The area (in sq.units) of the region enclosed by the curve $y = \cos x,\frac{-\pi}{2}≤ x ≤\frac{\pi}{2}$ and the x-axis is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The area enclosed by the curve $y = \cos x$ and the x-axis from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$ is given by:
$\displaystyle \text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx$
Since $\cos x$ is an even function, the integral over symmetric limits becomes:
$= 2 \int_{0}^{\frac{\pi}{2}} \cos x \, dx$
$= 2 [\sin x]_{0}^{\frac{\pi}{2}}$
$= 2 (\sin \frac{\pi}{2} - \sin 0) = 2(1 - 0) = 2$