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PROFESSOR: Hi.
Technically speaking, I suppose, if time permitted, we
should spend, or shall we say, give equal time to integral
calculus, the same time that we gave to differential
calculus, starting from scratch, building ideas slowly
and the like.
But there are two logistic reasons for not doing this,
one of which is that the basic definitions would be the same
regardless of which branch we started with.
The second one is, of course, that time becomes preciously
tight, and we have to make the best of what we can here.
So what we're going to do now is try to show, in terms of
hindsight being better than foresight, a motivation as to
how one would have invented differential calculus had it
been motivated by the existing integral calculus.
In other words, what we want to do today is to show the
beautiful interplay between differential
and integral calculus.
Since we have said in our last lecture that we would like to
begin with a study of integral calculus, assuming that
differential calculus had never been invented, I think
that we're obligated to try to show how, then, differential
calculus could have been invented in the environment of
the studying of areas.
For this reason, I have called today's lecture "The Marriage
of Differential and Integral Calculus." And essentially,
the idea goes something like this.
Last time, we were talking about finding
areas under a curve.
And to find the area under the curve, we saw that we could do
this as the limit of a certain sum.
In other words, an infinite sum having a particular limit.
Now, to motivate the concept of rate of change, one could
ask the following question.
A curve like 'y' equals 'f of x', and a starting point, 'x'
equals 'a'.
As I move out along the x-axis from 'a', I could study the
area under the curve, say, of the curve, 'y' equals 'f of
x', from 'x' equals 'a' to some value 'x1'.
And the question I could then ask is, I wonder how fast this
area is changing at the instant that 'x' equals 'x1'.
In other words, how fast is the area changing under the
curve as 'x' moves out along the positive x-axis this way?
You see?
Now, the idea is this.
I can then mimic the same definition that we gave for
instantaneous rate of change and differential calculus.
Namely, I can say, why can't we define the instantaneous
rate of change to be the average rate of change with
respect to 'x'?
And then we'll take the limit as 'delta x'
approaches 0, et cetera.
And if I do that, watch what happens over here.
You see what I do in terms of this picture, is I say, OK,
here's the area under the curve when
'x' is equal to 'x1'.
Now I'll let 'x1' change by some amount 'delta x', meaning
this'll be the point 'x1 + delta x'.
And this brings about a change in my area, which
I'll call 'delta A'.
See the change in area as 'x' goes from 'x1'
to 'x1 + delta x'.
Now, what I would like to find is 'delta A' divided by 'delta
x', taking the limit as 'delta x' approaches 0.
Remember, last time we showed our three
basic axioms for area.
We're obliged to still use them.
Let's do that here.
For example, the way I've drawn this curve by rising
this way, notice that the lowest point in this interval
occurs when 'x' equals 'x1'.
The highest point occurs when 'x' equals 'x1 + delta x'.
Consequently, the rectangle whose base is the closed
interval from 'x1' to 'x1 + delta x', the rectangle whose
height corresponds to 'x' equals 'x1' is too small in
area to be 'delta A'.
In other words, that rectangle is inscribed in 'delta A'.
And in the same way, the rectangle, with base 'x1' to
'x1 + delta x', whose height is given by the x-coordinate,
'x1 + delta x', that rectangle has an area which is too large
to be the exact area.
In other words, notice that the area of the rectangle,
which is too small to be the exact area, is 'f of x1'
times 'delta x'.
The area of the rectangle which is too big to be the
correct 'delta A' is 'f of 'x1 + delta x'' times 'delta x'.
And the area, 'delta A', is caught between these two.
In other words, we now have the inequality that 'f of x1'
times 'delta x' is less than 'delta A', which in turn is
less than 'f of 'x1 + delta x'' times 'delta x'.
Now, we'll assume that 'delta x' is positive, the way we've
drawn it in this diagram.
Adjustments have to be made if 'delta x' is negative, in
other words, if we let 'delta x' approach 0
through negative values.
This is taken care of in the textbook, as well as mentioned
in our notes.
But for the sake of just getting the main idea here,
we'll assume that 'delta x' is positive.
We divide through by 'delta x'.
