The area of `int_(-10)^(10) ` f(x) dx , where f(x) = min {x -[x] , -x-[-x]} , is

Options

A. (a)20
B. (b)40
C. (c)5
D. (d)30

Answer

To find the area of the integral \( \int_{-10}^{10} f(x) \, dx \) where \( f(x) = \min \{ x - [x], -x - [-x] \} \), we will follow these steps:

### Step 1: Understand the Function
The function \( f(x) \) involves the greatest integer function, denoted as \( [x] \). The term \( x - [x] \) represents the fractional part of \( x \), which we denote as \( \{ x \} \). The term \( -x - [-x] \) can also be simplified using the properties of the greatest integer function.

### Step 2: Simplify the Function
We know that:
- \( [x] \) is the greatest integer less than or equal to \( x \).
- Therefore, \( -x - [-x] = -x - (x - \{ x \}) = -2x + \{ x \} \).

Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \min \{ \{ x \}, -2x + \{ x \} \} \]

### Step 3: Determine the Behavior of \( f(x) \)
To analyze \( f(x) \), we need to consider the intervals defined by the integer values of \( x \). The function \( f(x) \) will have a periodic behavior with a period of 1.

### Step 4: Evaluate \( f(x) \) on the Interval [0, 1]
Within the interval \( [0, 1] \):
- \( \{ x \} = x \)
- \( -2x + \{ x \} = -2x + x = -x \)

Thus, in the interval \( [0, 1] \):
\[ f(x) = \min \{ x, -x \} \]
Since \( x \) is non-negative and \( -x \) is non-positive, we have:
\[ f(x) = -x \quad \text{for } x \in [0, 1] \]

### Step 5: Evaluate \( f(x) \) on the Interval [-1, 0]
In the interval \( [-1, 0] \):
- \( \{ x \} = x + 1 \)
- \( -2x + \{ x \} = -2x + (x + 1) = -x + 1 \)

Thus, in the interval \( [-1, 0] \):
\[ f(x) = \min \{ x + 1, -x + 1 \} \]
Here, \( x + 1 \) is non-positive and \( -x + 1 \) is non-negative, so:
\[ f(x) = -x + 1 \quad \text{for } x \in [-1, 0] \]

### Step 6: Calculate the Area from -10 to 10
Since \( f(x) \) is periodic with period 1, we can calculate the area from 0 to 1 and then multiply by 20 (the number of complete periods from -10 to 10).

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

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

3. **Total Area from -1 to 1**:
\[ \text{Total Area} = \left(-\frac{1}{2} + \frac{3}{2}\right) = 1 \]

4. **Total Area from -10 to 10**:
Since the area from -1 to 1 is 1, the area from -10 to 10 (which contains 20 such intervals) is:
\[ \text{Total Area} = 20 \times 1 = 20 \]

### Final Answer
The area of \( \int_{-10}^{10} f(x) \, dx \) is \( \boxed{20} \).