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: 6

Answer

$$F(x) = \frac{1}{3}x^3-5x^2+25x+C$$

Work Step by Step

Given $$f(x) =(x-5)^2=x^2-10x+25$$ Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$ \begin{align*} F(x) &=\frac{1}{3}x^3-\frac{10}{2}x^2+25x+C\\ &= \frac{1}{3}x^3-5x^2+25x+C \end{align*} To check \begin{align*} F'(x) &=x^2-10x+25\\ &=f(x) \end{align*}
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