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: 10


$$\int^{1}_{0}\frac{x^3}{\sqrt{x^4+9}}dx=\frac{1}{2}(\sqrt{10}-3)$$ $$\int^{0}_{-1}\frac{x^3}{\sqrt{x^4+9}}dx=-\frac{1}{2}(\sqrt{10}-3)$$

Work Step by Step

Part a) $$A=\int^{1}_{0}\frac{x^3}{\sqrt{x^4+9}}dx$$ Set $u=\sqrt{x^4+9}$, we then have $$du=\frac{(x^4+9)'}{2\sqrt{x^4+9}}dx=\frac{4x^3dx}{2\sqrt{x^4+9}}=\frac{2x^3dx}{\sqrt{x^4+9}}$$ $$\frac{x^3dx}{\sqrt{x^4+9}}=\frac{1}{2}du$$ For $x=0$, we have $u=3$ and for $x=1$, we have $u=\sqrt{10}$. Therefore, $$A=\frac{1}{2}\int^{\sqrt10}_3du=\frac{1}{2}u\Big]^{\sqrt10}_3$$ $$A=\frac{1}{2}(\sqrt{10}-3)$$ Part b) $$A=\int_{-1}^{0}\frac{x^3}{\sqrt{x^4+9}}dx$$ We carry out a similar substitution as in part a). For $x=0$, we have $u=3$ and for $x=-1$, we have $u=\sqrt{10}$. Therefore, $$A=\frac{1}{2}\int_{\sqrt10}^3du=-\frac{1}{2}\int^{\sqrt{10}}_3du=-\frac{1}{2}(\sqrt{10}-3)$$
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