University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 309: 39

Answer

$\frac{b}{3}$

Work Step by Step

Using equation (4), we get $\int^{\sqrt[3] b}_{0}x^{2}dx=\frac{(\sqrt[3] b)^{3}}{3}-\frac{0^{3}}{3}= \frac{b}{3}$
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