If {x} denotes the fractional part of x, then `overset(2)underset(0)int{x} dx` is equal to

Options

A. 1
B. 2
C. `(1)/(2)`
D. 4

Explanation

To solve the integral \(\int_{0}^{2} \{x\} \, dx\), where \(\{x\}\) denotes the fractional part of \(x\), we can follow these steps:

### Step 1: Understand the Fractional Part
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: Break the Integral into Intervals
Since we are integrating from \(0\) to \(2\), we can break this integral into two parts:
\[ \int_{0}^{2} \{x\} \, dx = \int_{0}^{1} \{x\} \, dx + \int_{1}^{2} \{x\} \, dx \]

### Step 3: Evaluate the Integral from 0 to 1
For \(0 \leq x < 1\), \(\lfloor x \rfloor = 0\). Therefore, the fractional part is:
\[ \{x\} = x - 0 = x \]
Thus, we have:
\[ \int_{0}^{1} \{x\} \, dx = \int_{0}^{1} x \, dx \]
Calculating this 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 4: Evaluate the Integral from 1 to 2
For \(1 \leq x < 2\), \(\lfloor x \rfloor = 1\). Therefore, the fractional part is:
\[ \{x\} = x - 1 \]
Thus, we have:
\[ \int_{1}^{2} \{x\} \, dx = \int_{1}^{2} (x - 1) \, dx \]
Calculating this integral:
\[ \int_{1}^{2} (x - 1) \, dx = \int_{1}^{2} x \, dx - \int_{1}^{2} 1 \, dx \]
Calculating each part:
\[ \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]
And,
\[ \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1 \]
Thus,
\[ \int_{1}^{2} (x - 1) \, dx = \frac{3}{2} - 1 = \frac{1}{2} \]

### Step 5: Combine the Results
Now, we can combine the results from both intervals:
\[ \int_{0}^{2} \{x\} \, dx = \int_{0}^{1} \{x\} \, dx + \int_{1}^{2} \{x\} \, dx = \frac{1}{2} + \frac{1}{2} = 1 \]

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