That gives us 'f of x1' is less than 'delta a' over
'delta x', which in turn is less than 'f
of 'x1 + delta x''.
In other words, had 'delta x' been negative, we would have
had to reverse the inequality signs, et cetera, but we're
not going to worry about that right now.
Remember, what is 'delta x'?
It's a non-zero number which is going to be made to become
arbitrarily close to 0.
Well, here's the idea.
We now have 'delta a' over 'delta x' squeezed between
these two values.
And we say, OK, let's let 'delta x' approach 0.
Now, here's the key point.
As 'delta x' approaches 0, certainly, 'x1 + delta x'
approaches 'x1'.
And here's where we use the fact that 'f' is continuous.
In other words, notice that the mere fact that 'x1 + delta
x' approaches 'x1' is not enough to say, therefore, that
'f of 'x1 + delta x'' approaches 'f of x1'.
As we saw before, that's only true if 'f' is continuous.
In other words, by continuity, if 'x1 + delta x' approaches
'x1', then 'f of 'x1 + delta x'' approaches 'f of x1'.
So assuming, then, that 'f' is continuous, what do we get we
put the squeeze on?
'x1' is what's picked to be a fixed point.
Consequently, as 'delta x' approaches 0, 'f of
x1' stays 'f of x1'.
'f of 'x1 + delta x'' approaches 'f of x1'.
In other words, 'delta A' divided by 'delta x' is caught
between two numbers, two sequences, two sets of bounds,
whichever way you want to say this, both of which converge
to the common limit, 'f of x1'.
In other words, since 'delta A' over 'delta x' is squeezed
between these two, the limit of 'delta a' over 'delta x',
as 'delta x' approaches 0, must equal this common limit.
In other words, 'dA/dx', the limit of 'delta A' over 'delta
x', as 'delta x' approaches 0, evaluated at 'x' equals 'x1',
is 'f of x1'.
Or another way of saying this, 'dA/dx' is 'f of x'.
Now, that result is partly intuitive and partly
remarkable.
What it says is, roughly speaking, is that the area
under the curve is changing at a rate equal instantaneously
to the height corresponding to the
x-coordinate at that point.
In other words, to define how fast the area is changing as
'x' moves, all you have to do is measure the height of the
curve corresponding to that particular value of 'x'.
Now, what does this have to do, then, with relating
integral and differential calculus?
Notice already we begin to get some sort of a hint.
What we've shown is now that the area is going to be
related somehow to the inverse derivative of 'f of x'.
In fact, this is precisely what is meant by a rather
important result.
In fact, it's important enough so it's called the first
fundamental theorem of integral calculus.
And it simply says this.
Suppose we know explicitly a function, capital 'G', such
that the derivative of capital 'G' is 'f', where the 'f' now
refers to the 'f' that we're talking about in our problem.
Now, look it, what we already know is that 'A', the area
function, has its derivative equal to 'f'.
Consequently, since the area function and G have the same
derivative, they must differ by, at most, a constant.
In other words, the area as a function of 'x' is this
function, 'G of x' plus 'c'.
And what is this 'G of x'?
It's not any old function.
It's what?
An inverse derivative of 'f of x'.
Since the area evaluated at 'x' equals 'a' is 0, see,
namely, the area under the curve, as 'x' goes from 'A' to
'a', see, that's just a line.
A line has no thickness, has no area.
This is 0.
We have, by plugging back in to here, that 'A' of
little 'a' is 0.
That in turn is 'G of a' plus 'c'.
And solving this for 'c', we find that 'c' is equal to
minus 'G of a'.
In other words, the area under the curve 'y' equals 'f of x',
as a function of 'x', is simply what?
It's 'G of x' minus 'G of a', where 'G' is any function
whose derivative is 'f'.
If we now let 'x' equal 'b', we're back to the situation of
finding the usual region that we're talking about here,
namely, what is 'A sub R'?
It's the region bounded above-- let's come back here
and take a look at this-- it's the region bounded above by
the curve, 'y' equals 'f of x', bounded on the left by 'x'
equals 'a', on the right by 'x' equals 'b', and below by
the x-axis.
