[music] We've already got an idea of how
we can approximate the area of a curved
region by replacing that curved region
with rectangles.
We've already done this.
If I want to approximate the area under
the curve, I'll just add up the areas of
these rectangles.
And that's a pretty cut approximation of
if these rectangles are thin enough.
If I make those rectangles thinner and
thinner, that gives me better and better
approximations to the area under the
curve.
In fact, more than just approximating the
area under the curve.
By making those rectangles thinner and
thinner, I can write down a definition of
the area under the curve.
So by taking those triangles to be thinner
and thinner.
I'm going to get a better and better
approximation to the area under this
curve.
And I'm going to denote that limit by
this, the integral, which is this long,
thin s, from a to b of the function dx,
right?
That's how I'm going to write down the
area Under the graph of this function,
above the x-axis, and between the line y
equals a and y equals b.
It's going to be by this notation.
Here is the precise definition of the
integral.
To precisely, the integral of this
function from a to b is a limit over
partitions of the Riemann sum, alright?
And it's a limit over partitions where
I've also chosen some sample points for
those partitions.
And I'm taking the limit as a maximum
width of a sub-interval in that partition
goes to 0.
In order to ensure that all of the widths
of the sub-intervals are going to 0, I
just demand that the maximum one, the
biggest one goes to 0, and that forces all
the other ones to be small as well.
I had to say that the maximum width goes
to 0 to prevent some sort of bizarre
scenario.
Well I've got one really big sub-interval
and then lots and lots of small
sub-intervals right?
I want to make sure that all of those
sub-intervals are getting very narrow so
that I'm getting a very fine partition in
the limit.
There's no guarantee whatsoever that, that
limit actually exists.
On the other hand, here's a theorem.
If a function f is continuous then it's
integrable, meaning that the integral,
right, the integral of the function from
say a to b actually exists.
So when I say integraable, what do I even
mean by integrable?
If I were to claim that the integral of f
from a to b is equal to some number, big
I, what that means is that no matter how I
partition the interval from a to b, as
long as that partition is fine enough.
Alright, as long as the maximum width of
any rectangle in my skyscraper picture is
small enough, right, as long as the widest
one Is thin enough.
Then no matter how I choose those sample
points within each of the sub-intervals in
my partition, the resulting Reimann sum
that I calculate is close to I.
How close?
Well it's as close as I want it to be.
Right?
To say that this is equal to something is
really to say something about a limit.
So that means that if I'm asserting that
this is equal to I, it means that I can
get the Reimann sum as close as I want to
I by simply demanding that my partition is
fine enough.
Regardless of how I exactly choose those
sample points.
And all this talk about Reimann sums is
reflected in the very notation in that
we've chosen for the integral.
So here, I've got the integral, the
integral of f from a to b.
And here, I've got a Riemann sum, right,
and this integral is defined to be a limit
of these Reimann sums.
And the notation really reflects their
common origin, right?
Here, I've got a summation symbol, a Greek
sigma, Greek letter s.
And here, I've got a long s.
Alright, these are both a kind of sum.
And the thing I'm summing is the function
evaluated somewhere times a change in x.
And the thing I'm summing here is the same
thing it's the function evaluated
somewhere times a change in x.
So I hope that even the way I write down
integrals really reminds you that there
are a limit of things that look like this
right.
The function evaluated somewhere times
some change in x being added up.