Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 847: 45

Answer

$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 y\sqrt {1 + xy} {\rm{d}}y{\rm{d}}x \simeq 0.575161$

Work Step by Step

We use Fubini's Theorem to change the order of integration and re-write the integral: $I = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 y\sqrt {1 + xy} {\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 0}^1 y\sqrt {1 + xy} {\rm{d}}x} \right){\rm{d}}y$ We evaluate first the inner integral with respect to $x$. $S\left( y \right) = \mathop \smallint \limits_{x = 0}^1 y\sqrt {1 + xy} {\rm{d}}x$ The antiderivative of $y\sqrt {1 + xy} $ with respect to $x$ is $\frac{2}{3}{\left( {1 + xy} \right)^{3/2}}$ because $\frac{\partial }{{\partial x}}\left( {\frac{2}{3}{{\left( {1 + xy} \right)}^{3/2}}} \right) = y\sqrt {1 + xy} $ So, $S\left( y \right) = \mathop \smallint \limits_{x = 0}^1 y\sqrt {1 + xy} {\rm{d}}x = \frac{2}{3}{\left( {1 + xy} \right)^{3/2}}|_{x = 0}^1$ $ = \frac{2}{3}{\left( {1 + y} \right)^{3/2}} - \frac{2}{3}$ Thus, $I = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 0}^1 y\sqrt {1 + xy} {\rm{d}}x} \right){\rm{d}}y$ $ = \mathop \smallint \limits_{y = 0}^1 \left( {\frac{2}{3}{{\left( {1 + y} \right)}^{3/2}} - \frac{2}{3}} \right){\rm{d}}y$ $ = \frac{2}{3}\mathop \smallint \limits_{y = 0}^1 {\left( {1 + y} \right)^{3/2}}{\rm{d}}y - \frac{2}{3}\mathop \smallint \limits_{y = 0}^1 {\rm{d}}y$ $ = \frac{4}{{15}}{\left( {1 + y} \right)^{5/2}}|_0^1 - \frac{2}{3}$ $ = \frac{{16}}{{15}}\sqrt 2 - \frac{4}{{15}} - \frac{2}{3}$ $ = \frac{{16}}{{15}}\sqrt 2 - \frac{{14}}{{15}} \simeq 0.575161$ So, $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 y\sqrt {1 + xy} {\rm{d}}y{\rm{d}}x \simeq 0.575161$.
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