Area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $2$ sq units
Required area enclosed by the curve $y = \cos x, x = 0$ and $x = \pi$ is
$= \int_{0}^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{\pi} \cos x \, dx \right| \quad [∵\int \cos x \, dx = \sin x]$
$= [\sin x]_{0}^{\pi/2} + |[\sin x]_{\pi/2}^{\pi}|$
$= \left[ \sin \frac{\pi}{2} - \sin 0 \right] + \left| \sin \pi - \sin \frac{\pi}{2} \right| = 1 + 1 = 2 \text{ sq. units}$