The area (in sq. units) bounded by the curve $y = \cos x$ and x-axis between $x = 0$ and $x = \frac{3\pi}{2}$ is

The curve $y = \cos x$ between $x = 0$ and $x = \frac{3\pi}{2}$ needs to be split into two regions:

From $x = 0$ to $x = \frac{\pi}{2}$: the curve is above the x-axis (positive)

From $x = \frac{\pi}{2}$ to $x = \frac{3\pi}{2}$: the curve is below the x-axis (negative)

When a curve is below the x-axis, the integral gives a negative value. Since area is always positive, absolute values must be taken.

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For the region from $x = 0$ to $x = \frac{\pi}{2}$:

$\text{Area}_1 = \int_{0}^{\pi/2} \cos x \, dx$

$= [\sin x]_{0}^{\pi/2}$

$= \sin\left(\frac{\pi}{2}\right) - \sin(0)$

$= 1 - 0$

$= 1$

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For the region from $x = \frac{\pi}{2}$ to $x = \frac{3\pi}{2}$:

$\text{Area}_2 = \left|\int_{\pi/2}^{3\pi/2} \cos x \, dx\right|$

$= \left|[\sin x]_{\pi/2}^{3\pi/2}\right|$

$= \left|\sin\left(\frac{3\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right|$

$= |(-1) - (1)|$

$= |-2|$

$= 2$

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Total Area $= \text{Area}_1 + \text{Area}_2$

$= 1 + 2$

$= 3$

Therefore, the area bounded by the curve $y = \cos x$ and the x-axis between $x = 0$ and $x = \frac{3\pi}{2}$ is $3$ square units.