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: 26

Answer

$f(x) = \frac{1}{56}x^{8} - \frac{2}{15}x^{6} + \frac{1}{6}x^{3} + \frac{1}{2}x^{2} + Cx+D$

Work Step by Step

$f''(x) = x^{6} -4x^{4} + x+ 1$ $f'(x) = \frac{x^{7}}{7} - 4(\frac{x^{5}}{5}) + \frac{x^{2}}{2} + x + C$ $f'(x) = \frac{1}{7}x^{7} - \frac{4}{5}x^{5} + \frac{1}{2}x^{2} + x +C$ $f'(x) = \frac{1}{7}x^{7} - \frac{4}{5}x^{5} + \frac{1}{2}x^{2} + x +C$ $f(x) = \frac{1}{7}(\frac{x^{8}}{8}) - \frac{4}{5}(\frac{x^{6}}{6}) + \frac{1}{2}(\frac{x^{3}}{3}) + \frac{x^{2}}{2} + Cx + D$ $f(x) = \frac{1}{56}x^{8} - \frac{4}{30}x^{6} + \frac{1}{6}x^{3} + \frac{1}{2}x^{2} + Cx+D$ $f(x) = \frac{1}{56}x^{8} - \frac{2}{15}x^{6} + \frac{1}{6}x^{3} + \frac{1}{2}x^{2} + Cx+D$
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