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: 48

Answer

7

Work Step by Step

$\int^{1}_{\frac{1}{2}}24u^{2}du= 24\int^{1}_{\frac{1}{2}}u^{2}du$ $=24\times[\frac{1^{3}}{3}-\frac{(\frac{1}{2})^{3}}{3}]$$= 24(\frac{1}{3}-\frac{1}{24})$ $=24\times\frac{7}{24}= 7$
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.