Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.2 Setting Up Integrals: Volume, Density, Average Value - Exercises - Page 298: 45



Work Step by Step

The average is given as follows $$ \frac{1}{3-(-1)} \int_{-1}^{3}\left(2 x^{3}-6 x^{2}\right) d x=\frac{1}{4} \int_{-1}^{3}\left(2 x^{3}-6 x^{2}\right) d x=\left.\frac{1}{4}\left(\frac{1}{2} x^{4}-2 x^{3}\right)\right|_{-1} ^{3}=\frac{1}{4}\left(-\frac{27}{2}-\frac{5}{2}\right)=-4 $$
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