Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - Exercises - Page 422: 16



Work Step by Step

$\int^1_{0}(\sqrt[4] u+1)^2du=\int^1_{0}(\sqrt u+2u^\frac{1}{4}+1)du$ $=[\frac{2}{3}u^\frac{3}{2}+\frac{8}{5}u^\frac{5}{4}+u] ^1_{0}$. Sub in the bounds and subtract the lower bound (0) from the upper bound (1): $= [\frac{2}{3}(1)^\frac{3}{2}+\frac{8}{5}(1)^\frac{5}{4}+(1)]-[\frac{2}{3}(0)^\frac{3}{2}+\frac{8}{5}(0)^\frac{5}{4}+(0)]$ $= \frac{2}{3}+\frac{8}{5}+1-0$ $=\frac{49}{15}$.
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