University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 74



Work Step by Step

$$A=\int^1_0(8s^3-12s^2+5)ds$$ $$A=\Big(\frac{8s^4}{4}-\frac{12s^3}{3}+5s\Big)\Big]^1_0$$ $$A=\Big(2s^4-4s^3+5s\Big)\Big]^1_0$$ $$A=(2-4+5)-0$$ $$A=3$$
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.