Let a function `f:R to R` be defined as `f (x) =x+ sin x.` The value of `int _(0) ^(2pi)f ^(-1)(x) dx` will be:

Options

A. `2pi^(2)`
B. `2pi^(2)-2`
C. `2x ^(2) +2`
D. `pi^(2)`

Answer

To solve the problem, we need to evaluate the integral \( \int_{0}^{2\pi} f^{-1}(x) \, dx \) where \( f(x) = x + \sin x \).

### Step-by-Step Solution:

1. **Determine the function and its limits:**
We have \( f(x) = x + \sin x \). First, we need to find the values of \( f(x) \) at the limits of integration:
- Calculate \( f(0) \):
\[ f(0) = 0 + \sin(0) = 0 \]
- Calculate \( f(2\pi) \):
\[ f(2\pi) = 2\pi + \sin(2\pi) = 2\pi + 0 = 2\pi \]
Thus, the function \( f(x) \) maps the interval \( [0, 2\pi] \) to itself.

2. **Using the property of inverse functions:**
We can use the property of integrals involving inverse functions:
\[ \int_{a}^{b} f^{-1}(x) \, dx + \int_{f^{-1}(a)}^{f^{-1}(b)} f(x) \, dx = b \cdot f^{-1}(b) - a \cdot f^{-1}(a) \]
Here, \( a = 0 \) and \( b = 2\pi \). Since \( f(0) = 0 \) and \( f(2\pi) = 2\pi \), we have \( f^{-1}(0) = 0 \) and \( f^{-1}(2\pi) = 2\pi \).

3. **Set up the equation:**
We can write:
\[ \int_{0}^{2\pi} f^{-1}(x) \, dx + \int_{0}^{2\pi} f(x) \, dx = 2\pi \cdot 2\pi - 0 \cdot 0 \]
This simplifies to:
\[ \int_{0}^{2\pi} f^{-1}(x) \, dx + \int_{0}^{2\pi} f(x) \, dx = 4\pi^2 \]

4. **Evaluate \( \int_{0}^{2\pi} f(x) \, dx \):**
We need to compute \( \int_{0}^{2\pi} (x + \sin x) \, dx \):
\[ \int_{0}^{2\pi} (x + \sin x) \, dx = \int_{0}^{2\pi} x \, dx + \int_{0}^{2\pi} \sin x \, dx \]
- Calculate \( \int_{0}^{2\pi} x \, dx \):
\[ \int_{0}^{2\pi} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2\pi} = \frac{(2\pi)^2}{2} - 0 = 2\pi^2 \]
- Calculate \( \int_{0}^{2\pi} \sin x \, dx \):
\[ \int_{0}^{2\pi} \sin x \, dx = [-\cos x]_{0}^{2\pi} = -\cos(2\pi) + \cos(0) = -1 + 1 = 0 \]
Thus,
\[ \int_{0}^{2\pi} f(x) \, dx = 2\pi^2 + 0 = 2\pi^2 \]

5. **Substitute back to find \( \int_{0}^{2\pi} f^{-1}(x) \, dx \):**
Now we substitute back into our equation:
\[ \int_{0}^{2\pi} f^{-1}(x) \, dx + 2\pi^2 = 4\pi^2 \]
This gives:
\[ \int_{0}^{2\pi} f^{-1}(x) \, dx = 4\pi^2 - 2\pi^2 = 2\pi^2 \]

### Final Answer:
Thus, the value of \( \int_{0}^{2\pi} f^{-1}(x) \, dx \) is \( 2\pi^2 \).