Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 245: 44



Work Step by Step

$\int^{5}_{0}(2f(x)-\frac{1}{3}g(x))dx=2\int^{5}_{0}f(x)dx-\frac{1}{3}\int^{5}_{0}g(x)dx $ $=2\times5-\frac{1}{3}\times12=10-4=6$
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