Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 27

Answer

$$\frac{{99}}{{10\ln 10}}$$

Work Step by Step

$$\eqalign{ & \int_{ - 1}^1 {{{10}^x}} dx \cr & {\text{by the formula }}\int {{a^x}} dx = \frac{{{a^x}}}{{\ln a}} + C \cr & {\text{letting }}a = 10 \cr & \int_{ - 1}^1 {{{10}^x}} dx = \left. {\left( {\frac{{{{10}^x}}}{{\ln 10}}} \right)} \right|_{ - 1}^1 \cr & = \left. {\frac{1}{{\ln 10}}\left( {{{10}^x}} \right)} \right|_{ - 1}^1 \cr & {\text{use the fundamental theorem}} \cr & = \frac{1}{{\ln 10}}\left( {{{10}^1} - {{10}^{ - 1}}} \right) \cr & = \frac{1}{{\ln 10}}\left( {\frac{{99}}{{10}}} \right) \cr & = \frac{{99}}{{10\ln 10}} \cr} $$
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