(Solved): EXAMPLE 2 For the region under f(x) = x2 on [0, 1], show that the sum of the areas of the upper ap ...

EXAMPLE 2 For the region under f(x) = x2 on [0, 1], show that the sum of the areas of the upper approximating rectangle approaches 1/3, that is 1 lim R = 3 = SOLUTION RIs the sum of the areas of the n rectangles in the figure. Each rectangle has width 1/n and the helghts are the values of the function f(x) = x at the points 1/n, 2/n, 3/n,...,n/n; that is, the heights are (1/n), , (3/n)-..... (n/n)". Thus, X 1 1 1 2 1 3 Rn=-(-) +-(-) +-(-) + --... +-(-)2 nn ? 1 1 =- n (12 + 2 + 3 + ... + n) (1 + 2 + 32 + ... + n) Here we need the formula for the sum of the squares of the first n positive integers: n(n + 1)(2n + 1) 1 + 2 + 3 + ... + n2 = 6 Perhaps you have seen this formula before. Putting this formula into our expression for Rar we get 1 Rr = =- n (n + 1)(2n + 1) 6n? Thus we have (n + 1) (2n +1) lim R, = lim 6n2 1 n = lim no 6 27 ("+-) (2^+) 2n +1 -- (1 + (2 + ) X X 1 lim 6 1 1 =_.1.2= 6 3