[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.