Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises - Page 521: 42

Answer

$$V = \pi \left( {\ln 4 - 1} \right)$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {e^{ - x}},{\text{ }}g\left( x \right) = 0{\text{ on the interval }}\left[ {0,\ln 2} \right],{\text{ revolved}} \cr & {\text{about the line }}x = \ln 2. \cr & {\text{The volume can be calculated by the Shell Method, then}} \cr & V = \int_0^{\ln 2} {\underbrace {2\pi \left( {\ln 2 - x} \right)}_{{\text{shell circumference}}}\underbrace {{e^{ - x}}}_{{\text{height}}}dx} \cr & V = 2\pi \int_0^{\ln 2} {\left( {\ln 2 - x} \right){e^{ - x}}} dx \cr & {\text{Distribute}} \cr & V = 2\pi \int_0^{\ln 2} {\left( {\left( {\ln 2} \right){e^{ - x}} - x{e^{ - x}}} \right)} dx \cr & V = 2\pi \int_0^{\ln 2} {\left( {\left( {\ln 2} \right){e^{ - x}}} \right)} dx - 2\pi \int_0^{\ln 2} {x{e^{ - x}}} dx \cr & V = 2\pi \ln 2\int_0^{\ln 2} {{e^{ - x}}dx} - 2\pi \int_0^{\ln 2} {x{e^{ - x}}} dx \cr & {\text{Integrating }}\int {x{e^{ - x}}dx{\text{ by parts we obtain }}} - x{e^{ - x}} - {e^{ - x}},{\text{ then}} \cr & V = 2\pi \ln 2\left[ { - {e^{ - x}}} \right]_0^{\ln 2} - 2\pi \left[ { - x{e^{ - x}} - {e^{ - x}}} \right]_0^{\ln 2} \cr & V = - 2\pi \ln 2\left[ {{e^{ - x}}} \right]_0^{\ln 2} + 2\pi \left[ {x{e^{ - x}} + {e^{ - x}}} \right]_0^{\ln 2} \cr & V = - 2\pi \ln 2\left[ {{e^{ - \ln 2}} - {e^0}} \right] + 2\pi \left[ {\ln 2{e^{ - \ln 2}} + {e^{ - \ln 2}}} \right] \cr & - 2\pi \left[ {0{e^0} + {e^0}} \right] \cr & V = - 2\pi \ln 2\left( { - \frac{1}{2}} \right) + 2\pi \left[ {\frac{1}{2}\ln 2 + \frac{1}{2}} \right] - 2\pi \cr & V = \pi \ln 2 + \pi \ln 2 + \pi - 2\pi \cr & V = 2\pi \ln 2 - \pi \cr & {\text{Factoring}} \cr & V = \pi \left( {2\ln 2 - 1} \right) \cr & V = \pi \left( {\ln 4 - 1} \right) \cr} $$
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