Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 13



Work Step by Step

$\int^{0}_{\frac{\pi}{2}}\frac{1+cos2t}{2}dt$ =$\int^{0}_{\frac{\pi}{2}}(\frac{1}{2}+\frac{1}{2}cos2t)dt$ =$[\frac{1}{2}t+\frac{1}{4}sin2t]^0_{\frac{\pi}{2}}$ =$(\frac{1}{2}0+\frac{1}{4}sin2(0))-(\frac{1}{2}(\frac{\pi}{2})+\frac{1}{4}sin2(\frac{\pi}{2}))$ =$-\frac{\pi}{4}$
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