Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 419: 55

Answer

$\frac{45}{28}$

Work Step by Step

$\int_{0}^{1}\sqrt[3] {1+7x}~dx$ Let $u = 1+7x$ $\frac{du}{dx} = 7$ $dx = \frac{du}{7}$ When $x = 0$, then $u = 1$ When $x = 1$, then $u = 8$ $\int_{1}^{8}\frac{1}{7}\sqrt[3] u~du$ $= \frac{1}{7}~[\frac{3}{4}(u)^{4/3}\vert_{1}^{8}~]$ $= \frac{3}{28}~[(u)^{4/3}\vert_{1}^{8}~]$ $= \frac{3}{28}~[(8)^{4/3}-(1)^{4/3}]$ $= \frac{3}{28}~(16-1)$ $= \frac{45}{28}$
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