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PROFESSOR: To begin today I want
to remind you, I need to
write it down on the board
at least twice, of the
fundamental theorem
of calculus.
We called it FTC 1 because it's
the first version of the
fundamental theorem.
We'll be talking about another
version, called the second
version, today.
And what it says is this: If F'
= f, then the integral from
a to b of f (x) dx =
F ( b) - F ( a).
So that's the fundamental
theorem of calculus.
And the way we used it last
time was, this was used to
evaluate integrals.
Not surprisingly, that's
how we used it.
But today, I want to reverse
that point of view.
We're going to read the equation
backwards, and we're
going to write it this way.
And we're going to use f to
understand capital F. Or in
other words, the derivative.
To understand the function.
So that's the reversal
of point of view
that I'd like to make.
And we'll make this point
in various ways.
So information about F, about
F', gives us information about
F. Now, since there were
questions about the mean value
theorem, I'm going to illustrate
this first by
making a comparison between
the fundamental theorem of
calculus and the mean
value theorem.
So we're going to compare this
fundamental theorem of
calculus with what we call
the mean value theorem.
And in order to do that,
I'm going to
introduce a couple of notations.
I'll write delta F as
F ( b ) - F ( a).
And another highly imaginative
notation, delta x = b - a.
So here's the change in f,
there's the change in x.
And then, this fundamental
theorem can be written, of
course, right up above
there is the formula.
And it's the formula for delta
F. So this is what we call the
fundamental theorem
of calculus.
I'm going to divide
by delta x, now.
And If I divide by delta x,
that's the same thing as 1 / b
- a times the integral from
a to b of f ( x) dx.
So I've just rewritten
the formula here.
And this expression here, on
the right-hand side, is a
fairly important one.
This is the average of f.
That's the average value of f.
Now, so this is going to
permit me to make the
comparison between the mean
value theorem, which we don't
have stated yet here.
And the fundamental theorem.
And I'll do it in the form
of inequalities.
So right in the middle here,
I'm going to put the
fundamental theorem.
It says that delta F in this
notation is equal to, well if
I multiply by delta x again, I
can write it as the average of
F. So I'm going to write it
as the average of F' here.
Times delta x.
So we have this factor here,
which is the average of F', or
the average of little f,
it's the same thing.
And then I multiplied
through again.
So I put the thing
in the red box.
Here.
Isn't what the average
of big F.
So the question is, why is this
the average of little f
rather than the average of big
F. So the average of a
function is the typical value.
If, for example, little f were
constant, little f were
constant, then this integral
would be, so the question is
why is this the average.
And I'll take a little second
to explain that.
But I think I'll explain
it over here.
Because I'm going to erase it.
So the idea of an average
is the following.
For example, imagine that
a = 0 and b = n,
let's say for example.
And so we might sum the
function from 1 to n.
Now, that would be the sum of
the values from 1 to n.
But the average is, we
divide by n here.
So this is the average.
And this is a kind of Riemann
sum, representing the integral
from 0 to n, of f (x) dx.
Where the increment,
delta x, is 1.
So this is the notion of
an average value here.
But in the continuum setting
as opposed to
the discrete setting.
Whereas what's on the
left-hand side is
the change in f.
The capital F. And this is the
average of the little f.
So an average is a sum.
And it's like an integral.
So, in other words what I have
here is that the change in f
is the average of its
infinitesimal change times the
amount of time elapsed,
if you like.
So this is the statement of
the fundamental theorem.
Just rewritten.
Exactly what I wrote there.
But I multiplied back
by delta x.
Now, let me compare this with
the mean value theorem.
The mean values theorem
also is an equation.
The mean value theorem says
that this is equal to
F' (c) delta x.
Now, I pulled a fast
one on you.
I used capital F's here to
make the analogy clear.
But the role of the letter
is important to make the
transition to this comparison.
We're talking about the function
capital F here.
And its derivative.
Now, this is true.
So now I claim that this thing
is fairly specific.
Whereas this, unfortunately,
is a little bit vague.
And the reason why it's vague is
that c is just somewhere in
the interval.
So some c - sorry, this is some
c, in between a and b.
So really, since we don't know
where this thing is, we don't
know which of the values it is,
we can't say what it is.
All we can do is say well for
sure it's less than the
largest value.
Say, (the maximum
of F' ) delta x.
And the only thing we can say
for sure on the other end is
that it's less than or equal to
- sorry, it's greater than
or equal to, (the minimum
of F' ) delta x.
