The area (in sq.units) of the region enclosed by the curve $y = \cos x$, $\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ and the x - axis is:

The region is enclosed by the curve $y = \cos x$, the x-axis, and the vertical lines $x = \frac{-\pi}{2}$ and $x = \frac{\pi}{2}$.

For this interval:

• At $x = \frac{-\pi}{2}$: $\cos(\frac{-\pi}{2}) = 0$
• At $x = 0$: $\cos(0) = 1$
• At $x = \frac{\pi}{2}$: $\cos(\frac{\pi}{2}) = 0$

The curve $y = \cos x$ is always positive (above the x-axis) in the interval $[\frac{-\pi}{2}, \frac{\pi}{2}]$.

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The area under the curve is given by:

Area $= \int_{-\pi/2}^{\pi/2} \cos x \, dx$

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The antiderivative of $\cos x$ is $\sin x$:

$\int \cos x \, dx = \sin x + C$

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Evaluating the definite integral:

Area $= [\sin x]_{-\pi/2}^{\pi/2}$

Area $= \sin(\frac{\pi}{2}) - \sin(\frac{-\pi}{2})$

Area $= 1 - (-1)$

Area $= 1 + 1$

Area $= 2$

Therefore, the area of the region is $2$ square units.