10 Dec 2008

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.



[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 = 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.


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