Area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is

$2$ sq units

The correct answer is Option (1) → $2$ sq units

Required area enclosed by the curve $y = \cos x, x = 0$ and $x = \pi$ is

$= \int_{0}^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx \right| \quad [∵\int \cos x \, dx = \sin x]$

$= [\sin x]_{0}^{\pi/2} + |[\sin x]_{\pi/2}^{\pi}|$

$= \left[ \sin \frac{\pi}{2} - \sin 0 \right] + \left| \sin \pi - \sin \frac{\pi}{2} \right| = 1 + 1 = 2 \text{ sq. units}$