,
we get the same answer.
00:14:52.440 --> 00:14:55.510
And secondly, notice this
two-line job over here.
00:14:55.510 --> 00:14:58.550
We not only get the correct
answer, but we get the answer
00:14:58.550 --> 00:15:01.450
very much more rapidly
than we did by the
00:15:01.450 --> 00:15:03.620
so-called limit process.
00:15:03.620 --> 00:15:06.640
Now, I'll come back to this in a
minute, because you may be a
00:15:06.640 --> 00:15:08.190
little bit angry at me now.
00:15:08.190 --> 00:15:10.810
Namely, you may be asking,
after having to do this
00:15:10.810 --> 00:15:14.220
problem the hard way, why did
I have to do the usual
00:15:14.220 --> 00:15:17.510
teacher's trick here and show
you the hard way of doing this
00:15:17.510 --> 00:15:20.000
and then waiting until the next
unit before I show you
00:15:20.000 --> 00:15:21.120
the easy way?
00:15:21.120 --> 00:15:23.780
Well, there happens to be a
catch here that I'll come back
00:15:23.780 --> 00:15:25.490
to in just a moment.
00:15:25.490 --> 00:15:29.340
But before I do that, I'd like
to make an aside, an aside
00:15:29.340 --> 00:15:30.410
that's rather important.
00:15:30.410 --> 00:15:34.220
Namely, you may recall, as I
started to work over here,
00:15:34.220 --> 00:15:38.290
that this reminded you of a
notation that we were using
00:15:38.290 --> 00:15:42.280
back when we introduced the
concept of the inverse of
00:15:42.280 --> 00:15:43.700
differentiation.
00:15:43.700 --> 00:15:48.070
Namely, we earlier used a
notation, integral from 'a' to
00:15:48.070 --> 00:15:53.160
'b', 'f of x' 'dx' to denote 'G
of b' minus 'G of a', where
00:15:53.160 --> 00:15:55.600
'G prime' equals 'f'.
00:15:55.600 --> 00:15:58.300
What we have shown by the first
fundamental theorem,
00:15:58.300 --> 00:16:02.770
then, using this notation, is
that the area of the region
00:16:02.770 --> 00:16:06.990
'R' is integral from 'a'
to 'b', 'f of x' 'dx'.
00:16:06.990 --> 00:16:08.480
In other words, this
means what?
00:16:08.480 --> 00:16:14.910
It means 'G of b' minus
'G of a', where 'G
00:16:14.910 --> 00:16:16.880
prime' equals 'f'.
00:16:16.880 --> 00:16:19.220
OK, let's pause here
for a moment.
00:16:19.220 --> 00:16:22.010
See, as we developed
our course, this is
00:16:22.010 --> 00:16:23.930
how this would evolve.
00:16:23.930 --> 00:16:28.470
The interesting thing is that,
historically, this notation
00:16:28.470 --> 00:16:33.400
was not used to define 'G of
b' minus 'G of a', where 'G
00:16:33.400 --> 00:16:34.750
prime' equaled 'f'.
00:16:34.750 --> 00:16:38.400
Historically, what happened was
that this notation, called
00:16:38.400 --> 00:16:40.970
the definite integral, the
integral from 'a' to 'b', 'f
00:16:40.970 --> 00:16:42.910
of x' 'dx', was--
00:16:42.910 --> 00:16:45.620
I don't know if it's proper to
say invented, but let me just
00:16:45.620 --> 00:16:47.650
say it in quotation marks
to play it safe--
00:16:47.650 --> 00:16:51.490
it was "invented" to denote
the limit of this sum.
00:16:51.490 --> 00:16:55.840
In fact, notice how much more
meaningful the symbol looks in
00:16:55.840 --> 00:16:56.930
this connotation.
00:16:56.930 --> 00:16:59.320
In other words, this is
the limit of a sum.
00:16:59.320 --> 00:17:02.100
You can think of sum as
beginning with 's', and the
00:17:02.100 --> 00:17:05.180
integral of sine as
an elongated 's'.
00:17:05.180 --> 00:17:08.690
In other words, when one first
wrote this symbol, the
00:17:08.690 --> 00:17:13.040
definite integral, it was meant
to denote this limit.
