The area bounded by the curve `y=|cos^(-1)(sinx)|-|sin^(-1)(cosx)|` and axis from `(3pi)/(2)lex le 2pi`

Options

A. `pi^(2)` sq. units
B. `pi^(2)//4` sq. units
C. `pi^(2)//2` sq. units
D. none of these

Explanation

To find the area bounded by the curve \( y = |\cos^{-1}(\sin x)| - |\sin^{-1}(\cos x)| \) and the x-axis from \( \frac{3\pi}{2} \) to \( 2\pi \), we will follow these steps:

### Step 1: Understand the Functions
We start with the functions involved:
- \( y = |\cos^{-1}(\sin x)| - |\sin^{-1}(\cos x)| \)

### Step 2: Determine the Range of x
We need to evaluate the area from \( x = \frac{3\pi}{2} \) to \( x = 2\pi \).

### Step 3: Evaluate the Functions at the Boundaries
1. At \( x = \frac{3\pi}{2} \):
- \( \sin\left(\frac{3\pi}{2}\right) = -1 \) and \( \cos\left(\frac{3\pi}{2}\right) = 0 \)
- Therefore, \( \cos^{-1}(-1) = \pi \) and \( \sin^{-1}(0) = 0 \)
- Thus, \( y\left(\frac{3\pi}{2}\right) = \pi - 0 = \pi \)

2. At \( x = 2\pi \):
- \( \sin(2\pi) = 0 \) and \( \cos(2\pi) = 1 \)
- Therefore, \( \cos^{-1}(0) = \frac{\pi}{2} \) and \( \sin^{-1}(1) = \frac{\pi}{2} \)
- Thus, \( y(2\pi) = \frac{\pi}{2} - \frac{\pi}{2} = 0 \)

### Step 4: Find the Expression for y
From the previous evaluations, we can simplify the expression for \( y \):
- For \( x \) in the interval \( \left[\frac{3\pi}{2}, 2\pi\right] \):
- \( y = \pi - \left(\frac{\pi}{2}\right) = \frac{3\pi}{2} - x \)

### Step 5: Set Up the Integral for Area
The area \( A \) under the curve from \( x = \frac{3\pi}{2} \) to \( x = 2\pi \) is given by:
\[ A = \int_{\frac{3\pi}{2}}^{2\pi} y \, dx = \int_{\frac{3\pi}{2}}^{2\pi} \left(\frac{3\pi}{2} - x\right) \, dx \]

### Step 6: Calculate the Integral
1. Compute the integral:
\[ A = \int_{\frac{3\pi}{2}}^{2\pi} \left(\frac{3\pi}{2} - x\right) \, dx \]
- This can be split into two parts:
\[ A = \int_{\frac{3\pi}{2}}^{2\pi} \frac{3\pi}{2} \, dx - \int_{\frac{3\pi}{2}}^{2\pi} x \, dx \]

2. Evaluate each integral:
- First integral:
\[ \int_{\frac{3\pi}{2}}^{2\pi} \frac{3\pi}{2} \, dx = \frac{3\pi}{2} \left(2\pi - \frac{3\pi}{2}\right) = \frac{3\pi}{2} \cdot \frac{\pi}{2} = \frac{3\pi^2}{4} \]
- Second integral:
\[ \int_{\frac{3\pi}{2}}^{2\pi} x \, dx = \left[\frac{x^2}{2}\right]_{\frac{3\pi}{2}}^{2\pi} = \frac{(2\pi)^2}{2} - \frac{\left(\frac{3\pi}{2}\right)^2}{2} = \frac{4\pi^2}{2} - \frac{\frac{9\pi^2}{4}}{2} = 2\pi^2 - \frac{9\pi^2}{8} = \frac{16\pi^2}{8} - \frac{9\pi^2}{8} = \frac{7\pi^2}{8} \]

3. Combine the results:
\[ A = \frac{3\pi^2}{4} - \frac{7\pi^2}{8} = \frac{6\pi^2}{8} - \frac{7\pi^2}{8} = -\frac{\pi^2}{8} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{\pi^2}{8} \]

### Final Result
The area bounded by the curve and the x-axis from \( \frac{3\pi}{2} \) to \( 2\pi \) is:
\[ \boxed{\frac{\pi^2}{8}} \]