The value of the integral `int_(0)^(100) sin(x-[x])pidx`, is

Options

A. `100//pi`
B. `200//pi`
C. `100pi`
D. `200pi`

Answer

To solve the integral \( \int_{0}^{100} \sin(x - [x]) \pi \, dx \), where \([x]\) is the greatest integer function (also known as the floor function), we can follow these steps:

### Step 1: Understand the Function
The expression \( \sin(x - [x]) \) can be simplified. The term \( [x] \) represents the greatest integer less than or equal to \( x \). Therefore, \( x - [x] \) gives us the fractional part of \( x \), which is denoted as \( \{x\} \). This means:
\[ \sin(x - [x]) = \sin(\{x\}) \]

### Step 2: Analyze the Integral
The integral can be rewritten as:
\[ \int_{0}^{100} \sin(\{x\}) \pi \, dx \]
Since the function \( \sin(\{x\}) \) is periodic with a period of 1, we can break the integral from 0 to 100 into 100 intervals of 1:
\[ \int_{0}^{100} \sin(\{x\}) \pi \, dx = \sum_{n=0}^{99} \int_{n}^{n+1} \sin(\{x\}) \pi \, dx \]

### Step 3: Evaluate One Period
Now, we can evaluate the integral over one period:
\[ \int_{0}^{1} \sin(\{x\}) \pi \, dx \]
Since \( \{x\} = x \) for \( x \in [0, 1) \), we have:
\[ \int_{0}^{1} \sin(x) \pi \, dx \]

### Step 4: Calculate the Integral
Now we can compute this integral:
\[ \int_{0}^{1} \sin(x) \pi \, dx = \pi \int_{0}^{1} \sin(x) \, dx \]
The integral of \( \sin(x) \) is:
\[ -\cos(x) \Big|_{0}^{1} = -\cos(1) + \cos(0) = 1 - \cos(1) \]
Thus, we have:
\[ \int_{0}^{1} \sin(x) \, dx = 1 - \cos(1) \]

### Step 5: Combine Results
Now substituting back, we get:
\[ \int_{0}^{1} \sin(x) \pi \, dx = \pi (1 - \cos(1)) \]
Since there are 100 such intervals from 0 to 100:
\[ \int_{0}^{100} \sin(\{x\}) \pi \, dx = 100 \cdot \pi (1 - \cos(1)) \]

### Step 6: Final Result
Thus, the value of the integral is:
\[ \int_{0}^{100} \sin(x - [x]) \pi \, dx = 100 \pi (1 - \cos(1)) \]