If {x} denotes the fractional part of x, then `int_(0)^(x)({x}-(1)/(2)) dx` is equal to

Options

A. `(1)/(2){x}({x}+1)`
B. `(1)/(2){x}({x}-1)`
C. `{x}({x}-1)`
D. `{x}({x}+1)`

Explanation

To solve the integral \( I = \int_{0}^{x} \left( \{x\} - \frac{1}{2} \right) dx \), where \(\{x\}\) denotes the fractional part of \(x\), we will follow these steps:

### Step 1: Understand the fractional part function
The fractional part of \(x\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\).

### Step 2: Split the integral based on the integer part
Let \(k = \lfloor x \rfloor\). The value of \(x\) can be expressed as \(k + f\), where \(f = \{x\} = x - k\) and \(0 \leq f < 1\). Therefore, we can split the integral into two parts:
\[ I = \int_{0}^{k} \left( \{x\} - \frac{1}{2} \right) dx + \int_{k}^{x} \left( \{x\} - \frac{1}{2} \right) dx \]

### Step 3: Evaluate the first integral
For \(0 \leq x < k\), \(\{x\} = x\), so:
\[ \int_{0}^{k} \left( \{x\} - \frac{1}{2} \right) dx = \int_{0}^{k} \left( x - \frac{1}{2} \right) dx \]
Calculating this integral:
\[ = \left[ \frac{x^2}{2} - \frac{x}{2} \right]_{0}^{k} = \left( \frac{k^2}{2} - \frac{k}{2} \right) - 0 = \frac{k^2 - k}{2} \]

### Step 4: Evaluate the second integral
For \(k \leq x < k + 1\), we have \(\{x\} = x - k\), thus:
\[ \int_{k}^{x} \left( \{x\} - \frac{1}{2} \right) dx = \int_{k}^{x} \left( (x - k) - \frac{1}{2} \right) dx \]
This simplifies to:
\[ = \int_{k}^{x} \left( x - k - \frac{1}{2} \right) dx = \int_{k}^{x} \left( x - k - \frac{1}{2} \right) dx \]
Calculating this integral:
\[ = \left[ \frac{(x - k)^2}{2} - \frac{(x - k)}{2} \right]_{k}^{x} = \left( \frac{(x - k)^2}{2} - \frac{(x - k)}{2} \right) - 0 \]
Evaluating this gives:
\[ = \frac{(x - k)^2}{2} - \frac{(x - k)}{2} \]

### Step 5: Combine the results
Now, combining both parts:
\[ I = \frac{k^2 - k}{2} + \left( \frac{(x - k)^2}{2} - \frac{(x - k)}{2} \right) \]
This can be simplified further, but we can also express it in terms of the fractional part:
\[ = \frac{1}{2} \left( (x - k)^2 - (x - k) + k^2 - k \right) \]
Recognizing that \(x - k = \{x\}\):
\[ I = \frac{1}{2} \left( \{x\}^2 - \{x\} + k^2 - k \right) \]

### Final Result
Thus, the integral evaluates to:
\[ I = \frac{1}{2} \{x\} (\{x\} - 1) \]