00:17:13.040 --> 00:17:16.160
The important point is that,
by the first fundamental
00:17:16.160 --> 00:17:19.355
theorem, the definite integral,
whether it's a limit
00:17:19.355 --> 00:17:23.230
or not, turns out to be 'G of
b' minus 'G of a', where 'G
00:17:23.230 --> 00:17:24.589
prime' equals 'f'.
00:17:24.589 --> 00:17:28.089
In other words, it really makes
no difference in terms
00:17:28.089 --> 00:17:30.540
of what answer you get, whether
you think of this
00:17:30.540 --> 00:17:34.010
definite integral as meaning
this, or whether you think of
00:17:34.010 --> 00:17:36.940
the definite integral
as meaning this.
00:17:36.940 --> 00:17:38.800
See, numerically, they'll
be the same.
00:17:38.800 --> 00:17:41.820
But conceptually, they're
quite different.
00:17:41.820 --> 00:17:46.260
And that's precisely the point
that leads to the most, I
00:17:46.260 --> 00:17:49.720
think, probably the most
difficult part all the course,
00:17:49.720 --> 00:17:51.090
at least up until now.
00:17:51.090 --> 00:17:53.670
It's something that more
people cause and have
00:17:53.670 --> 00:17:55.210
misinterpretation over.
00:17:55.210 --> 00:17:57.090
And so I'd like to present
this in a very,
00:17:57.090 --> 00:17:58.580
very gradual way.
00:17:58.580 --> 00:18:01.540
See, the result will ultimately
be known as the
00:18:01.540 --> 00:18:04.490
second fundamental theorem
of integral calculus.
00:18:04.490 --> 00:18:07.510
I've taken the liberty of
writing that over here.
00:18:07.510 --> 00:18:10.620
But it's going to be quite a
while before I actually say
00:18:10.620 --> 00:18:12.760
what the theorem
is explicitly.
00:18:12.760 --> 00:18:16.280
What I'd like to do is to pick
up where I left off when I
00:18:16.280 --> 00:18:18.120
made my aside about
the notation
00:18:18.120 --> 00:18:19.770
of a definite integral.
00:18:19.770 --> 00:18:21.820
And what I'd like to mention
now is, for the person who
00:18:21.820 --> 00:18:23.760
says, look it, why did
you make us find the
00:18:23.760 --> 00:18:24.900
areas the hard way?
00:18:24.900 --> 00:18:29.250
Why didn't you just say let us
compute 'G of b' minus 'G of
00:18:29.250 --> 00:18:32.200
a', where 'G prime' equals 'f',
and the heck with all
00:18:32.200 --> 00:18:34.090
this summation process?
00:18:34.090 --> 00:18:36.690
Now, I think the best way to
answer that question--
00:18:36.690 --> 00:18:38.510
and this is probably the
ultimate in teaching
00:18:38.510 --> 00:18:41.270
technique, when somebody thinks
that he has an easier
00:18:41.270 --> 00:18:44.410
way than the way that
he was taught--
00:18:44.410 --> 00:18:47.980
give them a problem that can't
be done by that easier way.
00:18:47.980 --> 00:18:51.040
In fact, as a sarcastic aside,
if you can't invent that kind
00:18:51.040 --> 00:18:53.990
of a problem, maybe the
so-called easier way is the
00:18:53.990 --> 00:18:55.330
better way of doing a thing.
00:18:55.330 --> 00:18:57.640
But let me just show you what
I'm driving at over here.
00:18:57.640 --> 00:19:00.620
Suppose I say to you, let's find
the area of the region
00:19:00.620 --> 00:19:03.220
'R' where 'R' is now
the following.
00:19:03.220 --> 00:19:07.130
It's bounded above by the
curve, 'y' equals '1/x'.
00:19:07.130 --> 00:19:09.950
It's bounded below
by the x-axis.
00:19:09.950 --> 00:19:13.220
It's bounded on the left by the
line, 'x' equals 1, and on
00:19:13.220 --> 00:19:15.630
the right by the line,
'x' equals 2.
00:19:15.630 --> 00:19:18.870
I would like to find the area
of this region 'R'.
