[MUSIC].
I want to begin with a statement of he fundamental theorem of calculus.
Suppose the function little f, from the closed interval a to b, to the real
numbers, is a continuous function. And let's define a function big F, to be
the accumulation function. Right, that's the integral.
From a to x of the function on the left well then the function big F turns out to
be continuous in the closed interval a to b differentiable on the open interval a
to b and this is the big point the derivative of big F is little f.
We've seen a bit of evidence for this already.
And suppose I got the accumulation function.
So big F of x is the integral from a to x of f of tdt.
And now I can ask, how is the accumulation function changing?
Well, the accumulation function is increasing, meaning it's derivative is
positive if the function that I'm integrating is positive.
And the accumulation function is decreasing, meaning the derivative is
negative. If the function that I'm integrating is
negative. And this suggests that the derivative of
the accumulation function and the function that I'm integrating might be
related in some way. Here's a graph that's particularly
compelling. So, I'll draw the coordinate plane,
here's the y axis, here's the t axis in this case and I'll draw some Random
looking graph. That'll be the graph y equals f of t.
And then I'll pick some point. I'll call it a and some other point x.
And then I'll look at the area under the graph between a and x, rigth?
And that's the integral from a to x of f of t d t.
Right that area is the accumulation function evaluated x.
Now what's the derivative of the accumulation function.
What got the accumulation function, right f(x) equals the interval from a to x that
f(t)dt and I don't know how that changes when x changes.
So lets make a small change in x. Lets go from x.
To say x plus h and then ask how does the area in this region compare to the area
in this much larger region. What's big F of x plus h minus big F of
x? Yeah, so what's F of x plus h right
that's the area. From a to x plus h, minus F of x, big F
of x. That's the area from a to x, and that
difference, right. That difference is just the area inside
here. That little leftover piece is practically
a rectangle of height, f of x, little f of x and width h.
I mean, this region is not quite a rectangle.
But it's awfully close to a rectangle of, width h.
And area f of x plus h minus f of x. Now, remember, by goal was to
differentiate the accumulation function. And all I've got is this rectangle.
Whose width is about h, and whose height, you think about it, it's about f of x.
So what's up with this rectangle? Alright?
I mean what I've really got here is a rectangle and its area is about f of x
plus h minus f of x, this accumulation function.
It's width is h, and it's height is about f of x.
Well I've got a rectangle with this area and this width and, and this height.
Alright, what do I know? I know its height times its width is its
area, so if I take its area and divide by its width, I should get its height.
Now all this is just approximate, but let's write that down.
I The area of this you know, rectangle is f of x plus h minus f of x.
And if I divide by the width, I should get, the height.
What does that mean after I take a limit? Oh of course this is only approximately
true, but if I take a limit, right. If I take a limit as h approaches 0.
This gives me the derivative of the accumulation function and that will in
fact be equal to the function's value. And that is the whole game.
The accumulation function F is an antiderivative of f.
Well this, I mean this is really remarkable.
For instance you might be wondering what's an anti-derivative of this
function, little f of x equals e to the negative x squared?
Well if a functions continuous then we can find an anti-derivative.
The anti-dirivitive is the accumulation function.
So an anti-derivative of either the negative x squared is the accumulation
function. I can define a function big f of x to be
the intregal from a to x, for some number a, of e to the negative t squared d t.
And if I differentiate this accumulation function, I get back e to the negative x
squared. So I found an anti-derivative.
For this function. Well I can't write it down any more
nicely. Because I can't write down an
anti-derivative for e to the negative x squared using the functions that I have
at hand. But nevertheless, the anti-derivative is
the accumulation function.