On-line Math 21

5.1  The definition of the Riemann Integral

Example 3

ó õ 1 0  x2dx = 1 3 .

Solution

Proceed as with the previous example, presuming that the limit exists. Then, we can just look at uniform subdivisions and circumscribed rectangles, as before. With n subintervals, then x_i = 0+i/n , and since we are using circumscribed rectangles and f(x) is increasing, then x_i^* = x_i in each case (the right-hand edge of the i^th subinterval gives f the largest value in that subinterval). So the Riemann sum is:

n å i = 1  f(xi)Dxi = n å i = 1  ( xi) 2 æ ç è 1 n ö ÷ ø = n å i = 1  æ ç è i n ö ÷ ø 2   æ ç è 1 n ö ÷ ø = 1 n2 n å i = 1  i2 = 1 n2 n(n+1)(2n+1) 6 .

Taking the limit as n® ¥, which also implies that the mesh of these subintervals goes to 0, we have

ó õ 1 0  x2dx = lim n® ¥  n å i = 1  f(xi)Dxi = lim n® ¥  n(n+1)(2n+1) 6n3 = 1 3 .

Copyright (c) 2000 by David L. Johnson.