## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 34

#### Answer

$$= 2\ln 9 - 2\ln 4 + 2$$

#### Work Step by Step

\eqalign{ & \int_4^9 {\frac{{2 + \sqrt t }}{t}} dt \cr & {\text{split the numerator}} \cr & = \int_4^9 {\left( {\frac{2}{t} + \frac{{\sqrt t }}{t}} \right)} dt \cr & {\text{writting }}\sqrt t {\text{ as }}{t^{1/2}} \cr & = \int_4^9 {\left( {\frac{2}{t} + \frac{{{t^{1/2}}}}{t}} \right)} dt \cr & = \int_4^9 {\left( {\frac{2}{t} + {t^{ - 1/2}}} \right)} dt \cr & {\text{integrating using the logarithmic rule and the power rule for integration}} \cr & \int_4^9 {\left( {\frac{2}{t} + {t^{ - 1/2}}} \right)} dt = \left. {\left( {2\ln \left| t \right| + \frac{{{t^{1/2}}}}{{1/2}}} \right)} \right|_4^9 \cr & = \left. {\left( {2\ln \left| t \right| + 2{t^{1/2}}} \right)} \right|_4^9 \cr & {\text{using The Fundamental Theorem}} \cr & = \left( {2\ln \left| 9 \right| + 2{{\left( 9 \right)}^{1/2}}} \right) - \left( {2\ln \left| 4 \right| + 2{{\left( 4 \right)}^{1/2}}} \right) \cr & {\text{simplify}} \cr & = 2\ln 9 + 6 - 2\ln 4 - 4 \cr & = 2\ln 9 - 2\ln 4 + 2 \cr}

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