Using the midpoint rule, find the area with two rectangles and 4 rectangles:

Problem: f(x) = x^3 btwn x = 0 and x = 1

My Process:

i) Two rectangles:

Δ x = (b-a)/n = (1-0)/2 = 1/2

A ˜ 1/2 [ f (1/4) + f (3/4) ]

A ˜ 1/2 [ (1/4)^3 + (3/4)^3 ] = . 21875

ii) Four rectangles:

Δ x = (b-a)/n = (1-0)/4 = 1/4

A ˜ 1/4 [ f (1/8) + f (3/8) + f (5/8) + f (7/8) ]

A ˜ 1/4 [ (1/8)^3 + (3/8)^3 + (5/8)^3 + (7/8)^3 ] = .2422

Is this right because when I tried doing the same process for the next problem, we got a different answer.