The value of `int_(-1)^(1)(x-[x])dx`, (where [.] denotes greatest integer function)is

Options

A. `0`
B. `1`
C. 3
D. `2`

Answer

To solve the integral \( \int_{-1}^{1} (x - [x]) \, dx \), where \([x]\) denotes the greatest integer function, we can break it down step by step.

### Step 1: Understanding the Function
The function \( x - [x] \) represents the fractional part of \( x \). This function behaves differently depending on the interval in which \( x \) lies. Specifically:
- For \( x \) in the interval \([-1, 0)\), \([x] = -1\), so \( x - [x] = x + 1 \).
- For \( x \) in the interval \([0, 1)\), \([x] = 0\), so \( x - [x] = x \).

### Step 2: Splitting the Integral
We can split the integral into two parts based on the intervals identified:
\[ \int_{-1}^{1} (x - [x]) \, dx = \int_{-1}^{0} (x + 1) \, dx + \int_{0}^{1} x \, dx \]

### Step 3: Calculating the First Integral
Now we calculate the first integral:
\[ \int_{-1}^{0} (x + 1) \, dx \]
To evaluate this, we find the antiderivative:
\[ \int (x + 1) \, dx = \frac{x^2}{2} + x \]
Now we evaluate it from \(-1\) to \(0\):
\[ \left[ \frac{x^2}{2} + x \right]_{-1}^{0} = \left( \frac{0^2}{2} + 0 \right) - \left( \frac{(-1)^2}{2} - 1 \right) = 0 - \left( \frac{1}{2} - 1 \right) = 0 - \left( -\frac{1}{2} \right) = \frac{1}{2} \]

### Step 4: Calculating the Second Integral
Next, we calculate the second integral:
\[ \int_{0}^{1} x \, dx \]
The antiderivative is:
\[ \int x \, dx = \frac{x^2}{2} \]
Now we evaluate it from \(0\) to \(1\):
\[ \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} - 0 = \frac{1}{2} \]

### Step 5: Adding the Results
Now we add the results of both integrals:
\[ \int_{-1}^{1} (x - [x]) \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

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