The value of `int_(0)^([x]) {x-[x]} dx` is

Options

A. [x]
B. 2[x]
C. `(1)/(2[x])`
D. `(1)/(2) [x]`

Answer

To solve the integral \( \int_{0}^{[x]} (x - [x]) \, dx \), where \([x]\) is the greatest integer function, we can follow these steps:

### Step 1: Understand the Function
The expression \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). This function is periodic with a period of 1.

### Step 2: Set Up the Integral
We can express the integral as:
\[ \int_{0}^{[x]} (x - [x]) \, dx = \int_{0}^{[x]} \{x\} \, dx \]
Since \(\{x\}\) is periodic with period 1, we can use the property of integrals over periodic functions.

### Step 3: Use the Periodicity Property
The property states that:
\[ \int_{0}^{n} f(x) \, dx = n \int_{0}^{1} f(x) \, dx \]
for a periodic function \( f(x) \) with period 1. Here, \( n = [x] \).

### Step 4: Calculate the Integral Over One Period
Now we need to compute:
\[ \int_{0}^{1} \{x\} \, dx \]
In the interval \( [0, 1] \), \( \{x\} = x \). Therefore:
\[ \int_{0}^{1} \{x\} \, dx = \int_{0}^{1} x \, dx \]

### Step 5: Evaluate the Integral
Calculating the integral:
\[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

### Step 6: Apply the Result to the Original Integral
Now, using the periodicity property:
\[ \int_{0}^{[x]} (x - [x]) \, dx = [x] \int_{0}^{1} \{x\} \, dx = [x] \cdot \frac{1}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{[x]} (x - [x]) \, dx = \frac{1}{2} [x] \]