Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 17


$$\int(4x^3-2x^2)dx= x^4-\frac{2}{3}x^3+c $$

Work Step by Step

We have $$\int(4x^3-2x^2)dx=\frac{4}{4}x^4-\frac{2}{3}x^3+c= x^4-\frac{2}{3}x^3+c.$$
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