Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 32

Answer

$f\left( x \right) = {x^3} - \frac{{{x^6}}}{{30}} + \frac{{{x^7}}}{{14}} + {C_1}x + {C_2}$

Work Step by Step

$$\eqalign{ & f''\left( x \right) = 6x - {x^4} + 3{x^5} \cr & {\text{Find the general antiderivative }}f'\left( x \right) \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right) \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr & {\text{we obtain}} \cr & f'\left( x \right) = 6\left( {\frac{{{x^2}}}{2}} \right) - \frac{{{x^5}}}{5} + 3\left( {\frac{{{x^6}}}{6}} \right) + {C_1} \cr & f'\left( x \right) = 3{x^2} - \frac{{{x^5}}}{5} + \frac{{{x^6}}}{2} + {C_1} \cr & {\text{Find the general antiderivative }}f\left( x \right) \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right) \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr & {\text{we obtain}} \cr & f\left( x \right) = 3\left( {\frac{{{x^3}}}{3}} \right) - \frac{{{x^6}}}{{5\left( 6 \right)}} + \frac{{{x^7}}}{{2\left( 7 \right)}} + {C_1}x + {C_2} \cr & {\text{Simplifying}} \cr & f\left( x \right) = {x^3} - \frac{{{x^6}}}{{30}} + \frac{{{x^7}}}{{14}} + {C_1}x + {C_2} \cr} $$
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