Let `I=int_(0)^(24pi){sinx}dx` then the value of 2I is equal to (where, `{.}` denotes the fractional part function)

Options

A. `10pi`
B. `24pi`
C. `12pi`
D. `4pi`

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{0}^{24\pi} \{\sin x\} \, dx \), where \( \{\cdot\} \) denotes the fractional part function. We will then find \( 2I \).

### Step-by-Step Solution:

1. **Understanding the Integral**:
The integral \( I = \int_{0}^{24\pi} \{\sin x\} \, dx \) involves the fractional part of \( \sin x \). The sine function oscillates between -1 and 1, and its periodicity is \( 2\pi \).

2. **Using the Property of the Fractional Part**:
The fractional part function \( \{x\} \) can be expressed as:
\[ \{x\} = x - \lfloor x \rfloor \]
Therefore, \( \{\sin x\} = \sin x \) when \( \sin x \geq 0 \) and \( \{\sin x\} = \sin x + 1 \) when \( \sin x < 0 \).

3. **Finding the Period of Integration**:
Since \( \sin x \) is periodic with a period of \( 2\pi \), we can break the integral from \( 0 \) to \( 24\pi \) into 12 intervals of \( 0 \) to \( 2\pi \):
\[ I = \int_{0}^{24\pi} \{\sin x\} \, dx = 12 \int_{0}^{2\pi} \{\sin x\} \, dx \]

4. **Evaluating the Integral from \( 0 \) to \( 2\pi \)**:
We need to evaluate \( \int_{0}^{2\pi} \{\sin x\} \, dx \). We can split this integral into two parts where \( \sin x \) is positive and negative:
- From \( 0 \) to \( \pi \), \( \sin x \geq 0 \) so \( \{\sin x\} = \sin x \).
- From \( \pi \) to \( 2\pi \), \( \sin x < 0 \) so \( \{\sin x\} = \sin x + 1 \).

Thus, we can write:
\[ \int_{0}^{2\pi} \{\sin x\} \, dx = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (\sin x + 1) \, dx \]

5. **Calculating Each Integral**:
- For \( \int_{0}^{\pi} \sin x \, dx \):
\[ \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]

- For \( \int_{\pi}^{2\pi} (\sin x + 1) \, dx \):
\[ \int_{\pi}^{2\pi} \sin x \, dx + \int_{\pi}^{2\pi} 1 \, dx = [-\cos x]_{\pi}^{2\pi} + [x]_{\pi}^{2\pi} = (-\cos(2\pi) + \cos(\pi)) + (2\pi - \pi) = (1 + 1) + \pi = 2 + \pi \]

6. **Combining the Results**:
Now we combine the results:
\[ \int_{0}^{2\pi} \{\sin x\} \, dx = 2 + (2 + \pi) = 4 + \pi \]

7. **Calculating \( I \)**:
Now substituting back into our expression for \( I \):
\[ I = 12(4 + \pi) = 48 + 12\pi \]

8. **Finding \( 2I \)**:
Finally, we compute \( 2I \):
\[ 2I = 2(48 + 12\pi) = 96 + 24\pi \]

9. **Finding the Fractional Part**:
The problem asks for \( \{2I\} \), which is the fractional part of \( 96 + 24\pi \). Since \( 24\pi \) is not an integer, we take:
\[ \{2I\} = 24\pi - \lfloor 24\pi \rfloor \]

### Final Answer:
The value of \( 2I \) is \( 96 + 24\pi \) and the fractional part is \( \{2I\} = 24\pi - \lfloor 24\pi \rfloor \).