The integral from 0 to 2 of (x+1)^2 dx when n=6?
I have no idea what n = 6 has to do with the problem ... ignoring that:
∫02(x + 1)2dx = ∫02(x2 + 2x + 1)dx = [x3/3 + x2 + x]02 = [23/3 + 22 + 2] - [03/3 + 02 + 0]
= 8/3 + 4 + 2 - 0 - 0 - 0 = 26/3
