Calculus with Applications (10th Edition)

Published by Pearson

Chapter 7 - Integration - Chapter Review - Review Exercises - Page 418: 49

Answer

$$3\ln 5 + \frac{{12}}{{25}}$$

Work Step by Step

\eqalign{ & \int_1^5 {\left( {3{x^{ - 1}} + {x^{ - 3}}} \right)} dx \cr & {\text{ integrate by using }}\int_a^b {{x^n}dx} = \left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right)_a^b{\text{ and }}\int_a^b {\frac{1}{x}} dx = \left. {\ln \left| x \right|} \right|_a^b + C \cr & = \left( {3\ln \left| x \right| + \frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right)_1^5 \cr & = \left( {3\ln \left| x \right| + \frac{{{x^{ - 2}}}}{{ - 2}}} \right)_1^5 \cr & = \left( {3\ln \left| x \right| - \frac{1}{{2{x^2}}}} \right)_1^5 \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 388}}} \right) \cr & = \left( {3\ln \left| 5 \right| - \frac{1}{{2{{\left( 5 \right)}^2}}}} \right) - \left( {3\ln \left| 1 \right| - \frac{1}{{2{{\left( 1 \right)}^2}}}} \right) \cr & {\text{simplifying}} \cr & = 3\ln 5 - \frac{1}{{50}} + \frac{1}{2} \cr & = 3\ln 5 + \frac{{12}}{{25}} \cr}

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