Area of the region bounded by the curve $y = \sin x$ and x-axis between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ is

The area between the curve $y = \sin x$ and the x-axis from $x = \frac{\pi}{2}$ to $x = \frac{3\pi}{2}$ needs to be calculated. Area is always positive, even if the curve goes below the x-axis.

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The behavior of $\sin x$ in the given interval:

At $x = \frac{\pi}{2}$: $\sin\left(\frac{\pi}{2}\right) = 1$

At $x = \pi$: $\sin(\pi) = 0$

At $x = \frac{3\pi}{2}$: $\sin\left(\frac{3\pi}{2}\right) = -1$

From $\frac{\pi}{2}$ to $\pi$, the curve is above the x-axis (positive).

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

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Since the curve crosses the x-axis at $x = \pi$, the area must be calculated in two parts:

Total Area $= \text{Area}_1 \left(\frac{\pi}{2} \text{ to } \pi\right) + \text{Area}_2 \left(\pi \text{ to } \frac{3\pi}{2}\right)$

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For Area$_1$ (above x-axis):

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

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

$= [-\cos \pi] - [-\cos(\pi/2)]$

$= [-(-1)] - [0]$

$= 1$

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For Area$_2$ (below x-axis):

$\text{Area}_2 = \int_{\pi}^{3\pi/2} \sin x \, dx$

$= [-\cos x]_{\pi}^{3\pi/2}$

$= [-\cos(3\pi/2)] - [-\cos \pi]$

$= [0] - [1]$

$= -1$

Since the curve is below the x-axis, taking absolute value: $|-1| = 1$

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Total Area $= 1 + 1$

$= 2$

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