Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 14


a)6 b)6

Work Step by Step

a)$\int_1^3h(r)dr=\int_{-1}^3h(r)dr -\int_{-1}^1 h(r)dr =6-0=6$ b)$-\int_3^1h(u) du=-[-\int_1^3h(u)du]=\int_1^3h(u) du=6$
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