00:19:18.870 --> 00:19:24.520
Now look it, from a purely
conceptual point of view, what
00:19:24.520 --> 00:19:27.810
difference is there between this
problem and the problem
00:19:27.810 --> 00:19:31.620
where the upper curve was
'y' equals 'x squared'?
00:19:31.620 --> 00:19:33.840
Conceptually, what's going
on is the same thing.
00:19:33.840 --> 00:19:37.280
We have a bounded region and we
want to compute the area.
00:19:37.280 --> 00:19:41.490
To this end, notice that, if I
use the precise definition of
00:19:41.490 --> 00:19:44.890
a definite integral, the area
of the region 'R' is what?
00:19:44.890 --> 00:19:49.370
The definite integral
from 1 to 2, 'dx/x'.
00:19:49.370 --> 00:19:50.970
What does that mean?
00:19:50.970 --> 00:19:52.900
It means a particular limit.
00:19:52.900 --> 00:19:53.890
What limit?
00:19:53.890 --> 00:19:56.990
Well, you partition this
integral from 1 to 2 into n
00:19:56.990 --> 00:20:02.200
equal parts, pick a point, 'c
sub k' in each partition, then
00:20:02.200 --> 00:20:05.880
'f of 'c sub k'' in this problem
is just '1/'c sub k''.
00:20:05.880 --> 00:20:08.230
In other words, what we want
to do is to compute this
00:20:08.230 --> 00:20:12.870
limit. 'Sigma k' goes from
1 to 'n', '1/'c sub k'',
00:20:12.870 --> 00:20:14.430
times 'delta x'.
00:20:14.430 --> 00:20:17.570
Now, the thing to notice is,
this thing may be a mess but
00:20:17.570 --> 00:20:18.990
it's computable.
00:20:18.990 --> 00:20:24.320
We could use specially designed
graph paper or
00:20:24.320 --> 00:20:27.650
measuring devices to count the
units of area under here.
00:20:27.650 --> 00:20:31.720
We can find all sorts of ways of
getting estimates, even the
00:20:31.720 --> 00:20:35.520
long hard way of the 'U sub
n's and the 'L sub n's to
00:20:35.520 --> 00:20:39.080
pinpoint the area of the region
'R' to as close a
00:20:39.080 --> 00:20:41.340
degree of accuracy as
we want, et cetera.
00:20:41.340 --> 00:20:44.650
The point is that, if we have
never heard of a derivative,
00:20:44.650 --> 00:20:50.070
the area of the region 'R' is
given precisely by this sum.
00:20:50.070 --> 00:20:53.970
And admittedly, the
sum is a mess.
00:20:53.970 --> 00:20:57.200
So let's try to do it the
so-called easier way.
00:20:57.200 --> 00:20:58.530
The skeptic looks at
this thing and
00:20:58.530 --> 00:21:00.630
says, who needs this?
00:21:00.630 --> 00:21:03.460
The area of the region 'R',
we just saw by the first
00:21:03.460 --> 00:21:08.370
fundamental theorem, is 'G of
2', 'G of 2', minus 'G of 1',
00:21:08.370 --> 00:21:11.360
where 'G' is any function of
'x' whose derivative with
00:21:11.360 --> 00:21:14.010
respect to 'x' is '1/x'.
00:21:14.010 --> 00:21:15.860
And the answer to that is yes.
00:21:15.860 --> 00:21:17.770
So far, so good.
00:21:17.770 --> 00:21:21.840
But do we know a function 'G'
whose derivative with respect
00:21:21.840 --> 00:21:23.610
to 'x' is '1/x'?
00:21:23.610 --> 00:21:25.870
Remember, this is calculus
revisited.
00:21:25.870 --> 00:21:27.790
For those of you who remember
calculus from
00:21:27.790 --> 00:21:29.100
the first time around--
00:21:29.100 --> 00:21:30.770
and I'll talk about
this later--
00:21:30.770 --> 00:21:33.570
it's going to turn out that
the required 'G' is a
00:21:33.570 --> 00:21:35.110
logarithmic function.
00:21:35.110 --> 00:21:37.570
For those of you don't remember
that, there's no harm
00:21:37.570 --> 00:21:38.770
in not remembering that.
