Evaluate `int_(-1)^(1)(x-[x])d x`, where `[.]`

Answer

To evaluate the integral \( I = \int_{-1}^{1} (x - [x]) \, dx \), where \([x]\) is the greatest integer function (also known as the floor function), we can follow these steps:

### Step 1: Understand the Function
The function \( [x] \) gives the greatest integer less than or equal to \( x \). For the interval \([-1, 1]\):
- For \( x \in [-1, 0) \), \([x] = -1\)
- For \( x \in [0, 1)\), \([x] = 0\)
- At \( x = 1\), \([x] = 1\)

### Step 2: Split the Integral
We can split the integral into two parts based on the behavior of the floor function:
\[ I = \int_{-1}^{0} (x - [x]) \, dx + \int_{0}^{1} (x - [x]) \, dx \]

### Step 3: Evaluate Each Integral

#### Integral from -1 to 0
For \( x \in [-1, 0) \):
\[ [x] = -1 \implies x - [x] = x - (-1) = x + 1 \]
Thus,
\[ \int_{-1}^{0} (x - [x]) \, dx = \int_{-1}^{0} (x + 1) \, dx \]

Calculating this integral:
\[ \int_{-1}^{0} (x + 1) \, dx = \int_{-1}^{0} x \, dx + \int_{-1}^{0} 1 \, dx \]

Calculating each part:
1. \(\int_{-1}^{0} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{0} = \frac{0^2}{2} - \frac{(-1)^2}{2} = 0 - \frac{1}{2} = -\frac{1}{2}\)
2. \(\int_{-1}^{0} 1 \, dx = [x]_{-1}^{0} = 0 - (-1) = 1\)

Combining these:
\[ \int_{-1}^{0} (x + 1) \, dx = -\frac{1}{2} + 1 = \frac{1}{2} \]

#### Integral from 0 to 1
For \( x \in [0, 1) \):
\[ [x] = 0 \implies x - [x] = x - 0 = x \]
Thus,
\[ \int_{0}^{1} (x - [x]) \, dx = \int_{0}^{1} x \, dx \]

Calculating this integral:
\[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

### Step 4: Combine the Results
Now, we combine the results of both integrals:
\[ I = \int_{-1}^{0} (x - [x]) \, dx + \int_{0}^{1} (x - [x]) \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{1} \]