According to this notation, the area of the region 'R' is
just the area as a function of 'x' when 'x1' is equal to 'b'.
In other words, the area of our region 'R' is just 'a' of
'b', which is 'G of b' minus 'G of a', where 'G' is any
function whose derivative is 'f'.
Now remember--
and here's the key point--
remember that the area of the region 'R' did not necessitate
us having to know a function whose derivative was 'f'.
In other words, remember, in our last lecture, how did we
find areas of regions like 'R'?
What we did was is we formed these 'U sub n's,
these 'L sub n's.
We inscribed and circumscribed networks of rectangles, put
the squeeze on by evaluating this particular limit.
And whatever that limit was, that was the area of the
region 'R'.
In other words, if I had never heard of the inverse
derivative, I could still find the area of the
region 'R' this way.
However, if I just happen to know a function 'G', whose
derivative is 'f', I have a much easier way of doing this.
In fact, let me summarize that.
You see, we can compute the limit as
'n' approaches infinity.
Summation 'k' goes from 1 to 'n', 'f of 'c sub k'' 'delta
x', by use of inverse derivatives.
See, again, the highlight being what?
You can still work with this the same way as we did in our
last lecture.
There is absolutely no need to have had to ever heard of a
derivative to solve this type of problem, even
though it may be messy.
Now, what's the best proof I have of this?
Well, I guess one of the best proofs is to specifically
refer to one of the exercises in the previous unit.
It was a tough exercise.
I did it as a learning exercise because I felt it was
something that was messy and that you had to be guided
through in order not to become hopelessly lost.
By way of review, the problem was this.
We took as our region 'R' the region bounded above by 'y'
equals 'sine x', below by the x-axis, on the right by the
line 'x' equals pi/2.
And the problem was, define the area of the region 'R'.
And we solved this problem in the last unit without recourse
to derivatives because, as of the time that we were in the
last unit, we had no results relating derivatives
to limits of sums.
How did we do this?
We partitioned this into n parts, and we formed the sum
limit as 'n' approaches infinity.
'Sigma k' goes from 1 to 'n'.
We broke this thing up into 'n' equal parts, so the size
of each piece was what?
pi/2 divided by 'n'.
That's 'pi/2n'.
We computed the sine of each of these n-points, et cetera.
And whatever that limit was, that was the area of the
region 'R'.
And as you recall, that homework problem, we found
that the answer to that problem was that the area of
the region 'R' was 1.
Now again, as messy as that was, it proved that we could
at least solve the problem with no knowledge of
derivatives.
How does the first fundamental theorem apply here?
The way the first fundamental theorem
applies is the following.
The first fundamental theorem says this.
Look it, to evaluate this particular sum, all we have to
do is find a function whose derivative is 'sine x'.
See, notice in this problem, the general 'a' and 'b' of the
above is now played by 'a' equals 0 and 'b' equals pi/2.
All we have to do is what?
Find the function 'G' whose derivative is 'sine x', and
the answer should then be 'G of pi/2' minus 'G of 0'.
Well, you see this happens to be one that, with our
knowledge of differential calculus, we
can do rather easily.
In other words, do we, at the tip of our tongues, have a
function whose derivative is 'sine x'?
And the answer is yes.
Since the derivative of 'cosine x' is 'minus sine x',
the derivative of 'minus cosine x' is 'sine x'.
In other words, we can choose for our 'G of x' here
'minus cosine x'.
In other words, that the area of the region
'R' should be what?
'G of pi/2' minus 'G of 0', where 'G of x' is
'minus cosine x'.
Well, you see, cosine pi/2 is 0.
Cosine of 0 is 1, so 0 minus minus 1 is equal to 1.
And notice that, first of all, we get the same answer.
And secondly, notice this two-line job over here.
We not only get the correct answer, but we get the answer
very much more rapidly than we did by the
so-called limit process.
Now, I'll come back to this in a minute, because you may be a
little bit angry at me now.
Namely, you may be asking, after having to do this
problem the hard way, why did I have to do the usual
teacher's trick here and show you the hard way of doing this
and then waiting until the next unit before I show you
the easy way?
Well, there happens to be a catch here that I'll come back
to in just a moment.
