Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises: 15

Answer

$-76$

Work Step by Step

$\int^9_{1}(\frac{\sqrt u-2u^2}{u})du=\int^9_{1}(\frac{1}{\sqrt u}-2u)du$ $=[2\sqrt u-u^2]^9_{1}$. Sub in the bounds and subtract the lower bound (1) from the upper bound (9): $=[2\sqrt {(9)}-(9)^2]-[2\sqrt {(1)}-(1)^2] $ $=6-81-(2-1)$ $=-76 $.
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