The Area Problem Revisited

We saw last time that we can approximate the area underneath the graph of the function f(x) over the interval [a,b] by dividing the interval up into N subintervals of width

and placing rectangles over each of these subintervals whose height are given by

where the point cn is a sample point that lives somewhere in the nth interval [a + (n-1) w, a + w]. The total area of all of the rectangles is then

(1)

Today I want to start by showing two more examples to demonstrate how this calculation works in practice. In both cases, the function f(x) is 2 x + 2 and the interval is [a , b].

Example One - Sample at the Left Endpoint

The only freedom given us by formula (1) is the choice of sample points and the number of subintervals N. Selecting different sample points will make a difference in the eventual outcome, so let's do a couple of examples that use differing choices for the sample points cn. For our first example, we will select the left endpoint of each subinterval as the sample point.

This leads to

To evaluate this sum we will start by expanding fully in the summand.

As far as the sum is concerned, everything but n is a constant. This leads to

Recalling that w = (b - a)/N we get

In the limit of large N we have

Example Two - Sample in the Interval

For our second example, let up place the sample points two-thirds of the way from the left endpoint to the right endpoint.

Here are the details of the calculation.

Substituting for w gives

It is easy to see that in this case the area goes to the same value in the limit as N gets large.

The Definite Integral

Up to now we have been estimating the area under the graph of between the vertical lines x = a and x = b by placing a collection of rectangles inside the area to be computed and adding up the areas of those rectangles. To construct the rectangles we subdivided the interval from a to b into N equal subintervals and used the value of the function over one endpoint of each subinterval to provide a height. The resulting approximation for the area looks like

where w = xn + 1 - xn is the width of the subinterval that a rectangle sits over.

More generally, if we don't subdivide the interval [a,b] into equal sized subdivisions, the area becomes

where cn is a sample point located somewhere in the sub-interval [xn , xn+1] and is the width of the nth subinterval.

In the limit as the number of subdivisions becomes very large and each subdivision becomes very small, the approximate area becomes an exact area, called the definite integral of f over the interval [a,b]. To represent the result of that limiting calculation we will introduce a new bit of notation:

Properties of the definite integral

Because the definite integral is an area, it has a number of simple properties. Here are some of them:

  1. : this follows from the fact that the area in question has no width.
  2. : areas are additive in a simple way.
  3. : areas also add this way.
  4. : areas subtract this way.
  5. If for then : areas above the x-axis are considered positive, while areas below the axis are considered negative.
  6. If for , then .
  7. : the area in question is a rectangle whose area is easily computed.
  8. : areas multiply in an obvious way when scaled by a constant.
  9. If for all x between a and b, then : this follows from the fact that area under f(x) is trapped between the areas of the two rectangles with height m and M.

Homework

Section 5.2: 4, 21, 22, 28