The value of `int_0^(2pi)[2 sin x] dx`, where `[.]` represents the greatest integral functions, is

Options

A. `(-5pi)/3`
B. `-pi`
C. `(5pi)/3`
D. `-2pi`

Explanation

To solve the integral \( \int_0^{2\pi} \lfloor 2 \sin x \rfloor \, dx \), where \( \lfloor . \rfloor \) represents the greatest integer function, we will follow these steps:

### Step 1: Analyze the function \( 2 \sin x \)
The function \( 2 \sin x \) oscillates between -2 and 2. We need to find the intervals where \( 2 \sin x \) takes specific integer values.

### Step 2: Identify the critical points
The sine function has specific points where it reaches its maximum and minimum values:
- \( \sin x = 1 \) at \( x = \frac{\pi}{2} \)
- \( \sin x = 0 \) at \( x = 0, \pi, 2\pi \)
- \( \sin x = -1 \) at \( x = \frac{3\pi}{2} \)

From this, we can see:
- \( 2 \sin x = 2 \) at \( x = \frac{\pi}{2} \)
- \( 2 \sin x = 0 \) at \( x = 0, \pi, 2\pi \)
- \( 2 \sin x = -2 \) at \( x = \frac{3\pi}{2} \)

### Step 3: Determine the intervals for \( \lfloor 2 \sin x \rfloor \)
1. **Interval \( [0, \frac{\pi}{2}] \)**:
- \( 2 \sin x \) ranges from 0 to 2.
- \( \lfloor 2 \sin x \rfloor = 0 \) for \( x \in [0, \frac{\pi}{6}) \) and \( \lfloor 2 \sin x \rfloor = 1 \) for \( x \in [\frac{\pi}{6}, \frac{\pi}{2}] \).

2. **Interval \( [\frac{\pi}{2}, \pi] \)**:
- \( 2 \sin x \) decreases from 2 to 0.
- \( \lfloor 2 \sin x \rfloor = 1 \) for \( x \in [\frac{\pi}{2}, \frac{5\pi}{6}) \) and \( \lfloor 2 \sin x \rfloor = 0 \) for \( x \in [\frac{5\pi}{6}, \pi] \).

3. **Interval \( [\pi, \frac{3\pi}{2}] \)**:
- \( 2 \sin x \) decreases from 0 to -2.
- \( \lfloor 2 \sin x \rfloor = 0 \) for \( x \in [\pi, \frac{7\pi}{6}) \) and \( \lfloor 2 \sin x \rfloor = -1 \) for \( x \in [\frac{7\pi}{6}, \frac{3\pi}{2}] \).

4. **Interval \( [\frac{3\pi}{2}, 2\pi] \)**:
- \( 2 \sin x \) increases from -2 to 0.
- \( \lfloor 2 \sin x \rfloor = -1 \) for \( x \in [\frac{3\pi}{2}, \frac{11\pi}{6}) \) and \( \lfloor 2 \sin x \rfloor = 0 \) for \( x \in [\frac{11\pi}{6}, 2\pi] \).

### Step 4: Set up the integral
Now we can break the integral into parts based on the intervals we found:
\[ \int_0^{2\pi} \lfloor 2 \sin x \rfloor \, dx = \int_0^{\frac{\pi}{6}} 0 \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 1 \, dx + \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 \, dx + \int_{\frac{5\pi}{6}}^{\pi} 0 \, dx + \int_{\pi}^{\frac{7\pi}{6}} 0 \, dx + \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} -1 \, dx + \int_{\frac{3\pi}{2}}^{\frac{11\pi}{6}} -1 \, dx + \int_{\frac{11\pi}{6}}^{2\pi} 0 \, dx \]

### Step 5: Calculate each integral
1. \( \int_0^{\frac{\pi}{6}} 0 \, dx = 0 \)
2. \( \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{3} \)
3. \( \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 \, dx = \frac{5\pi}{6} - \frac{\pi}{2} = \frac{5\pi}{6} - \frac{3\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \)
4. \( \int_{\frac{5\pi}{6}}^{\pi} 0 \, dx = 0 \)
5. \( \int_{\pi}^{\frac{7\pi}{6}} 0 \, dx = 0 \)
6. \( \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} -1 \, dx = -\left(\frac{3\pi}{2} - \frac{7\pi}{6}\right) = -\left(\frac{9\pi}{6} - \frac{7\pi}{6}\right) = -\frac{2\pi}{6} = -\frac{\pi}{3} \)
7. \( \int_{\frac{3\pi}{2}}^{\frac{11\pi}{6}} -1 \, dx = -\left(\frac{11\pi}{6} - \frac{3\pi}{2}\right) = -\left(\frac{11\pi}{6} - \frac{9\pi}{6}\right) = -\frac{2\pi}{6} = -\frac{\pi}{3} \)
8. \( \int_{\frac{11\pi}{6}}^{2\pi} 0 \, dx = 0 \)

### Step 6: Combine the results
Now we combine all the results:
\[ \int_0^{2\pi} \lfloor 2 \sin x \rfloor \, dx = 0 + \frac{\pi}{3} + \frac{\pi}{3} + 0 + 0 - \frac{\pi}{3} - \frac{\pi}{3} + 0 = 0 \]

### Final Answer
The value of \( \int_0^{2\pi} \lfloor 2 \sin x \rfloor \, dx \) is \( 0 \).