Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 57

Answer

$$A = \ln 2$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{1}{x}{\text{ on the interval }}\left[ { - 2,1} \right] \cr & {\text{so}}{\text{, the area is}} \cr & A = \int_{ - 2}^1 {\left( {\frac{1}{x}} \right)} dx \cr & {\text{using the logarithmic rule for integration}} \cr & A = \left. {\left( {\ln \left| x \right|} \right)} \right|_{ - 2}^1 \cr & {\text{Using The Fundamental Theorem}} \cr & A = \left( {\ln \left| 1 \right|} \right) - \left( {\ln \left| { - 2} \right|} \right) \cr & {\text{Simplify}} \cr & A = \left( 0 \right) - \left( {\ln 2} \right) \cr & A = - \ln 2 \cr & {\text{The area is under the x axis}}{\text{, then the absolute area is}} \cr & \left| A \right| = \left| { - \ln 2} \right| \cr & \left| A \right| = \ln 2 \cr & A = \ln 2 \cr} $$
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