Question

Let P = {0,1/4,1/2,3/4,1} be a partition of [0, 1]

and a function f : [0, 1] —> R be defined by

f(x) = x³ . Find U(P, f).

Answer 1

Given P = [0, 1/4, 1/2, 3/4, 1]. So $\delta x = 1/4$ .

U(P,f) = $[ f(\frac{1}{4})+f(\frac{1}{2})+f(\frac{3}{4})+f(1)].\delta x = [(1/4)^3 + (1/2)^3+(3/4)^3+1^3] (\frac{1}{4})$

= 1.5625/4 = 0.390625.