00:21:38.770 --> 00:21:40.850
All I'm trying to bring out is
that, whether you remember it
00:21:40.850 --> 00:21:44.000
or you don't, as far as we're
concerned so far in this
00:21:44.000 --> 00:21:47.090
course, as far as what we've
developed in this course, we
00:21:47.090 --> 00:21:50.720
do not know a function 'G' whose
derivative with respect
00:21:50.720 --> 00:21:52.710
to 'x' is '1/x'.
00:21:52.710 --> 00:21:56.450
That's what we mean by saying
we can't exhibit 'G'
00:21:56.450 --> 00:21:58.340
explicitly.
00:21:58.340 --> 00:21:59.490
What do I mean by that?
00:21:59.490 --> 00:22:01.150
Somebody says, I am thinking
of a function whose
00:22:01.150 --> 00:22:02.660
derivative is '1/x'.
00:22:02.660 --> 00:22:04.310
I say, well, that's simple.
00:22:04.310 --> 00:22:05.150
It's 'G of x'.
00:22:05.150 --> 00:22:06.470
He says, well, what's
'G of x'?
00:22:06.470 --> 00:22:08.600
I say, that's the function whose
derivative with respect
00:22:08.600 --> 00:22:10.340
to 'x' is '1/x'.
00:22:10.340 --> 00:22:13.210
Well, you see, that implicitly
tells me what 'G' is like.
00:22:13.210 --> 00:22:15.680
But in terms of concrete
measurements, I don't know
00:22:15.680 --> 00:22:16.630
anything about 'G'.
00:22:16.630 --> 00:22:18.390
I can't express it in terms of
00:22:18.390 --> 00:22:20.060
well-known, familiar functions.
00:22:20.060 --> 00:22:21.610
You see, I'm hung up now.
00:22:21.610 --> 00:22:23.620
Namely, this is precise.
00:22:23.620 --> 00:22:26.970
The answer to this problem will
be 'G of 2' minus 'G of
00:22:26.970 --> 00:22:29.790
1', where 'G prime' is '1/x'.
00:22:29.790 --> 00:22:33.850
But I don't know explicitly--
00:22:33.850 --> 00:22:35.020
yikes--
00:22:35.020 --> 00:22:37.470
I don't know, explicitly
such a 'G'.
00:22:37.470 --> 00:22:40.290
And I put the exclamation point
out here to emphasize
00:22:40.290 --> 00:22:42.690
that particular fact.
00:22:42.690 --> 00:22:43.980
That's the hang-up.
00:22:43.980 --> 00:22:47.130
The statement that says that the
area is 'G of b' minus 'G
00:22:47.130 --> 00:22:50.960
of a', where 'G prime' equals
'f' hinges on the fact that
00:22:50.960 --> 00:22:54.120
you can explicitly exhibit
such as 'G'.
00:22:54.120 --> 00:22:56.710
Certainly, if you can't exhibit
that 'G', you can
00:22:56.710 --> 00:22:58.910
still compute this
area as a limit.
00:22:58.910 --> 00:23:03.650
It would be tragic to say, oh,
the area doesn't exist because
00:23:03.650 --> 00:23:06.560
I don't know a 'G' whose
derivative is '1/x'.
00:23:06.560 --> 00:23:09.320
Certainly, this region
'R' has an area.
00:23:09.320 --> 00:23:13.550
In fact, because we can pinpoint
the area of the
00:23:13.550 --> 00:23:17.920
region 'R', we can actually
construct the 'G' such that 'G
00:23:17.920 --> 00:23:20.220
prime of x' equals '1/x'.
00:23:20.220 --> 00:23:22.580
And by the way, there's plenty
of drill on this.
00:23:22.580 --> 00:23:24.210
This is a hard concept.
00:23:24.210 --> 00:23:26.690
And as a result, you'll notice
that the exercises in this
00:23:26.690 --> 00:23:30.250
section hammer home on this
point, because it's a point
00:23:30.250 --> 00:23:33.090
that I'm positive that, if
you're having trouble at all
00:23:33.090 --> 00:23:35.580
with integral calculus, this
is certainly the most
00:23:35.580 --> 00:23:38.910
sophisticated part of what
we're doing right now.
