Question

The area bounded by the curve $y=\sin 2x+\cos 2x$ and the x-axis between the ordinate $x=0$, $x=\frac{\pi }{4}$ is

• A. $1$ sq.unit
• B. $\frac{1}{2}$ sq. unit
• C. $\frac{2}{3}$ sq. unit
• D. $0$ sq. units

Answer 1

The correct option is A $1$ sq.unit

Area $={\int }_{\pi /4}^0(\sin 2x+\cos 2x)dx$

$={[\frac{-\cos 2x}{2}+\frac{\sin 2x}{2}]}_0^{\pi /4}$

$=\left(\frac{-\cos \frac{\pi }{2}+\sin \frac{\pi }{2}}{2}\right)-\left(\frac{-\cos 0+\sin 0}{2}\right)$

$=-1\frac{1}{2}-\left(\frac{-1}{2}\right)$

$=1$