Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises - Page 251: 37



Work Step by Step

$h'(t)=8t^3+5\hspace{10mm}h(1)=-4$ $\int h'(t)dt=\int (8t^3+5)dt=2t^4+5t+C=h(t)$ $h(1)=2+5+C=7+C=-4$ $C=-11$ $h(t)=2t^4+5t-11$
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.