If `f (x) = min [ x ^(2), sin ""(x)/(2), (x -2pi) ^(2)],` the area bounded by the curve `y=f (x),` x-axis, `x=0 and x=2pi` is given by
Note: `x_(1)` is the point of intersection of the curves `x ^(2) and sin ""(x)/(2),x _(2)` is the point of intersection of the curves sin `""x/2 and (x-2pi)^(2))`

Options

A. `int_(0)^(x _(1))(sin ""(x)/(2)) dx + int _(x_(1))^(pi) x ^(2) dx + int _(pi)^(x _(2)) (x-2pi)^(2) dx + int _(x _(2))^(2pi) (sin ""(x)/(2)) dx`
B. `int _(0) ^(x_(1)) x ^(2) dx +int _(x _(1))^(x _(3))(sin""(x)/(2)) dx + int _( x_(2)) ^(2pi) (x-2pi)^(2) dx,` where `x _(1) in (0, (pi)/(3)) and x _(2) in ((5pi)/(3), 2pi)`
C. `int _(0)^(x_(1)) x ^(2) dx + int _(x _(1)) ^(x_(2))sin ((x )/(2)) dx + int _( x_(2))^(2pi) (x-2pi) ^(2)dx,` where `x _(1) in ((pi)/(3), (pi)/(2)) and x _(2) in ((3pi)/(2) , 2pi)`
D. `int _(0)^(x _(1))x ^(2) dx + int _(x _(1)) ^(x_(2)) sin ((x )/(2)) dx + int _( x_(2))^(2pi) (x-2pi) ^(2) dx`, where `x _(1) in ((pi)/(2), (2pi)/(3)) and x _(2) in (pi, 2pi)`

Answer

To find the area bounded by the curve \( y = f(x) \), the x-axis, \( x = 0 \), and \( x = 2\pi \), we need to analyze the function \( f(x) = \min \left[ x^2, \frac{\sin(x)}{2}, (x - 2\pi)^2 \right] \) and identify the points of intersection of the curves involved.

### Step 1: Identify the curves and their intersections
We have three curves:
- Curve 1: \( y = x^2 \)
- Curve 2: \( y = \frac{\sin(x)}{2} \)
- Curve 3: \( y = (x - 2\pi)^2 \)

We need to find the points of intersection:
1. **Intersection of Curve 1 and Curve 2**: Solve \( x^2 = \frac{\sin(x)}{2} \).
2. **Intersection of Curve 2 and Curve 3**: Solve \( \frac{\sin(x)}{2} = (x - 2\pi)^2 \).

Let \( x_1 \) be the intersection point of Curve 1 and Curve 2, and \( x_2 \) be the intersection point of Curve 2 and Curve 3.

### Step 2: Determine the intervals for integration
From the analysis of the curves:
- For \( 0 \leq x < x_1 \), \( f(x) = x^2 \).
- For \( x_1 < x < x_2 \), \( f(x) = \frac{\sin(x)}{2} \).
- For \( x_2 < x \leq 2\pi \), \( f(x) = (x - 2\pi)^2 \).

### Step 3: Set up the area integral
The area \( A \) can be expressed as:
\[ A = \int_0^{x_1} x^2 \, dx + \int_{x_1}^{x_2} \frac{\sin(x)}{2} \, dx + \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx \]

### Step 4: Calculate each integral
1. **Calculate \( \int_0^{x_1} x^2 \, dx \)**:
\[ \int_0^{x_1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{x_1} = \frac{x_1^3}{3} \]

2. **Calculate \( \int_{x_1}^{x_2} \frac{\sin(x)}{2} \, dx \)**:
\[ \int_{x_1}^{x_2} \frac{\sin(x)}{2} \, dx = \left[ -\frac{\cos(x)}{2} \right]_{x_1}^{x_2} = -\frac{\cos(x_2)}{2} + \frac{\cos(x_1)}{2} \]

3. **Calculate \( \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx \)**:
\[ \int_{x_2}^{2\pi} (x - 2\pi)^2 \, dx = \left[ \frac{(x - 2\pi)^3}{3} \right]_{x_2}^{2\pi} = \frac{(0)^3}{3} - \frac{(x_2 - 2\pi)^3}{3} = -\frac{(x_2 - 2\pi)^3}{3} \]

### Step 5: Combine the areas
Thus, the total area \( A \) is:
\[ A = \frac{x_1^3}{3} + \left( -\frac{\cos(x_2)}{2} + \frac{\cos(x_1)}{2} \right) - \frac{(x_2 - 2\pi)^3}{3} \]

### Final Result
The area bounded by the curve \( y = f(x) \), the x-axis, \( x = 0 \), and \( x = 2\pi \) is given by the expression above, where \( x_1 \) and \( x_2 \) are the points of intersection calculated earlier.