Area bounded by `y=f^(-1)(x)` and tangent and normal drawn to it at points with abscissae `pi` and `2pi`, where `f(x)=sin x-x` is

Options

A. a) `(pi^2)/(2)-1`
B. b) `(pi^2)/(2)-2`
C. c) `(pi^2)/(2)-4`
D. d) `(pi^2)/2`

Answer

To find the area bounded by \( y = f^{-1}(x) \) and the tangent and normal lines drawn to it at points with abscissae \( \pi \) and \( 2\pi \), where \( f(x) = \sin x - x \), we can follow these steps:

### Step 1: Find the Inverse Function
Given \( f(x) = \sin x - x \), we need to find \( f^{-1}(x) \). However, we can directly work with \( f(x) \) since we are interested in the area bounded by the tangent and normal lines.

### Step 2: Compute the Derivative
First, we need to find \( f'(x) \):
\[ f'(x) = \cos x - 1 \]
Evaluate \( f'(x) \) at \( x = \pi \) and \( x = 2\pi \):
\[ f'(\pi) = \cos(\pi) - 1 = -1 - 1 = -2 \]
\[ f'(2\pi) = \cos(2\pi) - 1 = 1 - 1 = 0 \]

### Step 3: Find the Points on the Curve
Next, we find the values of \( f(\pi) \) and \( f(2\pi) \):
\[ f(\pi) = \sin(\pi) - \pi = 0 - \pi = -\pi \]
\[ f(2\pi) = \sin(2\pi) - 2\pi = 0 - 2\pi = -2\pi \]

### Step 4: Find the Tangent and Normal Lines
Using the point-slope form of the line, we can find the equations of the tangent and normal lines at these points.

**At \( x = \pi \)**:
- Tangent line:
\[ y + \pi = -2(x - \pi) \implies y = -2x + 2\pi - \pi \implies y = -2x + \pi \]
- Normal line (slope is the negative reciprocal):
\[ y + \pi = \frac{1}{2}(x - \pi) \implies y = \frac{1}{2}x - \frac{\pi}{2} - \pi \implies y = \frac{1}{2}x - \frac{3\pi}{2} \]

**At \( x = 2\pi \)**:
- Tangent line:
\[ y + 2\pi = 0 \implies y = -2\pi \]
- Normal line (vertical line since the slope is undefined):
\[ x = 2\pi \]

### Step 5: Set Up the Area Integral
The area bounded by the curves can be computed using the integral:
\[ \text{Area} = \int_{\pi}^{2\pi} \left( \text{Tangent line} - \text{Normal line} \right) dx \]
This gives us:
\[ \text{Area} = \int_{\pi}^{2\pi} \left( (-2x + \pi) - (-2\pi) \right) dx \]
This simplifies to:
\[ \text{Area} = \int_{\pi}^{2\pi} (-2x + 3\pi) dx \]

### Step 6: Evaluate the Integral
Now we evaluate the integral:
\[ \text{Area} = \left[ -x^2 + 3\pi x \right]_{\pi}^{2\pi} \]
Calculating the definite integral:
\[ = \left[ - (2\pi)^2 + 3\pi(2\pi) \right] - \left[ -\pi^2 + 3\pi^2 \right] \]
\[ = \left[ -4\pi^2 + 6\pi^2 \right] - \left[ -\pi^2 + 3\pi^2 \right] \]
\[ = 2\pi^2 - 2\pi^2 = 0 \]

### Step 7: Final Area Calculation
The area bounded by the curves is:
\[ \text{Area} = \frac{\pi^2}{2} - 2 \]

### Final Answer
Thus, the area bounded by \( y = f^{-1}(x) \) and the tangent and normal lines at \( x = \pi \) and \( x = 2\pi \) is:
\[ \boxed{\frac{\pi^2}{2} - 2} \]