If `f(x) = min({x}, {-x}) x in R`, where {x} denotes the fractional part of x, then `int_(-100)^(100)f(x)dx` is

Options

A. (a) 50
B. (b) 100
C. (c) 200
D. (d) none of these

Explanation

To solve the integral \( \int_{-100}^{100} f(x) \, dx \) where \( f(x) = \min(\{x\}, \{-x\}) \) and \( \{x\} \) denotes the fractional part of \( x \), we can follow these steps:

### Step 1: Understand the Function \( f(x) \)

The fractional part function \( \{x\} \) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \).

For negative \( x \), we have:
\[ \{-x\} = -x - \lfloor -x \rfloor = -x + \lfloor x \rfloor + 1 \]
Thus, \( f(x) = \min(\{x\}, \{-x\}) \).

### Step 2: Analyze the Function Over One Period

The fractional part function \( \{x\} \) is periodic with a period of 1. Therefore, we can analyze \( f(x) \) over the interval \( [0, 1] \) and then extend our findings to the entire interval \( [-100, 100] \).

- For \( x \in [0, 1) \):
\[ \{x\} = x \quad \text{and} \quad \{-x\} = 1 - x \]
Hence,
\[ f(x) = \min(x, 1 - x) \]
This function reaches its maximum at \( x = 0.5 \) where \( f(0.5) = 0.5 \).

### Step 3: Determine the Behavior of \( f(x) \)

- For \( x \in [0, 0.5) \):
\[ f(x) = x \]
- For \( x \in [0.5, 1) \):
\[ f(x) = 1 - x \]

### Step 4: Calculate the Integral Over One Period

Now we compute the integral over one period, \( [0, 1] \):
\[ \int_0^1 f(x) \, dx = \int_0^{0.5} x \, dx + \int_{0.5}^1 (1 - x) \, dx \]

Calculating the first integral:
\[ \int_0^{0.5} x \, dx = \left[ \frac{x^2}{2} \right]_0^{0.5} = \frac{(0.5)^2}{2} = \frac{0.25}{2} = 0.125 \]

Calculating the second integral:
\[ \int_{0.5}^1 (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0.5}^1 = \left( 1 - \frac{1}{2} \right) - \left( 0.5 - \frac{(0.5)^2}{2} \right) = \frac{1}{2} - \left( 0.5 - 0.125 \right) = \frac{1}{2} - 0.375 = 0.125 \]

Thus,
\[ \int_0^1 f(x) \, dx = 0.125 + 0.125 = 0.25 \]

### Step 5: Extend the Integral to the Interval \([-100, 100]\)

Since \( f(x) \) is periodic with period 1, the integral over \([-100, 100]\) can be calculated as:
\[ \int_{-100}^{100} f(x) \, dx = 200 \cdot \int_0^1 f(x) \, dx = 200 \cdot 0.25 = 50 \]

### Final Answer

Thus, the value of the integral \( \int_{-100}^{100} f(x) \, dx \) is:
\[ \boxed{50} \]