Area under Graphs in Edwards & Penney

presentation of Edwards & Penney's work, by presentation of Edwards & Penney's work, by Corry Shores
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Edwards & Penney's Calculus is an incredibly-impressive, comprehensive, and understandable book. I highly recommend it.

[the following will not stray from Edwards & Penney's procedure, so it is largely quotation.]

We begin by considering the region determined by a continuous positive-valued function f defined on a closed interval [a, b]. We want to calculate the area A of the region R that lies below the curve y = f (x) and above the interval [a, b] on the x-axis.

The region R is bounded on the left side by the vertical line x = a and on the right side by the vertical line x = b.

We then divide the base interval [a, b] into subintervals, all having the same length. Above each subinterval is a vertical strip, and the area A is the sum of the strip's areas.

On each of these subintervals, we erect a rectangle that approximates the corresponding vertical strip. We may choose either an "inscribed" or a "circumscribed" rectangle

or even a rectangle that is intermediate between the two. These rectangles now make up a polygon that approximates the region R, and therefore the sum of the areas of these rectangles approximates the desired area A.

So for example, we want to approximate region R's area A lying below parabola

above the interval [0, 3] on the x-axis. The diagrams below show plottings for:

a) 5 inscribed and 5 circumscribed rectangles

b) 10 inscribed and 10 circumscribed rectangles

c) 20 inscribed and 20 circumscribed rectangles

d) 40 inscribed and 40 circumscribed rectangles

[as with all images, click on figure below for larger version]

The set of different versions of inscribed rectangles gives an underestimate of A, whereas the circumscribed rectangles give an overestimate of A. The "curvilinear triangles" (by which the rectangular polygons in the above figures undershoot or overshoot the region R) constitute the errors in these estimates. The more rectangles we use, the more accurate the approximation. Thus, to approximate accurately the area of such a region R, we need an effective way to calculate and sum the areas of rectangle collections like those in the above figures.

Example:

As in the figures above, we will say that R denotes the region lying below the graph of

and above interval [0, 3]. We will now calculate the underestimate as well as the overestimate of R's area A by using 5 rectangles each of width 3/5. Then we repeat the computations using 10 rectangles of widths 3/10.

We first suppose that n = 5 rectangles are being used. We then let

denote the underestimate and

denote the overestimate that we obtain by using 5 rectangles based on the 5 subintervals of length 3/5.

From the figure below,

we notice that the heights of the 5 inscribed rectangles (the first of which is degenerate -- its height is zero) are the values of the function

at the 5 left-hand endpoints

Because the base of each rectangle has length 3/5, we note also that

[A rectangle's area is found by multiplying its base times its height. The base for each one is 3/5. We square each term value because the function is

]. The heights of the 5 circumscribed rectangles are the values of

at the 5 right-hand endpoints

Thus the corresponding overestimate is

These are crude approximations to the actual area A. On the basis of this information alone, our best estimate of A might well be the average of the under- and overestimates:

We will see now if doubling the number of subintervals to n = 10 increases the accuracy. Considering the diagram below

we notice that the heights of the 10 inscribed rectangles are the values of

at the 10 left-hand endpoints

of the subintervals in the figure below

The base of each rectangle has length 3/10, so the resulting underestimate is

Likewise, the sum of the areas of the 10 circumscribed rectangles in the figure below

is the overestimate

At this stage, our best estimate for area A might be the average

If we calculate over and underestimates for 20, 40, 80, 160, and 320 rectangles, we find

Thus the average values in the final column suggest that

from Edwards & Penney: Calculus. New Jersey: Prentice Hall, 2002, p.287c-290a.