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: 13


$$F(x) = \frac{ -5}{4x^8} +C$$

Work Step by Step

Given $$f(x) =\frac{10}{x^9}= 10x^{-9}$$ Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$ \begin{align*} F(x) &= \frac{ -10}{8}x^{ -8}+C\\ &= \frac{ -5}{4x^8} +C \end{align*} To check \begin{align*} F'(x) &= \frac{10}{x^9}\\ &=f(x) \end{align*}
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