Question

The area bounded by the curves $y=cos x$ and $y=sin 2x,\forall x\in \left[\frac{\pi }{6} , \frac{\pi }{2}\right]$ is equal to

• A. $\frac{\pi }{2}$ sq. units
• B. $\frac{\pi }{3}$ sq. units
• C. $\frac{\pi }{6}$ sq. units
• D. $\frac{1}{4}$ sq. units

Answer 1

Answer: (D)

$A=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\sin 2 x-\cos x) d x=\left[-\frac{\cos 2 x}{2}-\sin x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$
$=\left[\frac{\cos 2 x}{2}+\sin x\right]_{\frac{\pi}{2}}^{\frac{\pi}{6}}$
$=\frac{1}{2} \times \frac{1}{2}+\frac{1}{2}+\frac{1}{2}-1=\frac{1}{4}$ sq. units