Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 86

Answer

$$\frac{{24}}{{\ln 5}}$$

Work Step by Step

$$\eqalign{ & \int_{ - 2}^0 {{5^{ - \theta }}} d\theta \cr & {\text{use substitution}}{\text{. Let }}u = - \theta,{\text{ so that }}du = - d\theta \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ }}u = \cr & \,\,\,\,\,\,{\text{If }}\theta = - 2,{\text{ }}u = 2 \cr & {\text{Then}} \cr & \int_{ - 2}^0 {{5^{ - \theta }}} d\theta = \int_2^0 {{5^u}} \left( { - du} \right) \cr & = \int_0^2 {{5^u}} du \cr & {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr & = \left( {\frac{{{5^u}}}{{\ln 5}}} \right)_0^2 \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = \frac{{{5^2} - {5^0}}}{{\ln 5}} \cr & {\text{simplifying}} \cr & = \frac{{25 - 1}}{{\ln 5}} \cr & = \frac{{24}}{{\ln 5}} \cr} $$
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