EXAMPLE 2 For the region under f(x) = 5x^2 on [0, 2], show that the sum of the areas of the upper approximating rectangle approaches 40/3, that is
lim n->infinity Rn = 40/3
SOLUTION Rn is the sum of the areas of the n rectangles in the figure. Each rectangle has width 2/n and the heights are the values of the function f(x) = 5x^2 at the points 2/n, 4/n, 6/n, ..., 2n/n; that is, the heights are 5(2/n)^2, 5(4/n)^2, 5(6/n)^2, ..., 5(2n/n)^2. Thus,
Rn = 10/n(2/n)^2 + 10/n(4/n)^2 + 10/n(6/n)^2 + ... + 10/n(2n/n)^2
= 40/n^3 * (1^2 + 2^2 + 3^2 + ... + n^2)
Here we need the formula for the sum of the squares of the first n positive integers:
1^2 + 2^2 + 3^2 + ... + n^2 = (n(n + 1)(2n + 1)) / 6
Perhaps you have seen this formula before. Putting this formula into our expression for Rn, we get
Rn = 40/n^3 * (n(n + 1)(2n + 1)) / 6
= (40(n + 1)(2n + 1)) / 6n^2
Thus we have
lim n->infinity Rn = lim n->infinity (40(n + 1)(2n + 1)) / 6n^2 = lim n->infinity 40/6 * ((n + 1) / n) * ((2n + 1) / n)
= lim n->infinity 40/6 * (1 + 1/n) * (2 + 1/n)
= 40/6 * 1 * 2 = 40/3