We've seen that we can *approximate* the area under a curve by
using a sum of rectangle areas. But suppose I'm trying to find the
*exact* area under a curve --- say from to . Since increasing the number of rectangles increases
the accuracy of the approximation, maybe I'd get the exact area if I
could use an "infinite number of rectangles". I will give a
rough description of how we might do this.

To do this, I start by breaking the interval up into n subintervals of lengths , , ..., . In the k-th subinterval, I pick some point , and use as the height of the k-th rectangle.

The sum of the areas of the rectangles approximates the area under the curve:

The more rectangles I take, the better the approximation. So it's
reasonable to suppose that the *exact* area would be given by
the *limit* of such sums, as n goes to infinity:

The expression on the right --- a limit of a sum, or a * Riemann sum* --- is called * the
definite integral of from a to b* and is denoted as
follows:

If the limit of the sum on the right side is defined, the function f
is * integrable* on the interval .

* Remark.* In the definition above, I'm passing
over a number of technical points for the sake of getting the
intuitive idea across. For example, what does it mean to let the
number of rectangles become infinite? Does it matter how the
rectangles' widths or the evaluation points are chosen?

A careful treatment of the * Riemann integral*
has to take these considerations into account.

It is possible to compute areas using the formula above, though it's not easy. Here's an example of a direct computation using the defintion.

* Example.* Use the limit of a sum to compute the
area under from to --- that is, the definite integral

I will need the following summation formulas:

Divide up the interval into n equal subintervals. Each has length . I'll evaluate the function at the right-hand endpoints. These are

The k-th point is , so the height of the k-th rectangle is

The sum of the rectangles' areas is

Applying the sum formulas above, I get

The (exact) area is

While this approach works, it's pretty complicated! I'll discuss better ways to compute definite integrals shortly.

Not all functions are integrable. The following result which I'll state without proof covers most of the cases we'll need:

* Theorem.* A bounded function with finitely many
discontinuities is integrable.

For example, consider the function

There is a single discontinuity at . Hence, f is integrable on any interval.

In particular, a continuous function is integrable.

On the other hand, is not bounded near , so it is not integrable on any interval containing 0.

In some cases, you can use the fact that the definite integral represents the area under a curve to evaluate the integral geometrically.

* Example.* Compute .

The integral represents the area of a triangle with height 4 and base 4, so

* Example.* Compute .

The area consists of the piece in the last problem, together with a
piece of area 2. But the second piece is below the x-axis, so it is
taken as *negative* in the definite integral:

* Example.* Compute .

This is half the area of a circle of radius 1:

Here are some properties of definite integrals. I'll present them without proofs.

1. If k is a number, then

This is another way of saying that the area of a rectangle is the base times the height.

* Example.* Compute .

2.

This says that the integral of a sum is the sum of the integrals.

3. If for , then

This says that bigger functions have bigger integrals.

* Example.* Use the inequality to estimate .

4.

That is, switching the limits of integration multiplies the integral by -1.

As a special case,

5.

That is, integrating from a to b and then from b to c is the same as integrating all the way from a to c:

* Example.* Simplify .

6. (* The Mean Value Theorem for Integrals*)
There is a number c, , such that

represents the height of a rectangle on the integral which has the same area as the area under the curve.

* Example.* Use the Mean Value Theorem for
Integrals to estimate .

Applying the Mean Value Theorem for Integrals, I find that there is a number c between 0 and 2 such that

Now

Therefore,

So

Copyright 2018 by Bruce Ikenaga