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

The area between the curve $y = \cos x$ and the x-axis from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$ needs to be found.

The curve $y = \cos x$ has the following values:

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

At $x = 0$: $\cos(0) = 1$

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

Between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$, the cosine curve stays above the x-axis (all values are positive).

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The area bounded by the curve and the x-axis is:

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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Applying the limits:

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

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

Area $= 1 - (-1)$

Area $= 2$

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Therefore, the area of the region bounded by the curve $y = \cos x$ between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ and the x-axis is $2$ square units.