If the value of definite integral `A=int_(0)^(10pi)[sinx]dx` is equal to `kpi`, then the absolute value of k is equal to (where, `[.]` is the greatest integer function)

Answer

To solve the problem, we need to evaluate the definite integral \( A = \int_{0}^{10\pi} \lfloor \sin x \rfloor \, dx \), where \( \lfloor \cdot \rfloor \) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Understanding the Function**:
The sine function, \( \sin x \), oscillates between -1 and 1. The greatest integer function \( \lfloor \sin x \rfloor \) will take values:
- \( 0 \) when \( \sin x \) is in the interval \( [0, 1) \)
- \( -1 \) when \( \sin x \) is in the interval \( [-1, 0) \)

2. **Identifying Intervals**:
The sine function completes one full cycle from \( 0 \) to \( 2\pi \). In each cycle:
- From \( 0 \) to \( \pi \), \( \sin x \) goes from \( 0 \) to \( 1 \) (where \( \lfloor \sin x \rfloor = 0 \)).
- From \( \pi \) to \( 2\pi \), \( \sin x \) goes from \( 0 \) to \( -1 \) (where \( \lfloor \sin x \rfloor = -1 \)).

This pattern repeats every \( 2\pi \).

3. **Calculating the Integral Over One Cycle**:
- From \( 0 \) to \( \pi \):
\[ \int_{0}^{\pi} \lfloor \sin x \rfloor \, dx = \int_{0}^{\pi} 0 \, dx = 0 \]
- From \( \pi \) to \( 2\pi \):
\[ \int_{\pi}^{2\pi} \lfloor \sin x \rfloor \, dx = \int_{\pi}^{2\pi} (-1) \, dx = -\left[ x \right]_{\pi}^{2\pi} = -\left( 2\pi - \pi \right) = -\pi \]

4. **Combining the Results**:
The integral over one full cycle \( [0, 2\pi] \) is:
\[ \int_{0}^{2\pi} \lfloor \sin x \rfloor \, dx = 0 - \pi = -\pi \]

5. **Extending to \( 10\pi \)**:
Since \( 10\pi \) consists of \( 5 \) full cycles of \( 2\pi \):
\[ A = 5 \cdot \int_{0}^{2\pi} \lfloor \sin x \rfloor \, dx = 5 \cdot (-\pi) = -5\pi \]

6. **Finding \( k \)**:
We have \( A = k\pi \). From our calculation, \( A = -5\pi \), thus:
\[ k = -5 \]

7. **Finding the Absolute Value**:
The absolute value of \( k \) is:
\[ |k| = |-5| = 5 \]

### Final Answer:
The absolute value of \( k \) is \( 5 \).