Let f be a real - valued function defined on R ( the set of real numbers) such that `f(x) = sin^(-1) ( sin x) + cos^(-1) ( cos x)`
The area bounded by curve `y = f(x)` and x- axis from `pi/2 le x le pi` is equal to

Options

A. `pi^(2)/4`
B. `pi^(2)/2`
C. `pi^(2)`
D. `pi^(2)/8`

Answer

To solve the problem step by step, we will analyze the function \( f(x) = \sin^{-1}(\sin x) + \cos^{-1}(\cos x) \) and find the area bounded by the curve \( y = f(x) \) and the x-axis from \( \frac{\pi}{2} \) to \( \pi \).

### Step 1: Determine the behavior of \( f(x) \) in the interval \( \left[\frac{\pi}{2}, \pi\right] \)

We know the properties of the inverse trigonometric functions:
- For \( x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( \sin^{-1}(\sin x) = x \).
- For \( x \in \left[\frac{\pi}{2}, \pi\right] \), \( \sin^{-1}(\sin x) = \pi - x \).
- For \( x \in [0, \pi] \), \( \cos^{-1}(\cos x) = x \).
- For \( x \in [\pi, 2\pi] \), \( \cos^{-1}(\cos x) = 2\pi - x \).

In the interval \( \left[\frac{\pi}{2}, \pi\right] \):
- \( \sin^{-1}(\sin x) = \pi - x \)
- \( \cos^{-1}(\cos x) = x \)

Thus, we can write:
\[ f(x) = \sin^{-1}(\sin x) + \cos^{-1}(\cos x) = (\pi - x) + x = \pi \]

### Step 2: Set up the integral for the area

The area \( A \) bounded by the curve \( y = f(x) \) and the x-axis from \( \frac{\pi}{2} \) to \( \pi \) can be calculated using the integral:
\[ A = \int_{\frac{\pi}{2}}^{\pi} f(x) \, dx \]
Since we have found that \( f(x) = \pi \) in this interval, we can substitute:
\[ A = \int_{\frac{\pi}{2}}^{\pi} \pi \, dx \]

### Step 3: Evaluate the integral

Calculating the integral:
\[ A = \pi \int_{\frac{\pi}{2}}^{\pi} 1 \, dx = \pi \left[ x \right]_{\frac{\pi}{2}}^{\pi} = \pi \left( \pi - \frac{\pi}{2} \right) = \pi \left( \frac{\pi}{2} \right) = \frac{\pi^2}{2} \]

### Final Answer

Thus, the area bounded by the curve \( y = f(x) \) and the x-axis from \( \frac{\pi}{2} \) to \( \pi \) is:
\[ \frac{\pi^2}{2} \]