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 271: 2


a) $3x^2+C$ b) $\dfrac{1}{8}x^8+C$ c) $\dfrac{1}{8}x^8-3x^2+8x+C$

Work Step by Step

a) The anti-derivative is: $\dfrac{6}{1+1}x^{1+1}+C=3x^2+C$ b) The anti-derivative is: $\dfrac{1}{7+1}x^{7+1}+C=\dfrac{1}{8}x^8+C$ c) The anti-derivative is: $\dfrac{1}{7+1}x^{7+1}-\dfrac{6}{1+1}x^{1+1}+(8)x^{0+1}+C=\dfrac{1}{8}x^8-3x^2+8x+C$
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.