If `k in N` and `I_(k)=int_(-2kp)^(2kpi) |sin x|[sin x]dx`, where [.] denotes the greatest integer function, then `int_(k=1)^(100) I_(k)` equal to

Options

A. -10100
B. -40400
C. -20200
D. none of these

Explanation

To solve the problem, we need to compute the integral \( I_k = \int_{-2k\pi}^{2k\pi} |\sin x| [\sin x] \, dx \), where \([ \cdot ]\) denotes the greatest integer function. We will then find the sum \( \sum_{k=1}^{100} I_k \).

### Step 1: Understanding the Integral
The integral \( I_k \) can be broken down into segments based on the periodicity of the sine function. The sine function has a period of \( 2\pi \), and we can analyze the behavior of \( |\sin x| \) and \([\sin x]\) over one period.

### Step 2: Analyzing the Interval
The integral can be split into intervals:
- From \( -2k\pi \) to \( -k\pi \)
- From \( -k\pi \) to \( 0 \)
- From \( 0 \) to \( k\pi \)
- From \( k\pi \) to \( 2k\pi \)

Due to the symmetry of the sine function, we can analyze the integral from \( 0 \) to \( 2k\pi \) and multiply the result appropriately.

### Step 3: Evaluating \( I_k \)
1. **From \( 0 \) to \( \pi \)**:
- Here, \( |\sin x| = \sin x \) and \([\sin x] = 0\) for \( x \in [0, \pi) \).
- Thus, \( \int_0^{\pi} |\sin x| [\sin x] \, dx = 0 \).

2. **From \( \pi \) to \( 2\pi \)**:
- Here, \( |\sin x| = -\sin x \) and \([\sin x] = -1\) for \( x \in [\pi, 2\pi) \).
- Thus, \( \int_{\pi}^{2\pi} |\sin x| [\sin x] \, dx = \int_{\pi}^{2\pi} -\sin x \cdot (-1) \, dx = \int_{\pi}^{2\pi} \sin x \, dx \).
- This evaluates to \( [-\cos x]_{\pi}^{2\pi} = -\cos(2\pi) + \cos(\pi) = -1 + 1 = 0 \).

3. **Combining the Results**:
- Since the contributions from \( [0, \pi] \) and \( [\pi, 2\pi] \) are both zero, we conclude that:
\[ I_k = 0 \quad \text{for all } k. \]

### Step 4: Summing Over k
Now, we need to compute:
\[ \sum_{k=1}^{100} I_k = \sum_{k=1}^{100} 0 = 0. \]

### Final Answer
Thus, the final answer is:
\[ \boxed{0}. \]