Over that same interval.
This is over 0 less than
- sorry, a < x < b.
So that means that the
fundamental theorem of
calculus is a much more
specific thing.
And indeed it gives the
same conclusion.
It's much stronger than the
mean value theorem.
It's way better than the
mean value theorem.
In fact, as soon as we have
integrals, we can abandon the
mean value theorem.
We don't want it.
It's too simple-minded.
And what we have is something
much more sophisticated, which
we can use.
Which is this.
So it's obvious that if this is
the average, the average is
less than the maximum.
So it's obvious that it works
just as well to draw this
conclusion.
And similarly over here
with the minimum.
OK, the average is always bigger
than the minimum and
smaller than the max.
So this is the connection,
if you like.
And I'm going to elaborate
just one step further by
talking about the problem that
you had on the exam.
So there was an Exam
2 problem.
And I'll show you how it works
using the mean value theorem
and how it works using
integrals.
But I'm going to have to use
this notation capital F. So
capital F', as opposed to the
little f, which was what was
the notation that was
on your exam.
So we had this situation here.
These were the givens
of the problem.
And then the question was, the
mean value theorem says, or
implies, if you like,
it doesn't say it,
but it implies it.
Implies a < capital F ( 4)
< b, for which a and b?
So let's take a look
at what it says.
Well, the mean value theorem
says that F( 4) - F (0) = F' (
c)( 4 - 0).
This is this (F') delta x,
this is the change in x.
And that's the same thing
as (1 / 1 + c )( 4).
And so the range of values of
this number here is from (1 /
1 plus 0)( 4), that's 4.
To, that's the largest value, to
the smallest that it gets,
which is (1 / 1 + 4)( 4).
That's the range.
And so the conclusion is that
F (4) - f ( 0 ) is between,
well, let's see.
It's between 4 and 4/5.
Which are those two numbers
down there.
And if you remember that F ( 0)
was 1, this is the same as
F ( 4), is between 5 and 9/5.
So that's the way that you were
supposed to solve the
problem on the exam.
On the other hand, let's compare
to what you would do
with the fundamental theorem
of calculus.
With the fundamentals theorem
of calculus, we have the
following formula.
F( 4) - F ( 0) = the integral
from 0 to 4 of dx / 1 + x.
That's what the fundamental
theorem says.
And now I claim that we can
get these same types of
results by a very elementary
observation.
It's really the same observation
that I made up
here, that the average is less
than or equal to the maximum.
Which is that the biggest this
can ever be is, let's see.
The biggest it is when
x is 0, that's 1.
So the biggest it ever
gets is this.
And that's equal to 4.
Right?
On the other hand, the smallest
it ever gets to be,
it's equal to this.
The smallest it ever gets to
be is the integral from
0 to 4 of 1/5 dx.
Because that's the
lowest value that
the integrand takes.
When x = 4, it's 1/5.
And that's equal to 4/5.
Now, there's a little tiny
detail which is that really we
know that this is the area of
some rectangle and this is
strictly smaller.
And we know that these
inequalities
are actually strict.
But that's a minor point.
And certainly not one that we'll
pay close attention to.
But now, let me show you what
this looks like geometrically.
So geometrically, we interpret
this as the
area under a curve.
Here's a piece of the
curve y = 1/1 + x.
And it's going up to 4 and
starting at 0 here.
And the first estimate that we
made; that is, the upper
bound, was by trapping this in
this big rectangle here.
We compared it to the constant
function, which was 1 all the
way across.
This is y = 1.
And then we also trapped it
from underneath by the
function which was
at the bottom.
And this was y = 1/5.
And so what this really is is,
these things are the simplest
possible Riemann sum.
Sort of a silly Riemann sum.
This is a Riemann sum
with one rectangle.
This is the simplest
possible one.
And so this is a very,
very crude estimate.
You can see it misses
by a mile.
The larger and the smaller
values are off
by a factor of 5.
But this one is called
the, this one is the
lower Riemann sum.
And that one is less than
our actual integral.
Which is less than the
upper Riemann sum.
And you should, by now, have
looked at those upper and
lower sums on your homework.
So it's just the rectangles
underneath and the
rectangles on top.
So at this point, we can
literally abandon the mean
value theorem.
Because we have a much better
way of getting at things.
If we chop things up into more
rectangles, we'll get much
better numerical
approximations.