But before I do that, I'd like to make an aside, an aside
that's rather important.
Namely, you may recall, as I started to work over here,
that this reminded you of a notation that we were using
back when we introduced the concept of the inverse of
differentiation.
Namely, we earlier used a notation, integral from 'a' to
'b', 'f of x' 'dx' to denote 'G of b' minus 'G of a', where
'G prime' equals 'f'.
What we have shown by the first fundamental theorem,
then, using this notation, is that the area of the region
'R' is integral from 'a' to 'b', 'f of x' 'dx'.
In other words, this means what?
It means 'G of b' minus 'G of a', where 'G
prime' equals 'f'.
OK, let's pause here for a moment.
See, as we developed our course, this is
how this would evolve.
The interesting thing is that, historically, this notation
was not used to define 'G of b' minus 'G of a', where 'G
prime' equaled 'f'.
Historically, what happened was that this notation, called
the definite integral, the integral from 'a' to 'b', 'f
of x' 'dx', was--
I don't know if it's proper to say invented, but let me just
say it in quotation marks to play it safe--
it was "invented" to denote the limit of this sum.
In fact, notice how much more meaningful the symbol looks in
this connotation.
In other words, this is the limit of a sum.
You can think of sum as beginning with 's', and the
integral of sine as an elongated 's'.
In other words, when one first wrote this symbol, the
definite integral, it was meant to denote this limit.
The important point is that, by the first fundamental
theorem, the definite integral, whether it's a limit
or not, turns out to be 'G of b' minus 'G of a', where 'G
prime' equals 'f'.
In other words, it really makes no difference in terms
of what answer you get, whether you think of this
definite integral as meaning this, or whether you think of
the definite integral as meaning this.
See, numerically, they'll be the same.
But conceptually, they're quite different.
And that's precisely the point that leads to the most, I
think, probably the most difficult part all the course,
at least up until now.
It's something that more people cause and have
misinterpretation over.
And so I'd like to present this in a very,
very gradual way.
See, the result will ultimately be known as the
second fundamental theorem of integral calculus.
I've taken the liberty of writing that over here.
But it's going to be quite a while before I actually say
what the theorem is explicitly.
What I'd like to do is to pick up where I left off when I
made my aside about the notation
of a definite integral.
And what I'd like to mention now is, for the person who
says, look it, why did you make us find the
areas the hard way?
Why didn't you just say let us compute 'G of b' minus 'G of
a', where 'G prime' equals 'f', and the heck with all
this summation process?
Now, I think the best way to answer that question--
and this is probably the ultimate in teaching
technique, when somebody thinks that he has an easier
way than the way that he was taught--
give them a problem that can't be done by that easier way.
In fact, as a sarcastic aside, if you can't invent that kind
of a problem, maybe the so-called easier way is the
better way of doing a thing.
But let me just show you what I'm driving at over here.
Suppose I say to you, let's find the area of the region
'R' where 'R' is now the following.
It's bounded above by the curve, 'y' equals '1/x'.
It's bounded below by the x-axis.
It's bounded on the left by the line, 'x' equals 1, and on
the right by the line, 'x' equals 2.
I would like to find the area of this region 'R'.
Now look it, from a purely conceptual point of view, what
difference is there between this problem and the problem
where the upper curve was 'y' equals 'x squared'?
Conceptually, what's going on is the same thing.
We have a bounded region and we want to compute the area.
To this end, notice that, if I use the precise definition of
a definite integral, the area of the region 'R' is what?
The definite integral from 1 to 2, 'dx/x'.
What does that mean?
It means a particular limit.
What limit?
Well, you partition this integral from 1 to 2 into n
equal parts, pick a point, 'c sub k' in each partition, then
'f of 'c sub k'' in this problem is just '1/'c sub k''.
In other words, what we want to do is to compute this
limit. 'Sigma k' goes from 1 to 'n', '1/'c sub k'',
times 'delta x'.
Now, the thing to notice is, this thing may be a mess but
it's computable.
We could use specially designed graph paper or
measuring devices to count the units of area under here.
