Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.9 Antiderivatives - 3.9 Exercises - Page 282: 5


$$F(x)= 6x^2+\frac{1}{3}x^3+C$$

Work Step by Step

Given $$f(x) =x(12+x)=12x+x^2$$ Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$ \begin{align*} F(x) &=\frac{12x^2}{2}+\frac{x^3}{3}+C\\ &= 6x^2+\frac{1}{3}x^3+C \end{align*} To check \begin{align*} F'(x) &=12x+x^2\\ &=f(x) \end{align*}
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