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: 15

Answer

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

Work Step by Step

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