Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.5 The Area Between Two Curves - 7.5 Exercises - Page 405: 24

Answer

$$A = - 4{e^2} + {e^4} - 1 + {e^3} + e$$

Work Step by Step

$$\eqalign{ & x = 0,\,\,\,\,\,x = 3,\,\,\,\,\,y = {e^x},\,\,\,\,\,\,y = {e^{4 - x}} \cr & {\text{Find the points of intersection of the graphs}} \cr & {e^x} = {e^{4 - x}} \cr & \ln {e^x} = \ln {e^{4 - x}} \cr & x = 4 - x \cr & 2x = 4 \cr & x = 2 \cr & {\text{Then with the given points and the critical value }}x = 0{\text{ we have the intervals}} \cr & \left( {0,2} \right){\text{ and }}\left( {2,3} \right) \cr & {\text{For the interval }}\left( {0,2} \right){\text{ we have that }}{e^{4 - x}} > {e^x}{\text{ and }} \cr & {\text{For the interval }}\left( {2,3} \right){\text{ }}{e^x} > {e^{4 - x}} \cr & \cr & {\text{The area is given by}}: \cr & A = \int_0^2 {\left( {{e^{4 - x}} - {e^x}} \right)} dx + \int_2^3 {\left( {{e^x} - {e^{4 - x}}} \right)} dx \cr & {\text{Integrating}}{\text{, use }}\int {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C \cr & A = \left( { - {e^{4 - x}} - {e^x}} \right)_0^2 + \left( {{e^x} - \left( { - {e^{4 - x}}} \right)} \right)_2^3 \cr & A = \left( { - {e^{4 - x}} - {e^x}} \right)_0^2 + \left( {{e^x} + {e^{4 - x}}} \right)_2^3 \cr & A = \left( { - {e^{4 - 2}} - {e^2}} \right) - \left( { - {e^{4 - 0}} - {e^0}} \right) + \left( {{e^3} + {e^{4 - 3}}} \right) - \left( {{e^2} + {e^{4 - 2}}} \right) \cr & A = - {e^2} - {e^2} + {e^4} - 1 + {e^3} + e - {e^2} - {e^2} \cr & A = - 4{e^2} + {e^4} - 1 + {e^3} + e \cr} $$
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