Answer
$1$
Work Step by Step
We know that $av(f)=\dfrac{1}{b-a}\int_{a}^{b} f(x) dx$
Therefore,
$av(f)=\dfrac{1}{3-0} \int_{0}^{3} (t-1)^2 \ dt \\=(\dfrac{1}{3}) \int_0^31 (t^2-2t+1) \ dt \\=(\dfrac{1}{3}) (\dfrac{3^0-0^3}{3}-(2) [\dfrac{3^2 -0^2}{2}]+1(3-0)] \\=\dfrac{1}{3} (9-9+3) \\=1$
