Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - True-False Quiz - Page 421: 18



Work Step by Step

Consider $f(x)=\frac{|x|}{x}$ Then $\int _{-1}^{1} f(x) dx=\int _{-1}^{1} \frac{|x|}{x}dx$ $\int _{-1}^{1} \frac{|x|}{x}dx=\int _{-1}^{0}(-1)dx+\int _{0}^{1} (1)dx$ $=(-x) _{0}^{1} +(x) _{0}^{1} $ $=-1+1$ $=0$ Here, the function $f(x)=\frac{|x|}{x}$ has only one jump discontinuity , that is at $x=0$
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