Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

$4 \text{ sq. units}$

The correct answer is Option (2) → $4 \text{ sq. units}$

Required area $= \int\limits_{0}^{2\pi} |\sin x| \, dx = \int\limits_{0}^{\pi} \sin x \, dx + \left| \int\limits_{\pi}^{2\pi} \sin x \, dx \right|$

$= -[\cos x]_{0}^{\pi} + |[-\cos x]_{\pi}^{2\pi}|$

$= -[\cos \pi - \cos 0] + |-\cos 2\pi + \cos \pi|$

$= -[-1 - 1] + |-1 - 1|$

$= 2 + 2 = 4 \text{ sq. units}$