Let `f(x) = {x}`, the fractional part of `x` then `int_(-1)^(1) f(x) dx` is equal to

Options

A. `1`
B. `2`
C. `0`
D. `1//2`

Answer

To solve the integral \( \int_{-1}^{1} f(x) \, dx \) where \( f(x) = \{x\} \) (the fractional part of \( x \)), we will break the integral into two parts: from \(-1\) to \(0\) and from \(0\) to \(1\).

### Step-by-Step Solution:

1. **Define the Function**:
The fractional part function \( f(x) = \{x\} \) is defined as:
\[ f(x) = x - \lfloor x \rfloor \]
For \( x \in [-1, 0) \), \( \lfloor x \rfloor = -1 \), hence:
\[ f(x) = x - (-1) = x + 1 \]
For \( x \in [0, 1) \), \( \lfloor x \rfloor = 0 \), hence:
\[ f(x) = x - 0 = x \]

2. **Split the Integral**:
We can write the integral as:
\[ \int_{-1}^{1} f(x) \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx \]

3. **Evaluate the Integral from \(-1\) to \(0\)**:
For \( x \in [-1, 0) \):
\[ \int_{-1}^{0} f(x) \, dx = \int_{-1}^{0} (x + 1) \, dx \]
Now, calculate this integral:
\[ \int (x + 1) \, dx = \frac{x^2}{2} + x \]
Evaluating from \(-1\) to \(0\):
\[ \left[ \frac{0^2}{2} + 0 \right] - \left[ \frac{(-1)^2}{2} + (-1) \right] = 0 - \left[ \frac{1}{2} - 1 \right] = 0 - \left( \frac{1}{2} - 1 \right) = 0 + \frac{1}{2} = \frac{1}{2} \]

4. **Evaluate the Integral from \(0\) to \(1\)**:
For \( x \in [0, 1) \):
\[ \int_{0}^{1} f(x) \, dx = \int_{0}^{1} x \, dx \]
Now, calculate this integral:
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from \(0\) to \(1\):
\[ \left[ \frac{1^2}{2} \right] - \left[ \frac{0^2}{2} \right] = \frac{1}{2} - 0 = \frac{1}{2} \]

5. **Combine the Results**:
Now, we combine the results of both 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:
\[ \int_{-1}^{1} f(x) \, dx = 1 \]