00:23:38.910 --> 00:23:41.730
You see, look it, suppose I want
to construct a function
00:23:41.730 --> 00:23:45.770
'G' whose derivative is '1/x'.
00:23:45.770 --> 00:23:47.470
The idea looks something
like this.
00:23:47.470 --> 00:23:48.990
I'm on the interval
from 1 to 2 here.
00:23:48.990 --> 00:23:52.970
What I do is, I plot the curve,
'y' equals '1/x'.
00:23:52.970 --> 00:23:57.450
And I compute the area of the
region formed by the curve,
00:23:57.450 --> 00:23:59.410
'y' equals '1/x'.
00:23:59.410 --> 00:24:00.740
'x' equals 1.
00:24:00.740 --> 00:24:06.500
The x-axis and the line,
'x' equals 'x1'.
00:24:06.500 --> 00:24:09.160
This is my region 'R'.
00:24:09.160 --> 00:24:11.210
And what I say is, look
it, what do I
00:24:11.210 --> 00:24:13.220
know about this area?
00:24:13.220 --> 00:24:16.120
Remember, what I started this
lecture with was the knowledge
00:24:16.120 --> 00:24:19.870
that a prime of 'x' is 'f
of x'. 'f of x', in
00:24:19.870 --> 00:24:21.360
this case, is '1/x'.
00:24:21.360 --> 00:24:24.460
In other words, what property
does this area function have?
00:24:24.460 --> 00:24:27.970
It has the property that its
derivative with respect to 'x'
00:24:27.970 --> 00:24:29.890
is going to be '1/x'.
00:24:29.890 --> 00:24:33.440
In other words, if I now say,
look it, boys, I can compute
00:24:33.440 --> 00:24:37.700
this area, if not exactly, at
least to as many decimal place
00:24:37.700 --> 00:24:41.500
accuracy as I wish, so that the
area function is something
00:24:41.500 --> 00:24:42.640
I can construct.
00:24:42.640 --> 00:24:48.100
Let me define 'G of x1' simply
to be the area under this
00:24:48.100 --> 00:24:50.150
curve when 'x' is
equal to 'x1'.
00:24:50.150 --> 00:24:51.940
In other words, written
now as a limit--
00:24:51.940 --> 00:24:53.600
see how this definite integral
comes in here--
00:24:53.600 --> 00:24:58.980
written as integral from 1 to 'x
sub 1', 'dx/x', which I now
00:24:58.980 --> 00:25:01.460
can compute as closely
as I want as a limit.
00:25:01.460 --> 00:25:02.840
The beauty is what?
00:25:02.840 --> 00:25:07.270
That 'G prime of x1' is, by
definition, 'A prime of 'x1'.
00:25:07.270 --> 00:25:11.120
And we have already seen that
the derivative of 'A' is 'f',
00:25:11.120 --> 00:25:13.350
where 'f' is the top curve.
00:25:13.350 --> 00:25:16.440
In this case, the derivative
of 'A' is '1/x'.
00:25:16.440 --> 00:25:20.090
In other words, the area under
the curve explicitly can be
00:25:20.090 --> 00:25:23.020
computed as a function of 'x'.
00:25:23.020 --> 00:25:25.930
And that function has the
property, but its derivative
00:25:25.930 --> 00:25:28.380
with respect to 'x', in
this case, is '1/x'.
00:25:30.910 --> 00:25:34.670
And by the way, let me make just
a quick aside over here.
00:25:34.670 --> 00:25:37.010
I happened to throw in the
word "logarithms" before.
00:25:37.010 --> 00:25:39.340
It's going to turn out, as we'll
talk about in our next
00:25:39.340 --> 00:25:43.030
block of material, that 'G of x'
turns out to be the natural
00:25:43.030 --> 00:25:46.480
log of 'x', which means that the
area of the region 'R' is
00:25:46.480 --> 00:25:49.760
precisely natural log 2.
