Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 25

Answer

$\displaystyle\int\limits_\frac{\pi}{6}^\pi \sin\theta d\theta=\dfrac{1+\sqrt{3}}{2}$

Work Step by Step

$\displaystyle\int\limits_\frac{\pi}{6}^\pi \sin\theta d\theta$ Evaluate the integral and apply the second part of the fundamental theorem of calculus: $\displaystyle\int\limits_\frac{\pi}{6}^\pi \sin\theta d\theta=-\cos\theta\Big|_\frac{\pi}{6}^\pi=-(\cos\pi-\cos\dfrac{\pi}{6})=-(-1-\dfrac{\sqrt{3}}{2})=...$ $...=\dfrac{\sqrt{3}}{2}+1=\dfrac{1+\sqrt{3}}{2}$
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