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: 52


$\frac{\pi\cdot b^3}{3}$

Work Step by Step

$\int_{0}^{b} \pi x^2dx=[ \frac{\pi\cdot x^3}{3}]_{0}^{b}=\frac{\pi\cdot b^3}{3}-\frac{\pi\cdot 0^3}{3}=\frac{\pi\cdot b^3}{3}-0=\frac{\pi\cdot b^3}{3}$
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