Evaluate `int_(0)^(2){x} d x `, where `{x}` denotes the fractional part of x.

Answer

To evaluate the integral \( I = \int_{0}^{2} \{x\} \, dx \), where \( \{x\} \) denotes the fractional part of \( x \), we can use the property of the fractional part function.

### Step 1: Rewrite the Fractional Part
The fractional part of \( x \) can be expressed as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). Thus, we can rewrite the integral as:
\[ I = \int_{0}^{2} (x - \lfloor x \rfloor) \, dx \]

### Step 2: Split the Integral
We can split the integral into two parts:
\[ I = \int_{0}^{2} x \, dx - \int_{0}^{2} \lfloor x \rfloor \, dx \]

### Step 3: Evaluate the First Integral
The first integral is straightforward:
\[ \int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2 \]

### Step 4: Evaluate the Second Integral
Next, we need to evaluate \( \int_{0}^{2} \lfloor x \rfloor \, dx \). We can break this integral into two intervals, from 0 to 1 and from 1 to 2:
\[ \int_{0}^{2} \lfloor x \rfloor \, dx = \int_{0}^{1} \lfloor x \rfloor \, dx + \int_{1}^{2} \lfloor x \rfloor \, dx \]

- For \( x \) in the interval \([0, 1)\), \( \lfloor x \rfloor = 0 \):
\[ \int_{0}^{1} \lfloor x \rfloor \, dx = \int_{0}^{1} 0 \, dx = 0 \]

- For \( x \) in the interval \([1, 2)\), \( \lfloor x \rfloor = 1 \):
\[ \int_{1}^{2} \lfloor x \rfloor \, dx = \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1 \]

### Step 5: Combine the Results
Now we can combine the results:
\[ I = 2 - (0 + 1) = 2 - 1 = 1 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{1} \]