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: 56



Work Step by Step

Since $av(f)=\dfrac{1}{B-A}\int_{A}^{B} f(x) dx$ Here, we have $av(f)=\dfrac{1}{ 3-0}\int_{0}^{3} (-x^2/2) dx$ This implies that $\dfrac{-1}{6}[\dfrac{(3)^3-(0)^3}{3}]=\dfrac{-3}{2}$
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