Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 46



Work Step by Step

Integrate with respect to s by taking the antiderivative of each term. Then plug in limits to evaluate: $=\frac{8s^4}{4}-\frac{12s^3}{3}+5s=2s^4-4s^3+5s$ with limits from 0 to 1 $=[2(1)^4-4(1)^3+5(1)]-[2(0)^4-4(0)^3+5(0)]$ $=(2-4+5)-0=3$
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.