Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 48

Answer

$$ \frac{7}{2}$$

Work Step by Step

Given $$\int_{\pi / 6}^{2 \pi} \cos x d x$$ Since \begin{aligned} \cos(x) &\geq 0\ \ \ \ \text{for }\ \ \pi/6\leq x\leq \pi/2 \\ \cos(x) &\leq 0\ \ \ \ \text{for }\ \ \pi/2\leq x\leq 3\pi/2\\ \cos(x) &\geq 0\ \ \ \ \text{for }\ \ 3\pi/2\leq x\leq 2\pi \end{aligned} Then area given by \begin{aligned} \text{Area}&= \int_{\pi/6}^{2\pi}\cos(x)dx\\ &= \int_{\pi/6}^{\pi/2}\cos(x)dx- \int_{\pi/2}^{3\pi/2}\cos(x)dx+ \int_{3\pi/2}^{2\pi}\cos(x)dx\\ &= \sin (x)\bigg|_{\pi/6}^{\pi/2} - \sin (x)\bigg|_{\pi/2}^{3\pi/2}+ \sin (x)\bigg|_{3\pi/2}^{2\pi}\\ &= \sin(\pi/2) -\sin(\pi/6) -[\sin(3\pi/2)-\sin(\pi/2)]+ \sin(2\pi)-\sin(3\pi/2)\\ &= \frac{7}{2} \end{aligned}
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