Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 25

Answer

$$\sqrt 2.$$

Work Step by Step

We have $$\int_{\pi / 4}^{3 \pi / 4} \sin \theta d \theta= -\cos\theta |_{\pi / 4}^{3 \pi / 4}\\ = -\cos (3 \pi / 4)+ \cos (\pi / 4)=\sqrt 2 /2 +\sqrt 2/2=\sqrt 2.$$
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