University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 2



Work Step by Step

Solve $\int_{-1}^{0}\dfrac{3dx}{3x-2}$ $\int_{-1}^{0}\dfrac{3dx}{3x-2}=[\ln |3x-2|]_{-1}^{0}$ This implies $[\ln |3x-2|]_{-1}^{0}=\ln |3(0)-2|-\ln |3(-1)-2|$ $=\ln |-2|-\ln |-3-2|$ (simplify) $=\ln 2-\ln 5$ Use logarithmic properties, $\ln m-\ln n=\ln(\dfrac{m}{n})$, we have Hence, $\int_{-1}^{0}\dfrac{3dx}{3x-2}=\ln(\dfrac{2}{5})$
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