Evaluate :
`(i) int_(-1)^(2){2x}dx` (where function`{*}` denotes fractional part function)

Answer

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

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

### Step 2: Analyze the function \(\{2x\}\)
We need to evaluate \(\{2x\}\) over the interval \([-1, 2]\).

- For \(x \in [-1, 0)\):
- \(2x \in [-2, 0)\)
- Thus, \(\{2x\} = 2x - (-2) = 2x + 2\)

- For \(x \in [0, 1)\):
- \(2x \in [0, 2)\)
- Thus, \(\{2x\} = 2x - 0 = 2x\)

- For \(x \in [1, 2]\):
- \(2x \in [2, 4)\)
- Thus, \(\{2x\} = 2x - 2 = 2x - 2\)

### Step 3: Break the integral into parts
Now, we can break the integral into three parts based on the intervals:
\[ I = \int_{-1}^{0} (2x + 2) \, dx + \int_{0}^{1} (2x) \, dx + \int_{1}^{2} (2x - 2) \, dx \]

### Step 4: Evaluate each integral
1. **First Integral:**
\[ \int_{-1}^{0} (2x + 2) \, dx = \int_{-1}^{0} 2x \, dx + \int_{-1}^{0} 2 \, dx \]
- \(\int_{-1}^{0} 2x \, dx = [x^2]_{-1}^{0} = 0 - 1 = -1\)
- \(\int_{-1}^{0} 2 \, dx = [2x]_{-1}^{0} = 0 - (-2) = 2\)
- Thus, the first integral is:
\[ -1 + 2 = 1 \]

2. **Second Integral:**
\[ \int_{0}^{1} 2x \, dx = [x^2]_{0}^{1} = 1 - 0 = 1 \]

3. **Third Integral:**
\[ \int_{1}^{2} (2x - 2) \, dx = \int_{1}^{2} 2x \, dx - \int_{1}^{2} 2 \, dx \]
- \(\int_{1}^{2} 2x \, dx = [x^2]_{1}^{2} = 4 - 1 = 3\)
- \(\int_{1}^{2} 2 \, dx = [2x]_{1}^{2} = 4 - 2 = 2\)
- Thus, the third integral is:
\[ 3 - 2 = 1 \]

### Step 5: Combine the results
Now, we can combine the results of the three integrals:
\[ I = 1 + 1 + 1 = 3 \]

### Final Answer:
Thus, the value of the integral is:
\[ \boxed{3} \]