Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.6 The Fundamental Theorem Of Calculus - Exercises Set 4.6 - Page 320: 19

Answer

$$\sqrt 2 $$

Work Step by Step

$$\eqalign{ & \int_{ - \pi /4}^{\pi /4} {\cos x} dx \cr & {\text{find antiderivative use integration formulas from table 4}}{\text{.2}}{\text{.1}} \cr & \int_{ - \pi /4}^{\pi /4} {\cos x} dx = \left( {\sin x} \right)_{ - \pi /4}^{\pi /4} \cr & {\text{part 1 of fundamental theorem of calculus}} \cr & = \sin \left( {\frac{\pi }{4}} \right) - \sin \left( { - \frac{\pi }{4}} \right) \cr & {\text{recall sin}}\left( { - \theta } \right) = - \sin \theta \cr & = \sin \left( {\frac{\pi }{4}} \right) + \sin \left( {\frac{\pi }{4}} \right) \cr & = 2\sin \left( {\frac{\pi }{4}} \right) \cr & {\text{simplify}} \cr & = 2\left( {\frac{{\sqrt 2 }}{2}} \right) \cr & = \sqrt 2 \cr} $$
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