Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.8 Improper Integrals - 7.8 Exercises - Page 578: 35

Answer

$$6$$

Work Step by Step

$$\eqalign{ & \int_0^8 {\frac{{dx}}{{\root 3 \of x }}} \cr & {\text{The integrand is not defined for }}x = 0,{\text{ then}} \cr & \int_0^8 {\frac{{dx}}{{\root 3 \of x }}} = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^8 {\frac{{dx}}{{\root 3 \of x }}} \cr & {\text{Integrating}} \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{{{x^{2/3}}}}{{2/3}}} \right]_a^8 \cr & = \frac{3}{2}\mathop {\lim }\limits_{a \to {0^ + }} \left[ {{x^{2/3}}} \right]_a^8 \cr & = \frac{3}{2}\mathop {\lim }\limits_{a \to {0^ + }} \left[ {{8^{2/3}} - {a^{2/3}}} \right] \cr & {\text{Evaluating the limit}} \cr & = \frac{3}{2}\left[ {4 - {{\left( 0 \right)}^{2/3}}} \right] \cr & = 6 \cr} $$
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