Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 303: 10


$a.\quad \displaystyle \frac{\sqrt{10}-3}{2}$ $ b.\quad \displaystyle \frac{3-\sqrt{10}}{2}$

Work Step by Step

$a.$ Applying th.7, let $u=g(x)=x^{4}+9$ (a continuous function). Then, $\displaystyle \quad du=g'(x)dx=4x^{3}dx,\qquad x^{3}dx=\frac{1}{4}\cdot du$ and, the borders change to $g(0)=9$ and $g(1)=10$. $\displaystyle \int_{0}^{1}\frac{x^{3}}{\sqrt{x^{4}+9}}dx=\int_{9}^{10 }\frac{\frac{1}{4}\cdot du}{u^{1/2}}=\frac{1}{4}\int_{9}^{10 }u^{-1/2}du$ $=\displaystyle \frac{1}{4}\left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{9}^{10}$ $=\displaystyle \frac{1}{4}\left[ \frac{u^{1/2}}{1/2} \right]_{9}^{10}$ $=\displaystyle \frac{1}{4}\left[ 2u^{1/2} \right]_{9}^{10}$ $=\displaystyle \frac{1}{2}(\sqrt{10}-3)$ $=\displaystyle \frac{\sqrt{10}-3}{2}$ $b.$ Using the same substitution, $u=x^{4}+9,\ du=4x^{3}dx$, the borders here change to $g(-1)=10$ and $g(0)=9$. $\displaystyle \int_{-1}^{0}\frac{x^{3}}{\sqrt{x^{4}+9}}dx=\int_{10}^{9 }\frac{\frac{1}{4}\cdot du}{u^{1/2}}=\frac{1}{4}\int_{10}^{9 }u^{-1/2}du$ and, if we exchange the borders, we can apply the Order of Integration rule: $=-\displaystyle \left(\frac{1}{4}\int_{9}^{10 }u^{-1/2}du \right)$ We evaluated this same integral in part (a) $=-\displaystyle \frac{\sqrt{10}-3}{2}$ $=\displaystyle \frac{3-\sqrt{10}}{2}$
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