University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 20



Work Step by Step

$$A=\int3y\sqrt{7-3y^2}dy$$ We set $u=7-3y^2$. Then $$du=-6ydy=-2\times3ydy$$ That means, $$3ydy=-\frac{1}{2}du$$ Therefore, $$A=-\frac{1}{2}\int\sqrt udu=-\frac{1}{2}\int u^{1/2}du$$ $$A=-\frac{1}{2}\times\frac{u^{3/2}}{\frac{3}{2}}+C=-\frac{u^{3/2}}{3}+C$$ $$A=-\frac{(7-3y^2)^{3/2}}{3}+C$$
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