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: 6


$$\int^{\sqrt7}_0t(t^2+1)^{1/3}dt=\frac{45}{8}$$ $$\int_{-\sqrt7}^0t(t^2+1)^{1/3}dt=-\frac{45}{8}$$

Work Step by Step

Part a) $$A=\int^{\sqrt7}_0t(t^2+1)^{1/3}dt$$ Set $u=t^2+1$, we then have $$du=2tdt$$ $$tdt=\frac{1}{2}du$$ For $t=0$, we have $u=1$ and for $t=\sqrt7$, we have $u=8$. Therefore, $$A=\frac{1}{2}\int^{8}_1u^{1/3}du$$ $$A=\frac{1}{2}\times\frac{3u^{4/3}}{4}\Big]^{8}_1=\frac{3u^{4/3}}{8}\Big]^{8}_1$$ $$A=\frac{3}{8}(8^{4/3}-1)$$ $$A=\frac{3}{8}(16-1)=\frac{45}{8}$$ Part b) $$A=\int_{-\sqrt7}^0t(t^2+1)^{1/3}dt$$ We carry out a similar substitution as in part a). For $t=0$, we have $u=1$ and for $t=-\sqrt7$, we have $u=8$. Therefore, $$A=\frac{1}{2}\int^{1}_8u^{1/3}du$$ $$A=-\frac{1}{2}\int^{8}_1u^{1/3}du=-\frac{45}{8}$$
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