For any `n in R^(+)`, the value of the integral
`int_(0)^(n[x]) (x-[x])dx` is

Options

A. n[x]
B. [x]
C. `(n)/(2)[x]`
D. none of these

Answer

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

### Step 1: Rewrite the integral
The expression \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). Therefore, we can rewrite the integral as:
\[ I = \int_{0}^{n} \{x\} \, dx \]

### Step 2: Understand the periodicity
The fractional part function \( \{x\} \) is periodic with a period of 1. This means that for any integer \( k \):
\[ \{x + k\} = \{x\} \]
Thus, we can break the integral from 0 to \( n \) into intervals of length 1.

### Step 3: Break the integral into intervals
If \( n \) is a positive real number, we can express \( n \) as \( n = m + f \), where \( m = [n] \) (the greatest integer less than or equal to \( n \)) and \( f = n - m \) (the fractional part of \( n \)). We can then express the integral as:
\[ I = \int_{0}^{m} \{x\} \, dx + \int_{m}^{n} \{x\} \, dx \]

### Step 4: Evaluate the integral over the integer intervals
For the first part, from 0 to \( m \):
\[ \int_{0}^{m} \{x\} \, dx = \sum_{k=0}^{m-1} \int_{k}^{k+1} \{x\} \, dx \]
Within each interval \( [k, k+1] \), \( \{x\} = x - k \). Therefore:
\[ \int_{k}^{k+1} \{x\} \, dx = \int_{k}^{k+1} (x - k) \, dx = \int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{2} \]
Thus, for \( m \) intervals:
\[ \int_{0}^{m} \{x\} \, dx = m \cdot \frac{1}{2} = \frac{m}{2} \]

### Step 5: Evaluate the integral from \( m \) to \( n \)
For the second part, from \( m \) to \( n \):
\[ \int_{m}^{n} \{x\} \, dx = \int_{m}^{n} (x - m) \, dx = \int_{0}^{f} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{f} = \frac{f^2}{2} \]

### Step 6: Combine the results
Now, combining both parts, we have:
\[ I = \frac{m}{2} + \frac{f^2}{2} \]
Since \( m = [n] \) and \( f = n - [n] \), we can express the final result as:
\[ I = \frac{[n]}{2} + \frac{(n - [n])^2}{2} \]

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