Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 356: 27


$f(x)=\displaystyle \frac{1}{3}x^3+3e^{x}+Cx+D$

Work Step by Step

$\left[\begin{array}{cc} \text{}f'' & \text{particular antiderivative,}f'\\ 2x^{1} & 2\cdot\dfrac{x^{2}}{2}=x^{2} \\ 3e^{x} & 3\cdot e^{x} \end{array}\right]$ $f'(x)=x^{2}+3e^{x}+C$ $\left[\begin{array}{cc} \text{}f' & \text{particular antiderivative,}f\\ x^{2} & \dfrac{x^{3}}{3}\\ 3e^{x} & 3\cdot e^{x} \\ C & Cx \end{array}\right]$ $f(x)=\displaystyle \frac{1}{3}x^3+3e^{x}+Cx+D$
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