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 i = 0, 1, 2, . . . , n. Thus,
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.
No comments:
Post a Comment