We can find all sorts of ways of getting estimates, even the
long hard way of the 'U sub n's and the 'L sub n's to
pinpoint the area of the region 'R' to as close a
degree of accuracy as we want, et cetera.
The point is that, if we have never heard of a derivative,
the area of the region 'R' is given precisely by this sum.
And admittedly, the sum is a mess.
So let's try to do it the so-called easier way.
The skeptic looks at this thing and
says, who needs this?
The area of the region 'R', we just saw by the first
fundamental theorem, is 'G of 2', 'G of 2', minus 'G of 1',
where 'G' is any function of 'x' whose derivative with
respect to 'x' is '1/x'.
And the answer to that is yes.
So far, so good.
But do we know a function 'G' whose derivative with respect
to 'x' is '1/x'?
Remember, this is calculus revisited.
For those of you who remember calculus from
the first time around--
and I'll talk about this later--
it's going to turn out that the required 'G' is a
logarithmic function.
For those of you don't remember that, there's no harm
in not remembering that.
All I'm trying to bring out is that, whether you remember it
or you don't, as far as we're concerned so far in this
course, as far as what we've developed in this course, we
do not know a function 'G' whose derivative with respect
to 'x' is '1/x'.
That's what we mean by saying we can't exhibit 'G'
explicitly.
What do I mean by that?
Somebody says, I am thinking of a function whose
derivative is '1/x'.
I say, well, that's simple.
It's 'G of x'.
He says, well, what's 'G of x'?
I say, that's the function whose derivative with respect
to 'x' is '1/x'.
Well, you see, that implicitly tells me what 'G' is like.
But in terms of concrete measurements, I don't know
anything about 'G'.
I can't express it in terms of
well-known, familiar functions.
You see, I'm hung up now.
Namely, this is precise.
The answer to this problem will be 'G of 2' minus 'G of
1', where 'G prime' is '1/x'.
But I don't know explicitly--
yikes--
I don't know, explicitly such a 'G'.
And I put the exclamation point out here to emphasize
that particular fact.
That's the hang-up.
The statement that says that the area is 'G of b' minus 'G
of a', where 'G prime' equals 'f' hinges on the fact that
you can explicitly exhibit such as 'G'.
Certainly, if you can't exhibit that 'G', you can
still compute this area as a limit.
It would be tragic to say, oh, the area doesn't exist because
I don't know a 'G' whose derivative is '1/x'.
Certainly, this region 'R' has an area.
In fact, because we can pinpoint the area of the
region 'R', we can actually construct the 'G' such that 'G
prime of x' equals '1/x'.
And by the way, there's plenty of drill on this.
This is a hard concept.
And as a result, you'll notice that the exercises in this
section hammer home on this point, because it's a point
that I'm positive that, if you're having trouble at all
with integral calculus, this is certainly the most
sophisticated part of what we're doing right now.
You see, look it, suppose I want to construct a function
'G' whose derivative is '1/x'.
The idea looks something like this.
I'm on the interval from 1 to 2 here.
What I do is, I plot the curve, 'y' equals '1/x'.
And I compute the area of the region formed by the curve,
'y' equals '1/x'.
'x' equals 1.
The x-axis and the line, 'x' equals 'x1'.
This is my region 'R'.
And what I say is, look it, what do I
know about this area?
Remember, what I started this lecture with was the knowledge
that a prime of 'x' is 'f of x'. 'f of x', in
this case, is '1/x'.
In other words, what property does this area function have?
It has the property that its derivative with respect to 'x'
is going to be '1/x'.
In other words, if I now say, look it, boys, I can compute
this area, if not exactly, at least to as many decimal place
accuracy as I wish, so that the area function is something
I can construct.
Let me define 'G of x1' simply to be the area under this
curve when 'x' is equal to 'x1'.
In other words, written now as a limit--
see how this definite integral comes in here--
written as integral from 1 to 'x sub 1', 'dx/x', which I now
can compute as closely as I want as a limit.
The beauty is what?
That 'G prime of x1' is, by definition, 'A prime of 'x1'.
And we have already seen that the derivative of 'A' is 'f',
where 'f' is the top curve.
In this case, the derivative of 'A' is '1/x'.
In other words, the area under the curve explicitly can be
computed as a function of 'x'.
