Find the area of the region bounded by the curve `y="sin"x,x=(pi)/(2)andx=(3pi)/(2)`

Answer

To find the area of the region bounded by the curve \( y = \sin x \) between \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \), we can follow these steps:

### Step 1: Understand the Function and the Limits
The function \( y = \sin x \) oscillates between -1 and 1. We need to find the area between the curve and the x-axis from \( x = \frac{\pi}{2} \) to \( x = \frac{3\pi}{2} \).

### Step 2: Identify the Regions
1. From \( x = \frac{\pi}{2} \) to \( x = \pi \), \( \sin x \) is positive.
2. From \( x = \pi \) to \( x = \frac{3\pi}{2} \), \( \sin x \) is negative.

### Step 3: Set Up the Integral
The area \( A \) can be calculated using the integral:
\[ A = \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx + \int_{\pi}^{\frac{3\pi}{2}} -\sin x \, dx \]
The negative sign in the second integral is used because the sine function is below the x-axis in that interval.

### Step 4: Calculate the First Integral
Calculate the first integral:
\[ \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx \]
The antiderivative of \( \sin x \) is \( -\cos x \). Thus,
\[ \int \sin x \, dx = -\cos x \]
Evaluating from \( \frac{\pi}{2} \) to \( \pi \):
\[ = \left[-\cos x\right]_{\frac{\pi}{2}}^{\pi} = -\cos(\pi) - (-\cos(\frac{\pi}{2})) = -(-1) - (0) = 1 \]

### Step 5: Calculate the Second Integral
Now calculate the second integral:
\[ \int_{\pi}^{\frac{3\pi}{2}} -\sin x \, dx \]
Using the same antiderivative:
\[ = -\left[-\cos x\right]_{\pi}^{\frac{3\pi}{2}} = \left[\cos x\right]_{\pi}^{\frac{3\pi}{2}} = \cos(\frac{3\pi}{2}) - \cos(\pi) = 0 - (-1) = 1 \]

### Step 6: Add the Areas
Now, add the areas from both integrals:
\[ A = 1 + 1 = 2 \]

### Final Answer
The area of the region bounded by the curve \( y = \sin x \), \( x = \frac{\pi}{2} \), and \( x = \frac{3\pi}{2} \) is \( 2 \) square units.

---