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: 44

Answer

We evaluate $I = \mathop \smallint \limits_1^3 \mathop \smallint \limits_0^1 y{{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x$ using two methods. Method 1. Hold $x$ constant and evaluate the inner integral with respect to $y$ Method 2. Change the order of integration: hold $y$ constant and integrate first with respect to $x$ The results are the same: $I = \mathop \smallint \limits_1^3 \mathop \smallint \limits_0^1 y{{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x \simeq 4.64356$. Using Fubini's Theorem to change the order of integration (Method 2) is easier.

Work Step by Step

Evaluate $I = \mathop \smallint \limits_1^3 \mathop \smallint \limits_0^1 y{{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x$. Method 1. Hold $x$ constant and evaluate the inner integral with respect to $y$ $I = \mathop \smallint \limits_{x = 1}^3 \left( {\mathop \smallint \limits_{y = 0}^1 y{{\rm{e}}^{xy}}{\rm{d}}y} \right){\rm{d}}x$ Write the inner integral: $S\left( x \right) = \mathop \smallint \limits_{y = 0}^1 y{{\rm{e}}^{xy}}{\rm{d}}y$ Let $u=y$ and $v = {{\rm{e}}^{xy}}$. So, $du = dy$ and $dv = x{{\rm{e}}^{xy}}dy$. Thus, $S\left( x \right) = \frac{1}{x}\mathop \smallint \limits_{y = 0}^1 u{\rm{d}}v$ Recall the Integration by Parts of Section 8.1, $\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$ Using the Integration by Parts, we obtain $S\left( x \right) = \frac{1}{x}\left( {uv|_{y = 0}^1 - \mathop \smallint \limits_{y = 0}^1 v{\rm{d}}u} \right)$ $ = \frac{1}{x}\left( {y{{\rm{e}}^{xy}}|_{y = 0}^1 - \mathop \smallint \limits_{y = 0}^1 {{\rm{e}}^{xy}}{\rm{d}}y} \right)$ $ = \frac{1}{x}\left( {{{\rm{e}}^x} - \frac{1}{x}{{\rm{e}}^{xy}}|_{y = 0}^1} \right)$ $ = \frac{1}{x}\left( {{{\rm{e}}^x} - \frac{1}{x}\left( {{{\rm{e}}^x} - 1} \right)} \right)$ $ = {{\rm{e}}^x}{x^{ - 1}} - {{\rm{e}}^x}{x^{ - 2}} + {x^{ - 2}}$ So, $I = \mathop \smallint \limits_{x = 1}^3 \left( {\mathop \smallint \limits_{y = 0}^1 y{{\rm{e}}^{xy}}{\rm{d}}y} \right){\rm{d}}x$ $ = \mathop \smallint \limits_{x = 1}^3 \left( {{{\rm{e}}^x}{x^{ - 1}} - {{\rm{e}}^x}{x^{ - 2}} + {x^{ - 2}}} \right){\rm{d}}x$ $ = \mathop \smallint \limits_{x = 1}^3 \left( {{{\rm{e}}^x}{x^{ - 1}} - {{\rm{e}}^x}{x^{ - 2}}} \right){\rm{d}}x + \mathop \smallint \limits_1^3 {x^{ - 2}}{\rm{d}}x$ Using the given formula: $\smallint {{\rm{e}}^x}\left( {{x^{ - 1}} - {x^{ - 2}}} \right){\rm{d}}x = {x^{ - 1}}{{\rm{e}}^x} + C$ yields $I = {x^{ - 1}}{{\rm{e}}^x}|_1^3 - {x^{ - 1}}|_1^3$ $ = \frac{1}{3}{{\rm{e}}^3} - {\rm{e}} - \frac{1}{3} + 1$ $ = \frac{1}{3}{{\rm{e}}^3} - {\rm{e}} + \frac{2}{3}$ $ \simeq 4.64356$ So, $I = \mathop \smallint \limits_1^3 \mathop \smallint \limits_0^1 y{{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x \simeq 4.64356$. Method 2. Change the order of integration: hold $y$ constant and integrate first with respect to $x$ Using Fubini's Theorem we change the order of integration and re-write the integral $I = \mathop \smallint \limits_{x = 1}^3 \mathop \smallint \limits_{y = 0}^1 y{{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x$ $I = \mathop \smallint \limits_{y = 0}^1 \left( {\mathop \smallint \limits_{x = 1}^3 y{{\rm{e}}^{xy}}{\rm{d}}x} \right){\rm{d}}y$ $ = \mathop \smallint \limits_{y = 0}^1 \left( {{{\rm{e}}^{xy}}|_1^3} \right){\rm{d}}y$ $ = \mathop \smallint \limits_{y = 0}^1 \left( {{{\rm{e}}^{3y}} - {{\rm{e}}^y}} \right){\rm{d}}y$ $ = \left( {\frac{1}{3}{{\rm{e}}^{3y}} - {{\rm{e}}^y}} \right)|_0^1$ $ = \frac{1}{3}{{\rm{e}}^3} - {\rm{e}} - \frac{1}{3} + 1$ $ = \frac{1}{3}{{\rm{e}}^3} - {\rm{e}} + \frac{2}{3}$ $ \simeq 4.64356$ So, $I = \mathop \smallint \limits_1^3 \mathop \smallint \limits_0^1 y{{\rm{e}}^{xy}}{\rm{d}}y{\rm{d}}x \simeq 4.64356$. Using Fubini's Theorem to change the order of integration method is easier.
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