Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 5 - Accumulating Change: Limits of Sums and the Definite Integral - 5.9 Activities - Page 409: 22

Answer

$$\left( {\bf{a}} \right)\frac{{\ln \left( {{2^x} + 2} \right)}}{{\ln 2}} + C,\,\,\,\,\left( {\bf{b}} \right)\frac{1}{{\ln 2}}\ln \left( {43} \right)$$

Work Step by Step

$$\eqalign{ & \int_2^8 {\frac{{{2^x}}}{{{2^x} + 2}}} dx \cr & \cr & \left( {\bf{a}} \right){\text{write the general antiderivative}} \cr & {\text{use substitution method}} \cr & u = {2^x} + 2,\,\,\,\,du = {2^x}\ln 2dx,\,\,\,\,dx = \frac{{du}}{{{2^x}\ln 2}} \cr & {\text{write in terms of }}u \cr & \int {\frac{{{2^x}}}{{{2^x} + 2}}} dx = \int {\frac{{{2^x}}}{u}} \left( {\frac{{du}}{{{2^x}\ln 2}}} \right) \cr & = \int {\frac{1}{u}} \left( {\frac{{du}}{{\ln 2}}} \right) \cr & = \frac{1}{{\ln 2}}\int {\frac{1}{u}} du \cr & {\text{integrating}} \cr & = \frac{1}{{\ln 2}}\ln \left| u \right| + C \cr & {\text{write in terms of }}x \cr & = \frac{{\ln \left| {{2^x} + 2} \right|}}{{\ln 2}} + C \cr & = \frac{{\ln \left( {{2^x} + 2} \right)}}{{\ln 2}} + C \cr & \cr & \left( {\bf{b}} \right){\text{evaluate the definite integral}} \cr & \int_2^8 {\frac{{{2^x}}}{{{2^x} + 2}}} dx = \left( {\frac{{\ln \left( {{2^x} + 2} \right)}}{{\ln 2}}} \right)_2^8 \cr & = \frac{{\ln \left( {{2^8} + 2} \right)}}{{\ln 2}} - \frac{{\ln \left( {{2^2} + 2} \right)}}{{\ln 2}} \cr & {\text{simplifying}} \cr & = \frac{1}{{\ln 2}}\ln \left( {258} \right) - \frac{1}{{\ln 2}}\ln \left( 6 \right) \cr & = \frac{1}{{\ln 2}}\left( {\ln \left( {258} \right) - \ln \left( 6 \right)} \right) \cr & = \frac{1}{{\ln 2}}\ln \left( {\frac{{258}}{6}} \right) \cr & = \frac{1}{{\ln 2}}\ln \left( {43} \right) \cr} $$
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