Area Sums and Areas as Limits in Edwards & Penney

presentation of Edwards & Penney's work, by Corry Shores

Edwards & Penney's Calculus is an incredibly-impressive, comprehensive, and understandable book. I highly recommend it.

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

The figure below 

shows the region R that lies below the graph of the positive-valued increasing function f and above the interval [a, b]. In order to approximate the area A of R, we have chosen here a fixed integer n and divided the interval [a, b] into n subintervals

all having the same length

On each of these subintervals is erected one inscribed rectangle and one circumscribed rectangle.

As we see below

the inscribed rectangle over the ith subinterval 

has height

whereas the ith circumscribed rectangle has height

Because the base of each rectangle has length Δx, the areas of the rectangle are 

respectively. When we sum the inscribed rectangles' areas for i = 1, 2, 3, ... , n, we obtain the underestimate

of A's actual area. Likewise, the sum of the circumscribed rectangles' areas is the overestimate

The inequality 

then yields

The inequalities in the above formula would be reversed if f (x)  were decreasing, rather than increasing, on [a, b].

Illustrations such as the one below

suggest that if the number n of subintervals is very large, so that Δx is small, then the areas 

of the inscribed and circumscribed polygons will differ only by a very little. Hence both will be very close to region R's actual area A. We may also observe this because if f either is increasing or is decreasing on the whole interval [a, b], then the small rectangles in the figure below (representing the difference between 

):

 

can be reassembled in a "stack," as indicated on the right in the figure. It thus follows that

But

Thus the difference between the left-hand and the right-hand sums in 

is approaching zero as

Whereas A does not change as

If follows that the area of the region R is given by

The meaning of these limits is simply that A can be found with any desired accuracy by calculating either sum in the above equation with a sufficiently large number n of subintervals.

When applying the above equation, keep in mind that

Also note that

because i = 0, 1, 2, . . . , n, because 

is i "steps" of length Δx to the right of

 

Example:

We will now calculate the exactly the area that we approximated in the first example: The area of the region under the graph of 

and over the interval [0, 3]. If we divide [0, 3] into n subintervals all of the same length, then equations

and

give

for i = 0, 1, 2, . . . , n. Thus,

[see first rule of summation]. Then the equation 

for

yields

When we take the limit as 

because the terms 1/(2n) and 

approach zero as 

Thus our earlier inference from the data in the chart below was correct: A = 9 exactly.

from Edwards & Penney: Calculus. New Jersey: Prentice Hall, 2002, p.290a; 292c-294c.