Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 11


$\int(4-18x)dx= 4x-9x^2 + C$

Work Step by Step

$\int(4-18x)dx= 4x-18\times\frac{1}{2}x^2 + C= 4x-9x^2 + C$
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.