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: 43


$$4x^5 - \frac{9}{4}x^4-x^2+c$$

Work Step by Step

\begin{align*} \int\left(20 x^{4}-9 x^{3}-2 x\right) d x&= 20\times \frac{1}{5}x^5- 9\times \frac{1}{4}x^4 -x^2+c\\ &= 4x^5 - \frac{9}{4}x^4-x^2+c \end{align*}
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