University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 28



Work Step by Step

Since, we have $\int (\dfrac{t^2}{2}+4t^3) dt=\dfrac{t^3}{6}+t^4+C$ and $\dfrac{d}{dt}(\dfrac{t^3}{6}+t^4+C)=\dfrac{t^2}{2}+4t^3$
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.