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: 14


$$\int y^2(4-y^3)^{2/3}dy=-\frac{1}{5}\sqrt[3]{(4-y^3)^5}+C$$

Work Step by Step

$$A=\int y^2(4-y^3)^{2/3}dy$$ Let $u=4-y^3$ Then $du=-3y^2dy$. So $y^2dy=-\frac{1}{3}du$ Substitute into $A$, we have $$A=-\frac{1}{3}\int u^{2/3}du$$ $$A=-\frac{1}{3}\frac{u^{5/3}}{\frac{5}{3}}+C$$ $$A=-\frac{\sqrt[3]{u^5}}{5}+C$$ $$A=-\frac{1}{5}\sqrt[3]{(4-y^3)^5}+C$$
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