Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 415: 33

Answer

$20 + \ln 3$

Work Step by Step

$$\eqalign{ & \int_1^3 {\left( {\frac{{3{x^2} + 4x + 1}}{x}} \right)} dx \cr & {\text{Distribute the integrand}} \cr & = \int_1^3 {\left( {3x + 4 + \frac{1}{x}} \right)} dx \cr & {\text{Using the basic integration rules}} \cr & = \left[ {\frac{{3{x^2}}}{2} + 4x + \ln \left| x \right|} \right]_1^3 \cr & {\text{Using the fundamental theorem of calculus}}{\text{, part 2}} \cr & = \left[ {\frac{{3{{\left( 3 \right)}^2}}}{2} + 4\left( 3 \right) + \ln \left| 3 \right|} \right] - \left[ {\frac{{3{{\left( 1 \right)}^2}}}{2} + 4\left( 1 \right) + \ln \left| 1 \right|} \right] \cr & {\text{Simplifying}} \cr & = \left[ {\frac{{27}}{2} + 12 + \ln 3} \right] - \left[ {\frac{3}{2} + 4 + \ln \left| 1 \right|} \right] \cr & = \frac{{27}}{2} + 12 + \ln 3 - \frac{3}{2} - 4 \cr & = 20 + \ln 3 \cr} $$
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