## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises - Page 480: 24

#### Answer

$$2$$

#### Work Step by Step

\eqalign{ & \int_{ - 2}^2 {\frac{{{e^{z/2}}}}{{{e^{z/2}} + 1}}} dz \cr & {\text{substitute }}u = {e^{z/2}} + 1,{\text{ }}du = \frac{1}{2}{e^{z/2}} \cr & {\text{express the limits in terms of }}u \cr & x = 2{\text{ implies }}u = {e^{2/2}} + 1 = e + 1 \cr & x = - 2{\text{ implies }}u = {e^{ - 2/2}} + 1 = {e^{ - 1}} + 1 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{ - 2}^2 {\frac{{{e^{z/2}}}}{{{e^{z/2}} + 1}}} dz = 2\int_{{e^{ - 1}} + 1}^{e + 1} {\frac{1}{u}} du \cr & {\text{find the antiderivative}} \cr & = 2\left. {\ln u} \right|_{{e^{ - 1}} + 1}^{e + 1} \cr & {\text{use the fundamental theorem}} \cr & = 2\left( {\ln \left( {e + 1} \right) - \ln \left( {{e^{ - 1}} + 1} \right)} \right) \cr & {\text{by logarithmic properties}} \cr & = 2\ln \left( {\frac{{e + 1}}{{{e^{ - 1}} + 1}}} \right) \cr & = 2\ln \left( {\frac{{e + 1}}{{\frac{{1 + e}}{e}}}} \right) \cr & = 2\ln \left( e \right) \cr & = 2 \cr}

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