Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.8 - Improper Integrals - Exercises 8.8 - Page 501: 6



Work Step by Step

Let $f(x)=\int_{-8}^{1} \dfrac{1}{x^{1/3}} dx$ This can be re-written as: $f(x)=\int_{-8}^{0} \dfrac{1}{x^{(1/3)}} dx +\int_{0}^{1} \dfrac{1}{x^{(1/3)}} dx$ Now, $\lim\limits_{k \to 0^{-}}\int_{-8}^{0} \dfrac{1}{x^{(1/3)}} dx +\lim\limits_{k \to 0^{+}}\int_{0}^{1} \dfrac{1}{x^{(1/3)}} dx=\lim\limits_{k \to 0^{-}} [\dfrac{3}{2}(k)^{2/3}-6]+\lim\limits_{k \to 0^{+}} [(\dfrac{3}{2})-(\dfrac{3}{2})k^{(2/3)}]$ Thus, $\int_{-8}^{1} \dfrac{1}{x^{(1/3)}} dx=-6+\dfrac{3}{2}= \dfrac{-9}{2}$
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