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: 5



Work Step by Step

Consider $f(x)=\int_{-1}^{1} \dfrac{1}{x^{2/3}} dx$ or, $f(x)=\int_{-1}^{0} \dfrac{1}{x^{2/3}} dx +\int_{0}^{1} \dfrac{1}{x^{2/3}} dx$ Thus, $\lim\limits_{a \to 0^{-}} f(x)= \lim\limits_{a \to 0^{-}} \int_{-1}^{a} \dfrac{1}{x^{2/3}} dx+ \lim\limits_{a \to 0^{+}} \int_{0}^{1} \dfrac{1}{x^{2/3}} dx=\lim\limits_{a \to 0^{-}} [3a^{1/3} -(-3)]+\lim\limits_{a \to 0^{+}} [3-3a^{1/3} ]$ or, $\lim\limits_{a \to 0^{-}} [3a^{1/3} +3]+\lim\limits_{a \to 0^{+}} [3-3a^{1/3} = 3+3=6$
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