Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 40

Answer

$= \ln \frac{1}{3}=-\ln{3}$

Work Step by Step

$\int^{3}_1 \frac{y^{3}-2y^{2}-y}{y^{2}} dy$ $= \int^{3}_1 \frac{y^{3}}{y^{2}}-\frac{2y^{2}}{y^{2}}-\frac{y}{y^{2}}dy$ $= \int^{3}_1 y-2-\frac{1}{y}dy$ $= \int^{3}_1 y-2-y^{-1}dy$ $= \frac{y^{2}}{2} - 2y - \ln |y| |^{3}_1$ $= [\frac{3^{2}}{2} - 2(3) - \ln |3|] - [\frac{1^{2}}{2} - 2(1) - \ln |1|]$ $= [\frac{9}{2} - 6 - \ln 3] - [\frac{1}{2} - 2 - \ln 1]$ $= [-\frac{3}{2} - \ln 3] - [-\frac{3}{2} - \ln 1]$ $= -\frac{3}{2} - \ln 3 +\frac{3}{2} + \ln 1$ $= - \ln 3 + \ln 1$ $= \ln 1 - \ln 3$ $= \ln \frac{1}{3}$
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