Evaluate :
(i) `int_(-1)^(2){2x}dx` (where function`{*}` denotes fractional part function)
(ii) `int_(0)^(10x)(|sinx|+|cosx|) dx`
(iii) `(int_(0)^(n)[x]dx)/(int_(0)^(n)[x]dx)` where [x] and `{x}` are integral and fractional parts of the x and `n in N`
(iv) `int_(0)^(210)(|sinx|-[|(sinx)/(2)|])dx` (where [] denotes the greatest integer function and `n in 1`)

Answer

To solve the given integrals step by step, let's break down each part of the question.

### (i) Evaluate \(\int_{-1}^{2} 2\{x\} \, dx\)

1. **Understanding 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 \(x\).

2. **Breaking the Integral into Intervals**: We need to evaluate the integral from \(-1\) to \(2\). The function \(\{x\}\) behaves differently in different intervals:
- From \(-1\) to \(0\), \(\{x\} = x + 1\).
- From \(0\) to \(1\), \(\{x\} = x\).
- From \(1\) to \(2\), \(\{x\} = x - 1\).

3. **Setting Up the Integral**:
\[ \int_{-1}^{2} 2\{x\} \, dx = \int_{-1}^{0} 2(x + 1) \, dx + \int_{0}^{1} 2x \, dx + \int_{1}^{2} 2(x - 1) \, dx \]

4. **Calculating Each Integral**:
- For \(\int_{-1}^{0} 2(x + 1) \, dx\):
\[ = \int_{-1}^{0} (2x + 2) \, dx = \left[x^2 + 2x\right]_{-1}^{0} = (0 + 0) - (1 - 2) = 1 \]
- For \(\int_{0}^{1} 2x \, dx\):
\[ = \left[x^2\right]_{0}^{1} = 1 - 0 = 1 \]
- For \(\int_{1}^{2} 2(x - 1) \, dx\):
\[ = \int_{1}^{2} (2x - 2) \, dx = \left[x^2 - 2x\right]_{1}^{2} = (4 - 4) - (1 - 2) = 1 \]

5. **Summing the Results**:
\[ \int_{-1}^{2} 2\{x\} \, dx = 1 + 1 + 1 = 3 \]