## Calculus (3rd Edition)

$\dfrac{9}{4}$
The area $(A)$ can be calculated as: $A=\int_{\pi/3}^{\pi} [\sin x -\sin 2x] \ dx \\= [-\cos x+\dfrac{\cos 2x}{2}]_{\pi/3}^{\pi}\\=-[\cos \pi-\cos (\pi/3)] +\dfrac{1}{2} (\cos 2 (\pi-\dfrac{\pi}{3})]\\= 1+\dfrac{1}{2}-(-\dfrac{1}{2}-\dfrac{1}{4}) \\=\dfrac{9}{4}$