On-line Math 21

5.1  Definition of the integral

How do we do this calculation?

As a reminder, we are finding the area area bounded by the graphs

y = x2, y = 0, and x = 3.

We first divide the interval [0,3] into 20 subintervals. For convenience, we can assume they are all the same width, 3/20. The endpoints of these intervals are at the values

xi: = 0+ 3i 20 ,

for i = 0,¼,20 . The function f(x) = x² is evaluated at each of those points, being careful to pick the left edge for inscribed rectangles, and the right edge for circumscribed rectangles. Then, you add up all the areas, with the area of the i^th inscribed rectangle given by

f(xi-1) 3 20 .

So, the lower value is

Area ³ 20 å i = 1  f(xi-1) 3 20 = 20 å i = 1  æ ç è 3(i-1) 20 ö ÷ ø 2   3 20 = 27 8000 19 å j = 0  j2 = 27 8000 (19)(20)(39) 6 ,

where j = i-1 . The upper value, for circumscribed rectangles, is

Area £ 20 å i = 1  f(xi) 3 20 = 20 å i = 1  æ ç è 3(i) 20 ö ÷ ø 2   3 20 = 27 8000 20 å i = 1  i2 = 27 8000 (20)(21)(41) 6 .

The real game is to see what happens as we take more and more rectangles. Using just the circumscribed rectangles, and taking more and more of them ( n of them) to get a better and better approximation of the area, gives, in the end:

Area = lim n® ¥  n å i = 1  f(xi) 3 n = lim n® ¥  n å i = 1  æ ç è 3i n ö ÷ ø 2   3 n = lim n® ¥  27 n3 n å i = 1  i2 = lim n® ¥  27(n(n+1)(2n+1)) n36 = 9.

Where did these sums come from?

The reality is that, for most curves, it doesn't matter whether you use the right edge, or the left, or anywhere in between. You also don't have to take the subintervals to all be the same length, just so they get smaller, to get a better and better approximation of the real area.

Copyright (c) 2000 by David L. Johnson.