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: 11

Answer

a) 5 b) $5\sqrt 3$ c) -5 d) -5

Work Step by Step

a) $\int^{2}_{1}f(u)du=\int^{2}_{1}f(x)dx=5$ b) $\int^{2}_{1}\sqrt {3}f(z)dz=\int^{2}_{1}\sqrt {3}f(x)dx $ $=\sqrt {3}\int^{2}_{1}f(x)dx=5\sqrt {3}$ c) $\int^{1}_{2}f(t)dt=\int^{1}_{2}f(x)dx $ $=-\int^{2}_{1}f(x)dx=-5$ d) $ \int^{2}_{1}[-f(x)]dx=-\int^{2}_{1}f(x)dx=-5$
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