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: 61

Answer

$$\sec v + 1$$

Work Step by Step

$$\eqalign{ & f\left( v \right) = \sec v\tan v \cr & {\text{find an antiderivative of }}f\left( t \right) \cr & F\left( v \right) = \sec v + C \cr & {\text{using the initial condition }}F\left( 0 \right) = 2 \cr & 2 = \sec 0 + C \cr & 2 = 1 + C \cr & C = 1 \cr & {\text{so}}{\text{,}} \cr & = \sec v + 1 \cr} $$
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