And if we use simpleminded
expressions with integrals,
we'll be able to figure out any
bound we could get using
the mean value theorem.
So that's not the relevance
of the mean value theorem.
I'll explain to you why we
talked about it, even, in a
few minutes.
OK, are there any questions
before we go on?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: I knew that the range
of c was from 0 to 4, I
should have said that
right here.
This is true for this theorem.
The mean value theorem
comes with an extra
statement, which I missed.
Which is that this is for
some c between 0 and 4.
So I know the range is
between 0 and 4.
The reason why it's between 0
and 4 is that 's part of the
mean value theorem.
We started at 0,
we ended at 4.
So the c has to be somewhere
in between.
That's part of the mean
value theorem.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is, do
you exclude any values that
are above 4 and below 0.
Yes, absolutely.
The point is that in order to
figure out how F changes,
capital F changes, between 0
and 4, you need only pay
attention to the values
in between.
You don't have to pay any
attention to what the function
is doing below 0 or above 4.
Those things are strictly
irrelevant.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, I mean
it's strictly in
between these two numbers.
I have to understand what the
lowest and the highest one is.
STUDENT: [INAUDIBLE]
PROFESSOR: It's approaching
that, so.
OK.
So now, the next thing that
we're going to talk about is,
since I've got that 1 up
there, that Fundamental
Theorem of Calculus 1, I need
to talk about version 2.
So here is the Fundamental
Theorem of Calculus version 2.
I'm going to start out with a
function little f, and I'm
going to assume that
it's continuous.
And then I'm going to define a
new function, which is defined
as a definite integral.
G ( x ) is the integral from
a to x of f ( t ) dt.
Now, I want to emphasize here
because it's the first time
that I'm writing something
like this, that this is a
fairly complicated gadget.
It plays a very basic and very
fundamental, but simple role
but it nevertheless is
a little complicated.
What's happening here is that
the upper limit I've now
called x, and the variable t
is ranging between a and x,
and that the a and the
x are fixed when I
calculate the integral.
And the t is what's called
the dummy variable.
It's the variable
of integration.
You'll see a lot of people who
will mix this x with this t.
And if you do that, you will
get confused, potentially
hopelessly confused,
in this class.
In 18.02 you will be completely
lost if you do that.
So don't do it.
Don't mix these two guys up.
It's actually done by many
people in textbooks, and it's
fairly careless.
Especially in old-fashioned
textbooks.
But don't do it.
So here we have this G ( x).
Now, remember, this G ( x)
really does make sense.
If you give me an a, and you
give me an x, I can figure out
what this is, because I can
figure out the Riemann sum.
So of course I need to know
what the function is, too.
But anyway, we have a numerical
procedure for
figuring out what the
function g is.
Now, as is suggested by this
mysterious letter x being in
the place where it is, I'm
actually going to vary x.
So the conclusion is that if
this is true, and this is just
a parenthesism not part
of the theorem.
It's just an indication of
what the notation means.
Then G' = f.
Let me first explain what the
significance of this theorem
is, from the point of view of
differential equations.
G ( x) solves the differential
equation y' = f ( x).
So y' = f, you put the x in
if I got it here, with the
condition y ( a) = 0.
So it solves this pair
of conditions here.
The rate of change, and
the initial position
is specified here.
Because when you integrate from
a to a, you get 0 always.
And what this theorem
says is you can
always solve that equation.
When we did differential
equations, I said that already.
I said we'll treat these
as always solved.
Well, here's the reason.
We have a numerical
procedure for
computing things like this.
We could always solve
this equation.
And the formula is a fairly
complicated gadget, but so far
just associated with
Riemann sums.
Alright, now.
Let's just do one example.
Unfortunately, not a complicated
example and maybe
not persuasive as to
why you would care
about this just yet.
But nevertheless
very important.
Because this is the quiz
question which everybody gets
wrong until they practice it.
So the integral from, say
1 to x of dt / t ^2.
Let's try this one here.
So here's an example of
this theorem, I claim.
Now, this is a question which
challenges your ability to
understand what the
question means.
Because it's got a
lot of symbols.
It's got the integration and
it's got the differentiation.
However, what it really is
an exercise in recopying.
You look at it and you write
down the answer.
And the reason is that, by
definition, this function in
here is a function of the
form G ( x) of the
theorem over here.
So this is the G ( x).
And by definition, we said
that G' ( x) = f(x).
Well, what's the f ( x)?
Look inside here.
It's what's called
the integrand.
