Answer
1
Work Step by Step
av(f)=$(\frac{1}{3-0})\int^{3}_{0}(t-1)^2dt$
=$\frac{1}{3}\int^{3}_{0}t^2dt-\frac{2}{3}\int^{3}_{0}tdt+\frac{1}{3}\int^{3}_{0}1dt$
=$\frac{1}{3}(\frac{3^3}{3})-\frac{2}{3}(\frac{(3)^2}{2}-\frac{0}{2})+\frac{1}{3}(3-0)$
=1
