Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 69

Answer

$$g\left( x \right) = \frac{{7{x^8}}}{8} - \frac{{{x^2}}}{2} + \frac{{13}}{8}$$

Work Step by Step

$$\eqalign{ & g'\left( x \right) = 7x\left( {{x^6} - \frac{1}{7}} \right) \cr & {\text{multiply}} \cr & g'\left( x \right) = 7{x^7} - x \cr & g\left( x \right) = \int {g'\left( x \right)} dx \cr & then \cr & g\left( x \right) = \int {\left( {7{x^7} - x} \right)} dx \cr & find{\text{ the general solution}} \cr & g\left( x \right) = \frac{{7{x^8}}}{8} - \frac{{{x^2}}}{2} + C \cr & {\text{using the initial condition }}g\left( 1 \right) = 2 \cr & 2 = \frac{{7{{\left( 1 \right)}^8}}}{8} - \frac{{{{\left( 1 \right)}^2}}}{2} + C \cr & 2 = \frac{3}{8} + C \cr & C = \frac{{13}}{8} \cr & {\text{the solution to the initial value problem is}} \cr & g\left( x \right) = \frac{{7{x^8}}}{8} - \frac{{{x^2}}}{2} + \frac{{13}}{8} \cr} $$
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