Let `f(x) = x-[x]`, for every real number x, where [x] is integral part of x. Then `int_(-1) ^1 f(x) dx` is

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To solve the integral \( \int_{-1}^{1} f(x) \, dx \) where \( f(x) = x - [x] \) (the fractional part of \( x \)), we can follow these steps:

### Step 1: Understand the function \( f(x) \)
The function \( f(x) = x - [x] \) represents the fractional part of \( x \). This function has a periodic behavior with a period of 1. Specifically, for any integer \( n \):
- For \( x \in [n, n+1) \), \( f(x) = x - n \).

### Step 2: Break the integral into intervals
Since \( f(x) \) is periodic, we can break the integral from \(-1\) to \(1\) into two parts:
\[ \int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx \]

### Step 3: Evaluate \( f(x) \) on the intervals
1. **For \( x \in [-1, 0) \)**:
- Here, \( [x] = -1 \), thus \( f(x) = x - (-1) = x + 1 \).
- Therefore, \( f(x) = x + 1 \) on this interval.

2. **For \( x \in [0, 1) \)**:
- Here, \( [x] = 0 \), thus \( f(x) = x - 0 = x \).
- Therefore, \( f(x) = x \) on this interval.

### Step 4: Calculate the integrals
Now we can compute each integral separately.

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

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

### Step 5: Combine the results
Now, we can combine the results of the two integrals:
\[ \int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

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