This is the integral from 0
to x of f ( t ) dt, right?
Where f ( t) = 1 / t^2.
So your ability is challenged.
You have to take that 1 / t ^2
and you have to plug in the
letter x, instead of t for it.
And then write it down.
As I say, this is an exercise
in recopying what's there.
So this is quite easy
to do, right?
I mean, you just look and
you write it down.
But nevertheless, it
looks like a long,
elaborate object here.
Pardon me?
STUDENT: [INAUDIBLE]
PROFESSOR: So the question
was, why did I integrate.
STUDENT: [INAUDIBLE]
PROFESSOR: Why did
I not integrate?
Ah.
Very good question.
Why did I not integrate.
The reason why I didn't
integrate is I didn't need to.
Just as when you take the -
sorry, the derivative of
something, you take the
antiderivative, you get back
to the thing.
So, in this case, we're taking
the antiderivative of
something and we're
differentiating.
So we end back in the same
place where we started.
We started with f ( t),
we're ending with f.
Little f.
So you integrate, and
then differentiate.
And you get back to
the same place.
Now, the only difference between
this and the other
version is, in this case when
you differentiate and
integrate you could be
off by a constant.
That's what that shift,
why there are two
pieces to this one.
But there's never an
extra piece here.
There's no + c here.
When you integrate and
differentiate, you kill
whatever the constant is.
Because the derivative
of a constant is 0.
So no matter what the constant
is hiding inside of g, you're
getting the same result.
So this is the basic idea.
Now, I just want to double-check
it, using the
Fundamental Theorem of
Calculus 1 here.
So let's actually evaluate
the integral.
So now I'm going to do what
you've suggested, which is I'm
just going to check
whether it's true.
No, no I am because I'm going
just double-check that it's
consistent.
It certainly is slower this way,
and we're not going to
want to do this all the
time, but we might
as well check one.
So this is our integral.
And we know how to do it.
No, I need to do it.
And this is - t ^ - 1,
evaluated at 1 and x.
Again, there's something
subliminally here for you to
think about.
Which is that, remember,
its t is ranging
between 1 and t = x.
And this is one of the big
reasons why this letter t has
to be different from x.
Because here it's 1
and there it's x.
It's not x.
So you can't put an x here.
Again, this is t = 1 and this
is t = x over there.
And now if I plug that
in, I get what?
I get - 1 / x, and then
I get - ( - 1).
So this is, let me get rid of
those little t's there.
This is a little
easier to read.
And so now let's check it.
It's d / dx.
So here's what G ( x) is.
G ( x) = 1 - 1 / x.
That's what G( x) is.
And if I differentiate that,
I get + 1 / x ^2.
That's it.
You see the constant
washed away.
So now, here's my job.
My job is to prove these
theorems. I never did prove
them for you.
So, I'm going to prove the
Fundamental Theorem of Calculus.
But I'm going to do 2 first.
And then I'm going to do 1.
And it's just going to take
me just one blackboard.
It's not that hard.
The proof is by picture.
And, using the interpretation
as area under the curve.
So if here's the value of a, and
this is the graph of the
function y equals f of x.
Then I want to draw three
vertical lines.
One of them is going
to be at x.
And one of them is going
to be at x + delta x.
So here I have the interval from
0 to x, and next I have
the interval from x to delta
x more than that.
And now the pieces that I've got
are the area of this part.
So this has area which
has a name.
It's called G ( x ).
By definition, G( x ), which is
sitting right over here in
the fundamental theorem, is the
integral from a to x of
this function.
So it's the area under
the curve.
So that area is G ( x ).
Now this other chunk here, I
claim that this is delta G.
This is the change in G. It's
the value of G ( x ) that is
the area of the whole business
all the way up to x + delta x
- the first part, G ( x ).
So it's what's left over.
It's the incremental amount
of area there.
And now I am going to carry
out a pretty standard
estimation here.
This is practically
a rectangle.
And it's got a base of delta
x, and so we need to figure
out what its height is.
This is delta G, and it's
approximately its base times
its height.
But what is the height?
Well, the height is maybe either
this segment or this
segment or something
in between.
But they're all about
the same.
So I'm just going to put in the
value at the first point.
That's the left end there.
So that's this height
here, is f ( x ).
So this is f (x), and so really
I approximate it by
that rectangle there.
And now if I divide and take the
limit, as delta x goes to
0 of delta G / delta x, it's
going to equal f ( x ).
