Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises: 6

Answer

$$ \frac{3\pi^{2}}{8} -1 $$
1517711694

Work Step by Step

The area enclosed between two curves $y=f(x)$ and $y = g(x)$ over $[ a,b]$ such that $f(x) \geq g(x)$ is $$ A = \int_{a}^{b} f(x) - g(x)dx$$ From the graph, we can see that $x\gt \sin x$ on the interval $\bigg[\frac{\pi}{2}, \pi \bigg]$ Therefore, the Area between the two curve is $\int_{\frac{\pi}{2}}^{\pi} x - \sin (x) dx$ $$=\left[\frac{x^{2}}{2}+ \cos (x) \right]^{\pi}_\frac{\pi}{2}$$ $$= \bigg[ \frac{\pi^{2}}{2} + \cos (\pi) \bigg] - \bigg[\frac{\pi^{2}}{8} + \cos (\frac{\pi}{2})\bigg]$$ $$ = \bigg[ \frac{\pi^{2}}{2} -1 \bigg] - \bigg[\frac{\pi^{2}}{8} + 0 \bigg]$$ $$ = \frac{\pi^{2}}{2} - \frac{\pi^{2}}{8} -1 $$ $$= \frac{3\pi^{2}}{8} -1 $$
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