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:
1. \( y = 2\sin x \)
2. \( y = 1 - \cos x \)
3. \( y = 1 \)
### Step 2: Find the points of intersection
To determine where these functions intersect, we need to set them equal to each other.
1. **Intersection of \( 2\sin x \) and \( 1 - \cos x \)**:
\[
2\sin x = 1 - \cos x
\]
Rearranging gives:
\[
2\sin x + \cos x = 1
\]
This can be solved using numerical or graphical methods.
2. **Intersection of \( 2\sin x \) and \( 1 \)**:
\[
2\sin x = 1 \implies \sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}
\]
3. **Intersection of \( 1 - \cos x \) and \( 1 \)**:
\[
1 - \cos x = 1 \implies \cos x = 0 \implies x = \frac{\pi}{2}
\]
### Step 3: Analyze the intervals
Now we have the critical points \( x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \). We will evaluate \( f(x) \) in the intervals:
- \( [0, \frac{\pi}{6}] \)
- \( [\frac{\pi}{6}, \frac{\pi}{2}] \)
- \( [\frac{\pi}{2}, \frac{5\pi}{6}] \)
- \( [\frac{5\pi}{6}, \pi] \)
### Step 4: Determine \( f(x) \) in each interval
1. **Interval \( [0, \frac{\pi}{6}] \)**:
- Here, \( 2\sin x \) is less than \( 1 - \cos x \) and \( 1 \).
- Thus, \( f(x) = 2\sin x \).
2. **Interval \( [\frac{\pi}{6}, \frac{\pi}{2}] \)**:
- At \( x = \frac{\pi}{6} \), \( 2\sin x = 1 \).
- For \( x \) in this interval, \( 1 - \cos x \) is less than \( 1 \) and \( 2\sin x \) is equal to \( 1 \) at \( x = \frac{\pi}{6} \).
- Thus, \( f(x) = 1 - \cos x \).
3. **Interval \( [\frac{\pi}{2}, \frac{5\pi}{6}] \)**:
- Here, \( 1 \) is less than both \( 2\sin x \) and \( 1 - \cos x \).
- Thus, \( f(x) = 1 \).
4. **Interval \( [\frac{5\pi}{6}, \pi] \)**:
- Here, \( 2\sin x \) is less than \( 1 - \cos x \) and \( 1 \).
- Thus, \( f(x) = 2\sin x \).
### Step 5: Set up the integral
Now we can set up the integral:
\[
\int_0^{\pi} f(x) \, dx = \int_0^{\frac{\pi}{6}} 2\sin x \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (1 - \cos x) \, dx + \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 \, dx + \int_{\frac{5\pi}{6}}^{\pi} 2\sin x \, dx
\]
### Step 6: Evaluate the integrals
1. **First integral**:
\[
\int_0^{\frac{\pi}{6}} 2\sin x \, dx = -2\cos x \bigg|_0^{\frac{\pi}{6}} = -2\left(\cos\frac{\pi}{6} - \cos 0\right) = -2\left(\frac{\sqrt{3}}{2} - 1\right) = 2 - \sqrt{3}
\]
2. **Second integral**:
\[
\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (1 - \cos x) \, dx = \left[x - \sin x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \left(\frac{\pi}{2} - 1\right) - \left(\frac{\pi}{6} - \frac{1}{2}\right) = \frac{\pi}{3} - \frac{1}{2}
\]
3. **Third integral**:
\[
\int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 \, dx = \left[x\right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} = \frac{5\pi}{6} - \frac{\pi}{2} = \frac{5\pi}{6} - \frac{3\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}
\]
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 the 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 final result is:
\[
\int_0^{\pi} f(x) \, dx = \frac{7}{2} + \frac{2\pi}{3} - 2\sqrt{3}
\]
