Let `I_n=int_(-n)^n({x+1}*{x^2+2}+{x^2+2}{x^3+4})dx`, where {*} denotet the fractional part of x. Find `I_1.`

Options

A. `-(1)/(3)`
B. `-(2)/(3)`
C. `(1)/(3)`
D. None of these

Answer

To solve the integral \( I_n = \int_{-n}^{n} \left( \{x+1\} \cdot \{x^2+2\} + \{x^2+2\} \cdot \{x^3+4\} \right) dx \), where \( \{x\} \) denotes the fractional part of \( x \), we will specifically find \( I_1 \).

### Step-by-Step Solution

1. **Set Up the Integral**:
We need to evaluate:
\[ I_1 = \int_{-1}^{1} \left( \{x+1\} \cdot \{x^2+2\} + \{x^2+2\} \cdot \{x^3+4\} \right) dx \]

2. **Evaluate the Fractional Parts**:
For \( x \in [-1, 1] \):
- \( \{x+1\} = x + 1 \) (since \( x + 1 \) ranges from 0 to 2)
- \( \{x^2 + 2\} = x^2 \) (since \( x^2 + 2 \) ranges from 2 to 3)
- \( \{x^3 + 4\} = x^3 \) (since \( x^3 + 4 \) ranges from 3 to 4)

3. **Substitute the Fractional Parts into the Integral**:
Thus, we can rewrite the integral as:
\[ I_1 = \int_{-1}^{1} \left( (x + 1) \cdot x^2 + x^2 \cdot x^3 \right) dx \]

4. **Simplify the Expression**:
Expanding the integrand:
\[ I_1 = \int_{-1}^{1} \left( x^3 + x^2 + x^5 \right) dx \]

5. **Break Down the Integral**:
We can split the integral:
\[ I_1 = \int_{-1}^{1} x^3 \, dx + \int_{-1}^{1} x^2 \, dx + \int_{-1}^{1} x^5 \, dx \]

6. **Evaluate Each Integral**:
- \( \int_{-1}^{1} x^3 \, dx = 0 \) (odd function)
- \( \int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \)
- \( \int_{-1}^{1} x^5 \, dx = 0 \) (odd function)

7. **Combine the Results**:
Thus, we have:
\[ I_1 = 0 + \frac{2}{3} + 0 = \frac{2}{3} \]

8. **Final Result**:
Therefore, the value of \( I_1 \) is:
\[ I_1 = \frac{2}{3} \]