Answer
$\dfrac{3}{2}$
Work Step by Step
We know that $av(f)=\dfrac{1}{b-a}\int_{a}^{b} f(x) dx$
Therefore,
$av(f)=\dfrac{1}{1-(-2)} \int_{-2}^{1} (t^2-t) \ dt \\=(\dfrac{1}{3}) \int_{-2}^1 (t^2) \ dt-\dfrac{1}{3} \int_{-2}^1 t \ dt \\=(\dfrac{1}{3}) \times [\dfrac{t^3}{3}]_{-2}^1-(\dfrac{1}{3}) \times [\dfrac{t^2}{2}]_{-2}^1 \\=(\dfrac{1}{3}) \times [\dfrac{1}{3}+\dfrac{8}{3}-\dfrac{1}{2}+\dfrac{4}{2}]\\=\dfrac{3}{2}$