00:25:49.760 --> 00:25:51.540
Now, the idea is--
00:25:51.540 --> 00:25:53.460
and this is exactly what's going
to happen in our last
00:25:53.460 --> 00:25:54.890
block in our course--
00:25:54.890 --> 00:25:56.790
that this is exactly how
the log tables are
00:25:56.790 --> 00:25:58.450
constructed, in fact.
00:25:58.450 --> 00:26:00.820
I mean, how do you think they
find the log of 2 to eight
00:26:00.820 --> 00:26:01.630
decimal places?
00:26:01.630 --> 00:26:04.070
Did they strand somebody on a
desert island and tell them to
00:26:04.070 --> 00:26:08.330
compute these things, measure
it with laser beams, what?
00:26:08.330 --> 00:26:08.740
No.
00:26:08.740 --> 00:26:11.490
The way we do it is, we know
it's an area, and knowing how
00:26:11.490 --> 00:26:14.710
to find the area as a limit to
as many decimal place accuracy
00:26:14.710 --> 00:26:17.910
as we wish, give or take a few
tricks of the trade, this is
00:26:17.910 --> 00:26:21.080
how we find logs, and later
on, trigonometric
00:26:21.080 --> 00:26:22.330
tables, and the like.
00:26:22.330 --> 00:26:25.050
But I just mention this as
an aside to show you how
00:26:25.050 --> 00:26:28.260
important knowing area
under a curve is.
00:26:28.260 --> 00:26:31.750
At any rate, to summarize what
we're saying over here, we
00:26:31.750 --> 00:26:32.670
have the following.
00:26:32.670 --> 00:26:36.520
In general, if 'f' is any
continuous function on the
00:26:36.520 --> 00:26:40.080
closed interval from 'a' to 'b',
we define a function 'G'
00:26:40.080 --> 00:26:41.320
as follows.
00:26:41.320 --> 00:26:46.180
'G of x1' is the definite
integral from 'a' to 'x1', 'f
00:26:46.180 --> 00:26:49.510
of x' 'dx', where 'x1' is any
point in the closed interval
00:26:49.510 --> 00:26:50.430
for 'a' to 'b'.
00:26:50.430 --> 00:26:52.670
What does this thing mean,
geometrically?
00:26:52.670 --> 00:26:57.200
It's the area under the curve,
'y' equals 'f of x' on top.
00:26:57.200 --> 00:27:00.940
'x' equals the x-axis on the
bottom, the line 'x' equals
00:27:00.940 --> 00:27:05.190
'a' on the left, the line 'x'
equals 'x1' on the right.
00:27:05.190 --> 00:27:06.800
And that area is what?
00:27:06.800 --> 00:27:08.750
It's the limit as 'n'
approaches infinity.
00:27:08.750 --> 00:27:12.730
'Sigma k' goes from 1 to 'n', 'f
of 'c sub k'' 'delta x', et
00:27:12.730 --> 00:27:16.060
cetera, et cetera, et cetera,
meaning we can go through this
00:27:16.060 --> 00:27:19.740
the same way as we did in
our previous lecture.
00:27:19.740 --> 00:27:21.370
This function, 'G of x1'--
00:27:21.370 --> 00:27:25.060
which we can now compute
explicitly as an area, because
00:27:25.060 --> 00:27:27.710
we have the curve 'f' drawn,
we can approximate it to as
00:27:27.710 --> 00:27:29.510
close a degree of accuracy
as we want--
00:27:29.510 --> 00:27:33.420
whatever that function 'G' is,
what property does it have?
00:27:33.420 --> 00:27:38.380
It has the property that 'G
prime' is equal to 'f'.
00:27:38.380 --> 00:27:41.090
And that's exactly what the
second fundamental theorem
00:27:41.090 --> 00:27:42.020
really means.
00:27:42.020 --> 00:27:45.430
The second fundamental theorem
says, look it, if you can
00:27:45.430 --> 00:27:49.480
compute the area, you can now
reverse this procedure that
00:27:49.480 --> 00:27:52.150
we're talking about in the first
fundamental theorem.
00:27:52.150 --> 00:27:54.890
See, in the first fundamental
theorem, what did we do?
00:27:54.890 --> 00:27:57.510
In the first fundamental
theorem, we found a quick way
00:27:57.510 --> 00:28:00.100
of computing the area as
a limit by knowing the
00:28:00.100 --> 00:28:01.980
appropriate inverse
derivative.
