# Thread: Average value of integrals?

1. ## Average value of integrals?

I figured it out, sorry for the bother!

Given f(x)=(x+1)^2
Find the average value of f on [-1,1].

is this it:

u=x+1
du=1dx

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

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

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

4/3