And this is where I'm using the
fact that f is continuous.
Because I need the values nearby
to be similar to the
value in the limit.
OK, that's the end.
This the end of the proof, so
I'll put a nice little q.e.d.
here.
So we've done Fundamental
Theorem of Calculus 2, and now
we're ready for Fundamental
Theorem of Calculus 1.
So now I still have it on the
blackboard to remind you.
It says that the integral
of the derivative is the
function, at least the
difference between the values
of the function at two places.
So the place where we
start is with this
property that F' = f.
That's the starting, that's
the hypothesis.
Now, unfortunately, I'm going
to have to assume something
extra in order to use the
Fundamental Theorem of
Calculus 2, which is I'm going
to assume that f is
continuous.
That's not really necessary, but
that's just a very minor
technical point, which I'm
just going to ignore.
So we're going to start
with F' = f.
And then I'm going to
go somewhere else.
I'm going to define a new
function, G (x), which is the
integral from a to
x of f ( t ) dt.
This is where we needed all of
the labor of Riemann's sums.
Because otherwise we don't have
a way of making sense out
of what this even means.
So hiding behind this one
sentence is the fact that we
actually have a number.
We have a formula for
such functions.
So there is a function g (x)
which, once you've produced a
little f for me, I can cook up
a function capital G for you.
Now, we're going to apply this
Fundamental Theorem of
Calculus 2, the one that
we've already checked.
So what does it say?
It says that G' = f.
And so now we're in the
following situation.
We know that F' (
x ) = G' ( x).
That's what we've got so far.
And now we have one last step
to get a good connection
between F and G. Which is that
we can conclude that F ( x ) =
G ( x) + c.
Now, this little step may seem
innocuous but I remind you
that this is the spot
that requires
the mean value theorem.
So in order not too lie to you,
we actually tell you what
the underpinnings of all
of calculus are.
And they're this: the fact,
if you like, that if two
functions have the
same derivative,
they differ by a constant.
Or that if a function
has derivative 0,
it's a constant itself.
Now, that is the fundamental
step that's needed, the
underlying step that's needed.
And, unfortunately, there aren't
any proofs of it that
are less complicated than using
the mean value theorem.
And so that's why we talk a
little bit about the mean
value theorem, because we don't
want to lie to you about
what's really going on.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is how
did I get from here, to here.
And the answer is that if G' is
little f, and we also know
that F' is little f,
then F' is G'.
OK.
Other questions?
Alright, so we're almost done.
I just have to work out
the arithmetic here.
So I start with F(b) - F ( a).
And that's equal to (G (
b) + c) - (G (a) + c).
And then I cancel the c's.
So I have here G(b) - G(a).
And now I just have to check
what each of these is.
So Remember the definition
of G here.
G ( b) is just what we want.
The integral from a
to b of f(x) dx.
Well I called it f ( t) dt,
that's the same as f(x) dx
now, because I have the limit
being b and I'm allowed to use
x as the dummy variable.
Now the other one,
I claim, is 0.
Because it's the integral
from a to a.
This one is the integral
from a to a.
Which gives us 0.
So this is just this - 0,
and that's the end.
That's it.
I started with F( b) - F( a),
I got to the integral.
Question?
STUDENT: [INAUDIBLE]
PROFESSOR: How did I get from
F ( b) - F (a), is (G ( b) +
c) - (G( a) + c), that's
the question.
STUDENT: [INAUDIBLE]
PROFESSOR: Oh, sorry this
is an equals sign.
Sorry, the second line
didn't draw.
OK, equals.
Because we're plugging in for
f (x) the formula for it.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: This step here?
Or this one? there's
STUDENT: [INAUDIBLE]
PROFESSOR: Right.
So that was a good question.
But the answer is that that's
the statement that we're
aiming for.
That's the Fundamental Theorem
of Calculus 1, which
we don't know yet.
So we're trying to prove it, and
that's why we haven't, we
can't assume it.
OK, so let me just notice that
in the example that we had,
before we go on to something
else here.
In the example above, what we
had was the following thing.
We had, say, F (
x ) = - 1 / x.
So F' (x) = 1 / x ^2.
And, say, G ( x)
= 1 - (1 / x).
And you can see that either
way you do that, if you
integrate from 1 to 2, let's
say, which is what we had over
there, dt / t ^2, you're going
to get either - 1 / t, 1 to 2
or, if you like, 1 -
(1 / t), 1 to 2.
So this is the F version,
this is the G version.
