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: 30

Answer

$$\frac{{900}}{{\ln 10}}$$

Work Step by Step

$$\eqalign{ & \int_{1/3}^{1/2} {\frac{{{{10}^{1/p}}}}{{{p^2}}}} dp \cr & {\text{substitute }}u = \frac{1}{p},{\text{ }}du = - \frac{1}{{{p^2}}}dp \cr & {\text{express the limits in terms of }}u \cr & x = 1/2{\text{ implies }}u = 1/\left( {1/2} \right) = 2 \cr & x = 1/3{\text{ implies }}u = 1/\left( {1/3} \right) = 3 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{1/3}^{1/2} {\frac{{{{10}^{1/p}}}}{{{p^2}}}} dp = - \int_3^2 {{{10}^u}du} \cr & {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr & {\text{letting }}a = 10 \cr & = - \left. {\left( {\frac{{{{10}^u}}}{{\ln 10}}} \right)} \right|_3^2 \cr & {\text{use the fundamental theorem}} \cr & = - \frac{1}{{\ln 10}}\left( {{{10}^2} - {{10}^3}} \right) \cr & {\text{simplify}} \cr & = \frac{{900}}{{\ln 10}} \cr} $$
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