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 309: 14


a) 6 b) 6

Work Step by Step

a) $\int^{3}_{-1}h(r)dr= \int^{1}_{-1}h(r)dr+\int^{3}_{1}h(r)dr$ $⇒ 6=0+\int^{3}_{1}h(r)dr$ $⇒ \int^{3}_{1}h(r)dr=6$ b) $-\int ^{1}_{3}h(u)du=\int^{3}_{1}h(u)du $ $= \int^{3}_{1}h(r)dr=6$
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