University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 31



Work Step by Step

Here, we can see that $f(x)=\dfrac{1}{|\sqrt x|}$ is not defined at $0$. However, $f(x)=\dfrac{1}{|\sqrt x|}=\dfrac{1}{-\sqrt x}$ for all $x \in [-1,0)$ and $f(x)=\dfrac{1}{|\sqrt x|}=\dfrac{1}{\sqrt x}$ for all $x \in (0,4]$ Thus, we have $\int_{-1}^{0} \dfrac{1}{|\sqrt x|}+\int_{0}^{4} \dfrac{1}{|\sqrt x|}=\lim\limits_{p \to 0^{-}} \int_{-1}^{p} \dfrac{1}{|\sqrt x|}+\lim\limits_{q \to 0^{+}} \int_{q}^{4} \dfrac{1}{|\sqrt x|}$ This implies that $\lim\limits_{p \to 0^{-}} (-2 \sqrt{-p}+2)+\lim\limits_{q \to 0^{+}} (4-2\sqrt q)=2+4=6$
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