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: 34



Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.