Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 39

Answer

$$1$$

Work Step by Step

$$\eqalign{ & \int_0^{\pi /4} {{{\sec }^2}\theta } d\theta \cr & {\text{recall that }}\int {{{\sec }^2}\theta } d\theta = \tan \theta + C \cr & \int_0^{\pi /4} {{{\sec }^2}\theta } d\theta = \left. {\left( {\tan \theta } \right)} \right|_0^{\pi /4} \cr & {\text{Using The Fundamental Theorem}} \cr & = \left( {\tan \frac{\pi }{4}} \right) - \left( {\tan 0} \right) \cr & {\text{simplify}} \cr & = \left( 1 \right) - \left( 0 \right) \cr & = 1 \cr} $$
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