Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.8 Improper Integrals - 7.8 Exercises - Page 574: 7

Answer

Divergent

Work Step by Step

Let \[I=\int_{-\infty}^{0}\frac{1}{3-4x}dx\;\;\;\ldots(1)\] \[I=\lim_{t\rightarrow -\infty}\int_{t}^{0}\frac{1}{3-4x}dx\;\;\;\ldots(2)\] \[I=\lim_{t\rightarrow -\infty}\left[-\frac{1}{4}\ln|3-4x|\right]_{t}^{0}\] \[I=\lim_{t\rightarrow -\infty}\left[-\frac{1}{4}\ln 3+\frac{1}{4}\ln|3-4t|\right]\] \[I=\infty\] Since limit on R.H.S. of (2) does not exist So given improper integral (1) is divergent.
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