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: 6



Work Step by Step

Consider $f(x)=\int_{-8}^{1} \dfrac{1}{x^{1/3}} dx$ or, $f(x)=\int_{-8}^{0} \dfrac{1}{x^{1/3}} dx +\int_{0}^{1} \dfrac{1}{x^{1/3}} dx$ Thus, $\lim\limits_{a \to 0^{-}}\int_{-8}^{0} \dfrac{1}{x^{1/3}} dx +\lim\limits_{a \to 0^{+}}\int_{0}^{1} \dfrac{1}{x^{1/3}} dx=\lim\limits_{a \to 0^{-}} [\dfrac{3}{2}a^{2/3}-6]+\lim\limits_{a \to 0^{+}} [\dfrac{3}{2}-\dfrac{3}{2}a^{2/3}]$ or, $=-6+\dfrac{3}{2}$ or, $= \dfrac{-9}{2}$
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