Area of the region bounded by the curve $y = \sin x$ and x-axis between $x =\frac{π}{2}$ and $x=\frac{3π}{2}$ is

2 sq. units

The correct answer is Option (4) → 2 sq. units

$\text{Area}=\int_{\pi/2}^{\pi}\sin x\,dx-\int_{\pi}^{3\pi/2}\sin x\,dx$

$=\left[-\cos x\right]_{\pi/2}^{\pi}-\left[-\cos x\right]_{\pi}^{3\pi/2}$

$=\big(-\cos\pi+ \cos\frac{\pi}{2}\big)-\big(-\cos\frac{3\pi}{2}+\cos\pi\big)$

$=(1+0)-\big(0-(-1)\big)=1-(-1)=2$

$\text{Area}=2\ \text{sq. units}$