00:28:01.980 --> 00:28:05.060
The second fundamental theorem,
in a sense, is the
00:28:05.060 --> 00:28:07.980
inverse of the first fundamental
theorem.
00:28:07.980 --> 00:28:11.110
Namely, it switches the roles.
00:28:11.110 --> 00:28:12.950
Namely, with the second
fundamental theorem, what we
00:28:12.950 --> 00:28:13.800
say is this.
00:28:13.800 --> 00:28:16.760
If we know how to find the
area under the curve, 'y'
00:28:16.760 --> 00:28:21.640
equals 'f of x', that gives us
a way of computing a function
00:28:21.640 --> 00:28:23.910
'G' whose derivative is 'f'.
00:28:23.910 --> 00:28:26.100
In fact, to summarize this--
00:28:26.100 --> 00:28:27.340
and I think this is
very important.
00:28:27.340 --> 00:28:28.860
Let me just summarize
this, then.
00:28:28.860 --> 00:28:32.560
First of all, the first
fundamental theorem allows us
00:28:32.560 --> 00:28:37.480
to compute this particular
limit, provided we can find a
00:28:37.480 --> 00:28:40.320
'G', such that 'G prime'
equals 'f'.
00:28:40.320 --> 00:28:43.690
In fact, in this case, the
limit, as 'n' approaches
00:28:43.690 --> 00:28:47.990
infinity, summation 'k' goes
from 1 to 'n', 'f of 'c sub
00:28:47.990 --> 00:28:52.660
k'' 'delta x', is given very,
very ingeniously, beautifully,
00:28:52.660 --> 00:28:57.110
and concisely by 'G of
b' minus 'G of a'.
00:28:57.110 --> 00:29:01.400
Secondly, the second fundamental
theorem allows us,
00:29:01.400 --> 00:29:05.010
given 'f', to construct
'G', such that 'G
00:29:05.010 --> 00:29:06.480
prime' equals 'f'.
00:29:06.480 --> 00:29:10.430
Namely, the required 'G' is
just the definite integral
00:29:10.430 --> 00:29:16.095
from 'a' to 'x1', 'f of x' 'dx',
which in turn is a limit
00:29:16.095 --> 00:29:20.120
of a network of rectangles,
namely the limit as 'n'
00:29:20.120 --> 00:29:24.420
approaches infinity, summation
'k' goes from 1 to 'n', 'f of
00:29:24.420 --> 00:29:27.040
'c sub k'' times 'delta x'.
00:29:27.040 --> 00:29:32.150
This then is perhaps the most
beautiful lecture in calculus,
00:29:32.150 --> 00:29:33.280
not necessarily the
way I've given it,
00:29:33.280 --> 00:29:34.620
but in terms of what?
00:29:34.620 --> 00:29:39.250
Relating two apparently diverse
branches of calculus,
00:29:39.250 --> 00:29:41.330
integral and differential
calculus.
00:29:41.330 --> 00:29:43.210
This is a difficult subject.
00:29:43.210 --> 00:29:45.980
I have taken great pains to try
to write this clearly in
00:29:45.980 --> 00:29:47.250
the supplementary notes.
00:29:47.250 --> 00:29:50.850
I have tried to pick typical
exercises that bring home
00:29:50.850 --> 00:29:52.550
these highlights.
00:29:52.550 --> 00:29:57.470
And in addition, our next two
lectures will emphasize the
00:29:57.470 --> 00:30:02.150
relationship between derivatives
and integrals as
00:30:02.150 --> 00:30:04.540
we study volume and
arc length.
00:30:04.540 --> 00:30:06.670
At any rate, until next
time, goodbye.
00:30:09.710 --> 00:30:12.240
MALE SPEAKER: Funding for the
publication of this video was
00:30:12.240 --> 00:30:16.960
provided by the Gabriella and
Paul Rosenbaum Foundation.
00:30:16.960 --> 00:30:21.130
Help OCW continue to provide
free and open access to MIT
00:30:21.130 --> 00:30:25.330
courses by making a donation
at ocw.mit.edu/donate.