I figured it out, sorry for the bother!

hi guys can you please help me with this:

Given f(x)=(x+1)^2

Find the average value of f on [-1,1].

is this it:

u=x+1

du=1dx

$\displaystyle \frac{1}{2} [\frac{u^3}{3}]^{2}_{0}$

$\displaystyle \frac{1}{2}[\frac{(2)^3}{3} - \frac{(0)^3}{3}] $

$\displaystyle \frac{1}{2}[\frac{8}{3}] $

4/3