# Definite Integrals

The area under a curve can be approximated by adding up the areas of rectangles.

Example. Approximate the area under from to using 20 equal subintervals and evaluating the function at the left-hand endpoints.

You can use a calculator to approximate this sum; it's around 0.84799.

Example. Note that the subintervals don't have to be the same size, and I don't have to choose the evaluation points systematically. These are both conveniences for the problems. For example, here is an approximation to the area under from to .

This gives an approximate area of 51.84. The actual area is 77.

In general, suppose I'm trying to find the area under from to . I break 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:

It is possible to compute areas using the formula above, though it's not easy.

Example. Use the limit of a sum to compute the area under from to .

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 left-hand endpoints. These are

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

I get the following expression for the area:

Apply the formulas from the beginning of the problem:

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

Since I'm taking a limit in computing the definite integral, it's possible for the limit (and hence, the definite integral) to be undefined. is integrable on the interval if is defined. The following fact says that many of the functions you'll use in calculus are integrable:

• A bounded function with finitely many discontinuities is integrable.

(A function on an interval is bounded if there is a number M such that for all x in the interval.)

For example,

is integrable on any interval.

In particular, a continuous function is integrable.

On the other hand, 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 .

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.

2. If f and g are integrable, then so is , and

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

3. If f and g are integrable and for , then

This says that bigger functions have bigger integrals.

Example. You can use the last rule to get estimates for integrals.

For example,

Therefore,

4. If f is integrable, then

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

5. If f is integrable on the interval and , then

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.

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 graph of .

Example.

for some c between 0 and 2.

Now

Therefore,

So

I've gotten a rough estimate for the value of the integral.

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