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 - Page 329: 111

Answer

$$F\left( x \right) = - \cos x + 3x + 3 - 3\pi $$

Work Step by Step

$$\eqalign{ & F''\left( x \right) = \cos x,{\text{ }}F'\left( 0 \right) = 3,{\text{ }}F\left( \pi \right) = 4,{\text{ }} \cr & F'\left( x \right) = \int {F''\left( x \right)} dx \cr & {\text{Substitute }}F''\left( x \right) \cr & F'\left( x \right) = \int {\cos xdx} \cr & F'\left( x \right) = \sin x + C \cr & {\text{Use the initial condition }}F'\left( 0 \right) = 3 \cr & 3 = \sin \left( 0 \right) + C \cr & C = 3,{\text{ then}} \cr & F'\left( x \right) = \sin x + 3 \cr & \cr & F\left( x \right) = \int {F'\left( x \right)} dx \cr & {\text{Substitute }}F'\left( x \right) \cr & F\left( x \right) = \int {\left( {\sin x + 3} \right)dx} \cr & F\left( x \right) = - \cos x + 3x + C \cr & {\text{Use the initial condition }}F\left( \pi \right) = 4 \cr & 4 = - \cos \left( \pi \right) + 3\left( \pi \right) + C \cr & 4 = 1 + 3\pi + C \cr & C = 3 - 3\pi ,{\text{ then}} \cr & F\left( x \right) = - \cos x + 3x + 3 - 3\pi \cr} $$
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