On-line Math 21

5.1  The definition of the Riemann Integral

Example 2 If f(x) = 3x , then

ó õ 2 1  f(x)dx = 9/2.

Solution

This is not trivial to see. If you trust that the limit exists, which is not really obvious, but believable, then you can always use the circumscribed rectangles, that is, take c_i = x_i in the sum. Also, take a regular partition, with each subinterval of the same width (which would be (2-1)/n ), so that x_i = 1+i/n , then the Riemann sum of that partition, with those sampling points, is

n å i = 1  f(xi)Dxi = n å i = 1  3(1+i/n)(1/n) = n å i = 1  3(1+i/n)(1/n) = (3/n) n å i = 1  1+(3/(n2)) n å i = 1  i = (3/n)×n+(3/(n2)) n(n+1) 2 ,

so that the limit, when the mesh goes to 0 (here meaning that n® ¥), goes to

3+ 3 2 = 9 2 ,

as advertised.

One approach to the more serious question of whether or not the limit exists is to change to the inscribed rectangles for the same partition above. Since the height of the inscribed rectangle over the t^th subinterval is

f(xi-1) = f(1+(i-1)/n) = 3(1+(i-1)/n).

Then, the Riemann sum, using inscribed rectangles at each stage, is

n å i = 1  f(xi-1)Dxi = n å i = 1  3(1+(i-1)/n)(1/n) = n å i = 1  3(1+(i-1)/n)(1/n) = (3/n) n å i = 1  1+(3/(n2)) n å i = 1  (i-1) = (3/n) n å i = 1  1+(3/(n2)) n-1 å j = 1  j, substituting j = i-1 = (3/n)×n+(3/(n2)) (n-1)(n) 2 ,

which has the same limit as n® ¥.

Then, since any Riemann sum with this partition, for any choice of sampling points, lies in between these two, any such sum would tend to the same limit. This does not quite cover all possibilities, though, since the partitions could be quite different from these. We'll take care of that last detail in the next section.

Copyright (c) 2000 by David L. Johnson.