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: 71

Answer

$$f\left( u \right) = 4\sin u + 2\cos 2u - 3$$

Work Step by Step

$$\eqalign{ & f'\left( u \right) = 4\left( {\cos u - \sin 2u} \right) \cr & {\text{multiply}} \cr & f'\left( u \right) = 4\cos u - 4\sin 2u \cr & f\left( u \right) = \int {f'\left( u \right)} du \cr & then \cr & f\left( u \right) = \int {\left( {4\cos u - 4\sin 2u} \right)} du \cr & find{\text{ the general solution}} \cr & f\left( u \right) = 4\sin u + 2\cos 2u + C \cr & {\text{using the initial condition }}f\left( {\pi /6} \right) = 0 \cr & 0 = 4\sin \left( {\pi /6} \right) + 2\cos 2\left( {\pi /6} \right) + C \cr & 0 = 4\left( {1/2} \right) + 2\left( {1/2} \right) + C \cr & 0 = 2 + 1 + C \cr & C = - 3 \cr & {\text{the solution to the initial value problem is}} \cr & f\left( u \right) = 4\sin u + 2\cos 2u - 3 \cr} $$
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