`int_(-2)^2min(x-[x],-x-[-x])dx` equals, where [ x ] represents greatest integer less than or equal to x.

A. 2
B. 1
C. 4
D. 0

Options

A. 2
B. 1
C. 4
D. 0

Answer

To solve the integral \( \int_{-2}^{2} \min(x - [x], -x - [-x]) \, dx \), where \([x]\) represents the greatest integer less than or equal to \(x\), we will follow these steps:

### Step 1: Understand the Functions
The functions involved are:
1. \( f_1(x) = x - [x] \) (the fractional part of \(x\))
2. \( f_2(x) = -x - [-x] \)

We can express \(f_2(x)\) as follows:
- Since \([-x] = -[x] - 1\) when \(x\) is not an integer, we have:
\[ f_2(x) = -x + [x] + 1 = 1 + [x] - x \]
Thus, \(f_2(x) = 1 - (x - [x]) = 1 - f_1(x)\).

### Step 2: Identify the Interval and Behavior of Functions
We need to evaluate the integral from \(-2\) to \(2\). We will analyze the behavior of both functions in this interval.

- For \(x \in [-2, -1)\):
- \([x] = -2\) so \(f_1(x) = x + 2\) and \(f_2(x) = 1 - (x + 2) = -x - 1\).
- For \(x \in [-1, 0)\):
- \([x] = -1\) so \(f_1(x) = x + 1\) and \(f_2(x) = 1 - (x + 1) = -x\).
- For \(x \in [0, 1)\):
- \([x] = 0\) so \(f_1(x) = x\) and \(f_2(x) = 1 - x\).
- For \(x \in [1, 2)\):
- \([x] = 1\) so \(f_1(x) = x - 1\) and \(f_2(x) = 1 - (x - 1) = 2 - x\).

### Step 3: Determine the Minimum Function
Now we will determine which function is smaller in each interval:
- For \(x \in [-2, -1)\):
- \(f_1(x) = x + 2\) and \(f_2(x) = -x - 1\).
- The minimum is \(f_1(x)\) since \(x + 2 < -x - 1\) in this interval.

- For \(x \in [-1, 0)\):
- \(f_1(x) = x + 1\) and \(f_2(x) = -x\).
- The minimum is \(f_2(x)\) since \(x + 1 < -x\) in this interval.

- For \(x \in [0, 1)\):
- \(f_1(x) = x\) and \(f_2(x) = 1 - x\).
- The minimum is \(f_1(x)\) since \(x < 1 - x\) in this interval.

- For \(x \in [1, 2)\):
- \(f_1(x) = x - 1\) and \(f_2(x) = 2 - x\).
- The minimum is \(f_2(x)\) since \(x - 1 < 2 - x\) in this interval.

### Step 4: Set Up the Integral
Now we can set up the integral based on the intervals:
\[ \int_{-2}^{-1} (x + 2) \, dx + \int_{-1}^{0} (-x) \, dx + \int_{0}^{1} (x) \, dx + \int_{1}^{2} (2 - x) \, dx \]

### Step 5: Calculate Each Integral
1. **First Integral**:
\[ \int_{-2}^{-1} (x + 2) \, dx = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{-1} = \left( \frac{1}{2} - 2 \right) - \left( 2 - 4 \right) = -\frac{3}{2} + 2 = \frac{1}{2} \]

2. **Second Integral**:
\[ \int_{-1}^{0} (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^{0} = 0 - \frac{1}{2} = -\frac{1}{2} \]

3. **Third Integral**:
\[ \int_{0}^{1} (x) \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2} \]

4. **Fourth Integral**:
\[ \int_{1}^{2} (2 - x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2} = \left( 4 - 2 \right) - \left( 2 - \frac{1}{2} \right) = 2 - \frac{3}{2} = \frac{1}{2} \]

### Step 6: Combine the Results
Now we combine all the results:
\[ \frac{1}{2} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-2}^{2} \min(x - [x], -x - [-x]) \, dx = 1 \]