The area (in sq. units) of the region enclosed by the curve \(y = \cos x, \displaystyle\frac{-\pi}{2} \le x \le \frac{\pi}{2}\) and the x-axis is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
On \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), \(\cos x \ge 0\), so the curve lies on or above the x-axis and
\( \displaystyle \text{Area} = \int_{-\pi/2}^{\pi/2} \cos x\,dx \)
Since cos x is an even function,
\( \displaystyle = 2\int_{0}^{\pi/2} \cos x\,dx \)
\( \displaystyle = 2\Big[\sin x\Big]_{0}^{\pi/2} \)
\( = 2(1 - 0) = 2 \) sq. units