University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 79



Work Step by Step

$$A=\int^{1}_0\frac{36dx}{(2x+1)^3}$$ We set $a=2x+1$, which means $$da=2dx$$ $$dx=\frac{1}{2}da$$ For $x=0$, $a=1$ and for $x=1$, $a=3$ Therefore, $$A=\frac{1}{2}\int^{3}_1\frac{36}{a^3}da=18\int^{3}_1a^{-3}da$$ $$A=\Big(\frac{18a^{-2}}{-2}\Big)\Big]^{3}_1=\Big(-\frac{9}{a^2}\Big)\Big]^{3}_1$$ $$A=-1-(-9)=-1+9$$ $$A=8$$
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