University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 271: 9

Answer

a) $x^{\frac{2}{3}}+C$ b) $x^{\frac{1}{3}}+C$ c) $x^{\frac{-1}{3}}+C$

Work Step by Step

a) The anti-derivative is: $\dfrac{2}{3}\dfrac{x^{-1/3+1}}{-1/3+1}+C=x^{\frac{2}{3}}+C$ b) The anti-derivative is: $\dfrac{1}{3}\dfrac{x^{-2/3+1}}{-2/3+1}+C=x^{\frac{1}{3}}+C$ c) The anti-derivative is: $\dfrac{-1}{3}\dfrac{x^{-4/3+1}}{-4/3+1}+C= x^{\frac{-1}{3}}+C$
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