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: 34

Answer

$0.009$

Work Step by Step

Because $0\lt0.3$, by Equation (4): $\int_{0}^{0.3} s^2 ds=\frac{(0.3)^3}{3}-\frac{0^3}{3}=\frac{0.027}{3}-\frac{0}{3}=\frac{0.027}{3}=0.009$
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