Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 14 - Multiple Integration - 14.2 Exercises - Page 984: 55

Answer

$${\left( {e - 1} \right)^2}$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = {e^{x + y}} \cr & {\text{From the image of the rectangle shown below we obtain}} \cr & {\text{the region }}R \cr & R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 1,{\text{ }}x \leqslant y \leqslant 1} \right\} \cr & {\text{The area }}A{\text{ of the region is:}} \cr & A = \frac{1}{2}\left( 1 \right)\left( 1 \right) \cr & A = \frac{1}{2} \cr & {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr & {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr & {\text{Average value}} = \frac{1}{{1/2}}\int_0^1 {\int_x^1 {{e^{x + y}}} dydx} \cr & = 2\int_0^1 {\left[ {\int_x^1 {{e^{x + y}}} dy} \right]dx} \cr & {\text{Integrate with respect to }}y \cr & = 2\int_0^1 {\left[ {{e^{x + y}}} \right]_x^1dx} \cr & = 2\int_0^1 {\left( {{e^{x + 1}} - {e^{2x}}} \right)dx} \cr & {\text{Integrate}} \cr & = 2\left[ {{e^{x + 1}} - \frac{1}{2}{e^{2x}}} \right]_0^1 \cr & = 2\left[ {{e^{1 + 1}} - \frac{1}{2}{e^2}} \right] - 2\left[ {{e^{0 + 1}} - \frac{1}{2}{e^0}} \right] \cr & = 2{e^2} - {e^2} - 2\left( {e - \frac{1}{2}} \right) \cr & = {e^2} - 2e + 1 \cr & = {\left( {e - 1} \right)^2} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.