[music] We can get a much better sense for
what the integral's actually computing by
converting the integral into a function.
So the way that I'm going to package
together a whole bunch of integrals is
into a function, the accumulation
function.
What is the accumulation function going to
do?
Well, here, I've graphed y equals f of t
on the t y plane.
And I picked a fixed point a, and I
imagine I've got another point that I can
vary labeled x.
And the accumulation function, is the
integral from a to x.
So it computes the area between a and x
under the graph of this function.
And it's called the accumulation function
because I imagine that as I move x to the
side, I'm kind of accumulating more and
more area.
I'm determining how much area I've
accumulated between a and whatever I plug
into the accumulation function.
When is this accumulation function
increasing?
As long as this function little f is
positive, the accumulation function is
increasing.
Look, to determine whether a function's
increasing, I'd have to plug in bigger
inputs and see if I get bigger outputs.
That's what increasing means.
So if I were to plug in some bigger input
right, the accumulation function just
figures out how much area is between a and
this bigger input.
And as long as my function's positive
there's now more area between a and this
bigger input than there was between a and
x.
So I can summarize this in saying.
That f is positive, and that makes my
accumulation function increasing.
When is this accumulation function
decreasing?
Is it ever possible to integrate over a
longer interval and yet get a smaller
value?
Yes, it sounds counter-intuitive, but
that's entirely possible.
And it really comes down to this issue
about what these integrals are really
measuring.
The integrals are not exactly measuring
area, they're measuring signed area.
So here's an example where the function
that I'm graphing, y equals f of t, passes
below the, I'm calling it here, the t
axis.
And when the graph is below and I'm
calculating the integral, this area here
counts as negative in the integral.
So this area which is above the axis here
is calculated as honest area, but this
area here is really contributing negative
area to the integral, right?
I mean, I'm taking this area and I'm
subtracting this area to figure out the
integral from a to x.
What it comes down to is how this integral
is defined.
The integral is defined as a Riemann sum.
It's the functions value evaluated at
various sample points times the widths of
those tiny rectangles and if I'm
evaluating the function and the functions
value is negative then that's not really
an area.
That's a signed area and it could be
negative.
All right.
Well in light of that, what happens in
this particular place?
Right.
What happens if I plug in a bigger input
to my accumulation function.
Is it possible that I could integrate over
a longer interval, and yet get a smaller
value.
And yeah, this is an exact picture of the
sort of situation where that happens.
If I plug in a bigger input here, then
when I integrate from a to this bigger
input, I'm not only subtracting this red
negative area, I'm subtracting this larger
red negative area.
And in that case my accumulation function
really is smaller, right?
As I take x and drag it over here.
I'm gathering up more negative area, and
so my accumulation function is going down.
I'll summarize that like this, if
function's negative the accumulation
function is decreasing.
What does this sound like?
Well what that sounds like is that the
derivative of the accumulation function is
related to the thing I'm integrating, the
so called integrand.
A, if I'm integrating a positive function,
accumulation functions increasing.
And another way to say that the
accumulation function is increasing would
be to say that the derivative is positive.
So the integrands positive, the derivative
of accumulation function is positive.
The function that I'm integrating is
negative, the accumulation function is
decreasing.
The derivative of the accumulation
function is negative.
Integrand negative.
Derivative of accumulation function also
negative.
This is our first hint at the fundamental
theorem of calculus.
Some sort of relationship between
derivatives.
And integrals.