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: 5



Work Step by Step

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