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