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: 8

Answer

$F(x) = \frac{x^{4.4}}{4.4} - \frac{2}{\sqrt 2}x^{\sqrt 2} + C$

Work Step by Step

$f(x) = x^{3.4} - 2x^{\sqrt 2 -1}$ $F(x) = \frac{x^{4.4}}{4.4} - 2(\frac{x^{\sqrt 2 -1+1}}{\sqrt 2 -1+1})$ $F(x) = \frac{x^{4.4}}{4.4} - 2(\frac{x^{\sqrt 2}}{\sqrt 2})$ $F(x) = \frac{x^{4.4}}{4.4} - \frac{2}{\sqrt 2}x^{\sqrt 2}$ Check: $F(x) = \frac{1}{4.4}x^{4.4} - \frac{2}{\sqrt 2}x^{\sqrt 2}$ $F'(x) = \frac{4.4}{4.4}x^{4.4-1}- \frac{2\sqrt 2}{\sqrt 2}x^{\sqrt 2-1}$ $F'(x) = 1x^{3.4}- \frac{2}{1}x^{\sqrt 2-1}$ $F'(x) = x^{3.4}- 2x^{\sqrt 2-1}$
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