University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 310: 55



Work Step by Step

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 $\dfrac{1}{\sqrt 3-0}[\dfrac{(\sqrt 3)^3-(0)^3}{3}-(\sqrt 3-0)]=0$
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