And that's what plays
itself out here, in
this general proof.
Alright.
So now I want to go back to the
theme for today, which is
using little f to understand
capital F. In other words,
using the derivative of f to
understand capital F. And I
want to illustrate it by some
more complicated examples.
So I guess I just erased it,
but we just took the
antiderivative of 1 / t ^2.
And there's all of the
powers work easily.
But one, and the tricky one
is the power 1 / x.
So let's consider the
differential equation
L' ( x) = 1 / x.
And say, with the initial
value L (1) = 0.
The solution, so the Fundamental
Theorem of
Calculus 2 tells us
the solution is
this function here.
L( x) equals the integral
from 1 to x, dt / t.
That's how we solve all
such equations.
We just integrate, take
the definite integral.
And I'm starting at 1 because
I insisted that L( 1 ) be 0.
So that's the solution
to the problem.
And now the thing that's
interesting here is that we
started from a polynomial.
Or we started from a rational,
a ratio of polynomials; that
is, 1 / t or 1 / x.
And we get to a function which
is actually what's known as a
transcendental function.
It's not an algebraic
function.
Yeah, question.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is why
is this equal to that.
And the answer is, it's for the
same reason that this is
equal to that.
It's the same reason as this.
It's that the 1's cancel.
We're taken the value of
something at 2 minus
the value at 1.
The value at 2 minus
the value at 1.
And you'll get a 1 in the one
case, and you get a 1 in the
other case.
And you subtract them and
they will cancel.
They'll give you 0.
These two things really
are equal.
This is not a function evaluated
at one place, it's
the difference between the
function evaluated at 2 and
the value at 1.
And whenever you subtract
two things like that,
constants drop out.
STUDENT: [INAUDIBLE]
PROFESSOR: That's right.
If I put 2, here if I
put c here, it would
have been the same.
It would just have
dropped out.
It's not there.
And that's exactly this
arithmetic right here.
It doesn't matter which
antiderivative you take.
When you take the differences,
the c's will cancel.
You always get the same
answer in the end.
That's exactly why I wrote
this down, so that
you would see that.
It doesn't matter which
one you do.
So, we still have a couple
of minutes left here.
This is actually, so
let me go back.
So here's the antiderivative of
1 / x, with value 1 at 0.
Now, in disguise, we know
what this function is.
We know this function is
the logarithm function.
But this is actually a better
way of deriving all of the
formulas for the logarithm.
This is a much quicker
and more efficient
way of doing it.
We had to do it by very
laborious processes.
This will allow us to
do it very easily.
And so, I'm going to
do that next time.
But rather than do that now, I'm
going to point out to you
that we can also get truly
new functions.
OK, so there are all kinds
of new functions.
So this is the first example
of this kind would be, for
example, to solve the equation
y' = e ^ - x^2 with y( 0 ) =
0, let's say.
Now, the solution to that is a
function which again I can
write down by the fundamental
theorem.
It's the integral from 0
to x of e ^ - t^2 dt.
This is a very famous
function.
This shape here is known
as the bell curve.
And it's the thing
that comes up in
probability all the time.
This shape e ^ - x^2.
And our function is
geometrically just the area
under the curve here.
This is F (x).
If this place is x.
So I have a geometric
definition, I have a way of
constructing what it is by
Riemann's sums. And I have a
function here.
But the curious thing about F
( x ) is that F ( x ) cannot
be expressed in terms of any
function you've seen
previously.
So logs, exponentials, trig
functions, cannot be.
It's a totally new function.
Nevertheless, we'll be able to
get any possible piece of
information we would want to,
out of this function.
It's perfectly acceptable
function, it will work just
great for us.
Just like any other function.
Just like the ln.
And what this is analogous
to is the
following kind of thing.
If you take the circle, the
ancient Greeks, if you like,
already understood that if you
have a circle of radius 1,
then its area is pi.
So that's a geometric
construction of what you could
call a new number.
Which is outside of the realm
of what you might expect.
And the weird thing about this
number pi is that it is not
the root of an algebraic
equation with rational
coefficients.
It's what's called
transcendental.
Meaning, it's just completely
outside of
the realm of algebra.
And, indeed, the logarithm
function is called a
transcendental function, because
it's completely out of
the realm of algebra.
It's only in calculus that
you come up with
this kind of thing.
So these kinds of functions
will have access to a huge
class of new functions here,
all of which are important
tools in science and
engineering.
So, see you next time.
