Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 14 - Section 14.5 - Improper Integrals and Applications - Exercises - Page 1059: 11

Answer

$$3{\left( 5 \right)^{2/3}}$$

Work Step by Step

$$\eqalign{ & \int_0^5 {\frac{2}{{{x^{1/3}}}}} dx \cr & {\text{The integrand }}\frac{2}{{{x^{1/3}}}}{\text{ is not continuous at }}x = 0,{\text{ then by the}} \cr & {\text{definition of improper integrals}} \cr & \int_0^5 {\frac{2}{{{x^{1/3}}}}} dx = \mathop {\lim }\limits_{r \to {0^ + }} \int_r^5 {\frac{2}{{{x^{1/3}}}}} dx \cr & = 2\mathop {\lim }\limits_{r \to {0^ + }} \int_r^5 {{x^{ - 1/3}}} dx \cr & {\text{Integrating }} \cr & = 2\mathop {\lim }\limits_{r \to {0^ + }} \left[ {\frac{{{x^{2/3}}}}{{2/3}}} \right]_r^5 \cr & = 3\mathop {\lim }\limits_{r \to {0^ + }} \left[ {{x^{2/3}}} \right]_r^5 \cr & = 3\mathop {\lim }\limits_{r \to {0^ + }} \left[ {{{\left( 5 \right)}^{2/3}} - {r^{2/3}}} \right] \cr & {\text{Evaluate the limit when }}r \to {0^ + } \cr & = 3\left[ {{{\left( 5 \right)}^{2/3}} - {{\left( {{0^ + }} \right)}^{2/3}}} \right] \cr & = 3{\left( 5 \right)^{2/3}} \cr} $$
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