Question

Find the area bounded by the curve $y=\sin x,0\le x\le \pi$ and line $y=\frac{1}{\sqrt{2}}$.

Answer 1

Given:

$y=\sin x$

$y=\frac{1}{\sqrt{2}}$

$\therefore \sin x=\frac{1}{\sqrt{2}}$

$\therefore x=\frac{\pi }{4},\frac{3\pi }{4}$

$\therefore Area={\int }_{\pi /4}^{3\pi /4}(\sin x-\frac{1}{\sqrt{2}})dx$

$={[-\cos x-\frac{x}{\sqrt{2}}]}_{\pi /4}^{3\pi /4}$

$=(\frac{1}{\sqrt{2}}-\frac{3\pi }{4\sqrt{2}})-(-\frac{1}{\sqrt{2}}-\frac{\pi }{4\sqrt{2}})$

$=\frac{4-\pi }{2\sqrt{2}}$