## University Calculus: Early Transcendentals (3rd Edition)

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Since $av(f)=\dfrac{1}{B-A}\int_{A}^{B} f(x) dx$ Here, we have $av(f)=\dfrac{1}{\sqrt 3-0}\int_{0}^{\sqrt 3} (x^2-1) dx$ This implies that $\dfrac{1}{\sqrt 3-0}[\dfrac{(\sqrt 3)^3-(0)^3}{3}-(\sqrt 3-0)]=0$