University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 278: 98



Work Step by Step

Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Now, $\int (8t^3-\dfrac{t^2}{2}+t) dt=\dfrac{8t^{3+1}}{3+1}- \dfrac{t^{2+1}}{2+1}+\dfrac{t^2}{2}+c$ or, $=\dfrac{8t^4}{4}+\dfrac{t^3}{6}+\dfrac{t^2}{2}+c$ or, $=2t^4+\dfrac{t^3}{6}+\dfrac{t^2}{2}+c$
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.