And that function has the property, but its derivative
with respect to 'x', in this case, is '1/x'.
And by the way, let me make just a quick aside over here.
I happened to throw in the word "logarithms" before.
It's going to turn out, as we'll talk about in our next
block of material, that 'G of x' turns out to be the natural
log of 'x', which means that the area of the region 'R' is
precisely natural log 2.
Now, the idea is--
and this is exactly what's going to happen in our last
block in our course--
that this is exactly how the log tables are
constructed, in fact.
I mean, how do you think they find the log of 2 to eight
decimal places?
Did they strand somebody on a desert island and tell them to
compute these things, measure it with laser beams, what?
No.
The way we do it is, we know it's an area, and knowing how
to find the area as a limit to as many decimal place accuracy
as we wish, give or take a few tricks of the trade, this is
how we find logs, and later on, trigonometric
tables, and the like.
But I just mention this as an aside to show you how
important knowing area under a curve is.
At any rate, to summarize what we're saying over here, we
have the following.
In general, if 'f' is any continuous function on the
closed interval from 'a' to 'b', we define a function 'G'
as follows.
'G of x1' is the definite integral from 'a' to 'x1', 'f
of x' 'dx', where 'x1' is any point in the closed interval
for 'a' to 'b'.
What does this thing mean, geometrically?
It's the area under the curve, 'y' equals 'f of x' on top.
'x' equals the x-axis on the bottom, the line 'x' equals
'a' on the left, the line 'x' equals 'x1' on the right.
And that area is what?
It's the limit as 'n' approaches infinity.
'Sigma k' goes from 1 to 'n', 'f of 'c sub k'' 'delta x', et
cetera, et cetera, et cetera, meaning we can go through this
the same way as we did in our previous lecture.
This function, 'G of x1'--
which we can now compute explicitly as an area, because
we have the curve 'f' drawn, we can approximate it to as
close a degree of accuracy as we want--
whatever that function 'G' is, what property does it have?
It has the property that 'G prime' is equal to 'f'.
And that's exactly what the second fundamental theorem
really means.
The second fundamental theorem says, look it, if you can
compute the area, you can now reverse this procedure that
we're talking about in the first fundamental theorem.
See, in the first fundamental theorem, what did we do?
In the first fundamental theorem, we found a quick way
of computing the area as a limit by knowing the
appropriate inverse derivative.
The second fundamental theorem, in a sense, is the
inverse of the first fundamental theorem.
Namely, it switches the roles.
Namely, with the second fundamental theorem, what we
say is this.
If we know how to find the area under the curve, 'y'
equals 'f of x', that gives us a way of computing a function
'G' whose derivative is 'f'.
In fact, to summarize this--
and I think this is very important.
Let me just summarize this, then.
First of all, the first fundamental theorem allows us
to compute this particular limit, provided we can find a
'G', such that 'G prime' equals 'f'.
In fact, in this case, the limit, as 'n' approaches
infinity, summation 'k' goes from 1 to 'n', 'f of 'c sub
k'' 'delta x', is given very, very ingeniously, beautifully,
and concisely by 'G of b' minus 'G of a'.
Secondly, the second fundamental theorem allows us,
given 'f', to construct 'G', such that 'G
prime' equals 'f'.
Namely, the required 'G' is just the definite integral
from 'a' to 'x1', 'f of x' 'dx', which in turn is a limit
of a network of rectangles, namely the limit as 'n'
approaches infinity, summation 'k' goes from 1 to 'n', 'f of
'c sub k'' times 'delta x'.
This then is perhaps the most beautiful lecture in calculus,
not necessarily the way I've given it,
but in terms of what?
Relating two apparently diverse branches of calculus,
integral and differential calculus.
This is a difficult subject.
I have taken great pains to try to write this clearly in
the supplementary notes.
I have tried to pick typical exercises that bring home
these highlights.
And in addition, our next two lectures will emphasize the
relationship between derivatives and integrals as
we study volume and arc length.
At any rate, until next time, goodbye.
MALE SPEAKER: Funding for the publication of this video was
provided by the Gabriella and Paul Rosenbaum Foundation.
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