University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Practice Exercises - Page 278: 106


$-\dfrac{5}{8} (2-x)^{8/5}+c$

Work Step by Step

Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Substitute $k=2-x \implies dx =-dk$ Thus, $k^{3/5} du=[\dfrac{k^{3/5+1}}{3/5+1}]+c$ or, $=-\dfrac{5}{8} k^{8/5}+c$ plug back in: $k=2-x$, we have $=-\dfrac{5}{8} (2-x)^{8/5}+c$
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