University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 21



Work Step by Step

$$A=\int^1_0(4y-y^2+4y^3+1)^{-2/3}(12y^2-2y+4)dy$$ We set $u=4y-y^2+4y^3+1$, which means $$du=(4-2y+12y)dy=(12y-2y+4)dy$$ For $y=1$, we have $$u=4-1+4+1=8$$ For $y=0$, we have $$u=0-0+0+1=1$$ Therefore, $$A=\int^8_1u^{-2/3}du$$ $$A=3u^{1/3}\Big]^8_1=3\sqrt[3]u\Big]^8_1$$ $$A=3(\sqrt[3]8-\sqrt[3]1)$$ $$A=3(2-1)=3$$
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