`f(x) = min{2 sinx, 1- cos x, 1}` then `int_0^(pi)f(x) dx` is equal to

Options

A. `(pi)/(3)+1-sqrt(3)`
B. `(2pi)/(3)-1+sqrt(3)`
C. `(2pi)/(3)-1-sqrt(3)`
D. `(5pi)/(6)+1-sqrt(3)`

Answer

To solve the problem, we need to evaluate the integral \( \int_0^{\pi} f(x) \, dx \) where \( f(x) = \min\{2 \sin x, 1 - \cos x, 1\} \).

### Step 1: Identify the functions
We have three functions to consider:
1. \( f_1(x) = 2 \sin x \)
2. \( f_2(x) = 1 - \cos x \)
3. \( f_3(x) = 1 \)

### Step 2: Find intersection points
We need to find the points where these functions intersect within the interval \( [0, \pi] \).

1. Set \( 2 \sin x = 1 \):
\[ \sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6} \]
(Only \( \frac{\pi}{6} \) is in \( [0, \pi] \))

2. Set \( 1 - \cos x = 1 \):
\[ -\cos x = 0 \implies \cos x = 0 \implies x = \frac{\pi}{2} \]

3. Set \( 2 \sin x = 1 - \cos x \):
\[ 2 \sin x + \cos x = 1 \]
This equation can be solved numerically or graphically to find the intersection points.

### Step 3: Analyze the intervals
We will evaluate \( f(x) \) in the intervals defined by the intersection points:
- \( [0, \frac{\pi}{6}] \)
- \( [\frac{\pi}{6}, \frac{\pi}{2}] \)
- \( [\frac{\pi}{2}, \frac{5\pi}{6}] \)
- \( [\frac{5\pi}{6}, \pi] \)

### Step 4: Determine which function is minimum in each interval
1. **Interval \( [0, \frac{\pi}{6}] \)**:
- \( 2 \sin x \) is increasing and \( 1 - \cos x \) is also increasing.
- At \( x = 0 \), \( f_1(0) = 0 \) and \( f_2(0) = 0 \).
- Thus, \( f(x) = 2 \sin x \) in this interval.

2. **Interval \( [\frac{\pi}{6}, \frac{\pi}{2}] \)**:
- At \( x = \frac{\pi}{6} \), \( f_1(\frac{\pi}{6}) = 1 \) and \( f_2(\frac{\pi}{6}) = 1 - \frac{\sqrt{3}}{2} \).
- \( f(x) = 1 - \cos x \) dominates in this interval.

3. **Interval \( [\frac{\pi}{2}, \frac{5\pi}{6}] \)**:
- \( f(x) = 1 \) since \( 1 - \cos x \) is maximum.

4. **Interval \( [\frac{5\pi}{6}, \pi] \)**:
- \( f(x) = 2 \sin x \) again since it is decreasing and \( 1 - \cos x \) is also decreasing.

### Step 5: Set up the integral
Now we can set up the integral based on the intervals:
\[ \int_0^{\pi} f(x) \, dx = \int_0^{\frac{\pi}{6}} 2 \sin x \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (1 - \cos x) \, dx + \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 \, dx + \int_{\frac{5\pi}{6}}^{\pi} 2 \sin x \, dx \]

### Step 6: Calculate each integral
1. **First Integral**:
\[ \int_0^{\frac{\pi}{6}} 2 \sin x \, dx = -2 \cos x \bigg|_0^{\frac{\pi}{6}} = -2 \left( \cos\left(\frac{\pi}{6}\right) - \cos(0) \right) = -2 \left( \frac{\sqrt{3}}{2} - 1 \right) = 2 - \sqrt{3} \]

2. **Second Integral**:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (1 - \cos x) \, dx = \left( x - \sin x \right) \bigg|_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \left( \frac{\pi}{2} - 1 \right) - \left( \frac{\pi}{6} - \frac{1}{2} \right) = \frac{\pi}{3} - \frac{1}{2} \]

3. **Third Integral**:
\[ \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 \, dx = \frac{5\pi}{6} - \frac{\pi}{2} = \frac{5\pi}{6} - \frac{3\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

4. **Fourth Integral**:
\[ \int_{\frac{5\pi}{6}}^{\pi} 2 \sin x \, dx = -2 \cos x \bigg|_{\frac{5\pi}{6}}^{\pi} = -2 \left( -1 - \left( -\frac{\sqrt{3}}{2} \right) \right) = 2 \left( 1 - \frac{\sqrt{3}}{2} \right) = 2 - \sqrt{3} \]

### Step 7: Combine results
Now we combine all the results:
\[ \int_0^{\pi} f(x) \, dx = (2 - \sqrt{3}) + \left( \frac{\pi}{3} - \frac{1}{2} \right) + \frac{\pi}{3} + (2 - \sqrt{3}) \]
\[ = 4 - 2\sqrt{3} + \frac{2\pi}{3} - \frac{1}{2} \]
\[ = \frac{8 - 1}{2} + \frac{2\pi}{3} - 2\sqrt{3} = \frac{7}{2} + \frac{2\pi}{3} - 2\sqrt{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^{\pi} f(x) \, dx = \frac{7}{2} + \frac{2\pi}{3} - 2\sqrt{3} \]