`int_(0)^(pi) [2sin x]dx=`

Options

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

Answer

To solve the integral \( \int_{0}^{\pi} \lfloor 2 \sin x \rfloor \, dx \), where \( \lfloor x \rfloor \) denotes the greatest integer function (or floor function), we will follow these steps:

### Step 1: Analyze the function \( 2 \sin x \)

The function \( 2 \sin x \) oscillates between 0 and 2 as \( x \) varies from 0 to \( \pi \). Specifically:
- At \( x = 0 \), \( 2 \sin(0) = 0 \)
- At \( x = \frac{\pi}{2} \), \( 2 \sin\left(\frac{\pi}{2}\right) = 2 \)
- At \( x = \pi \), \( 2 \sin(\pi) = 0 \)

### Step 2: Determine the intervals for the floor function

Next, we will find the intervals where \( \lfloor 2 \sin x \rfloor \) takes constant integer values.

1. **From \( 0 \) to \( \frac{\pi}{6} \)**:
- \( 2 \sin x \) ranges from \( 0 \) to \( 1 \).
- Thus, \( \lfloor 2 \sin x \rfloor = 0 \).

2. **From \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \)**:
- \( 2 \sin x \) ranges from \( 1 \) to \( 2 \).
- Thus, \( \lfloor 2 \sin x \rfloor = 1 \).

3. **From \( \frac{5\pi}{6} \) to \( \pi \)**:
- \( 2 \sin x \) ranges from \( 1 \) to \( 0 \).
- Thus, \( \lfloor 2 \sin x \rfloor = 0 \).

### Step 3: Set up the integral based on the intervals

Now we can break the integral into three parts based on the intervals we determined:

\[ \int_{0}^{\pi} \lfloor 2 \sin x \rfloor \, dx = \int_{0}^{\frac{\pi}{6}} 0 \, dx + \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 1 \, dx + \int_{\frac{5\pi}{6}}^{\pi} 0 \, dx \]

### Step 4: Evaluate each part of the integral

1. **First integral**:
\[ \int_{0}^{\frac{\pi}{6}} 0 \, dx = 0 \]

2. **Second integral**:
\[ \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 1 \, dx = \left[ x \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

3. **Third integral**:
\[ \int_{\frac{5\pi}{6}}^{\pi} 0 \, dx = 0 \]

### Step 5: Combine the results

Combining all parts, we have:

\[ \int_{0}^{\pi} \lfloor 2 \sin x \rfloor \, dx = 0 + \frac{2\pi}{3} + 0 = \frac{2\pi}{3} \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{0}^{\pi} \lfloor 2 \sin x \rfloor \, dx = \frac{2\pi}{3} \]

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