If the definite integral `I=int_(0)^(pi)sin[x] dx=sum_(epsilon=0)^(n)a_(epsilon)sin epsilon`. (where, `[.]` is the greatest integer function), then the vlaue of `(a_(n)+n)/(pi)` is equal to

Answer

To solve the given problem, we need to evaluate the definite integral and relate it to the sum involving the greatest integer function. Let's break down the solution step by step.

### Step 1: Evaluate the definite integral
We start with the integral:
\[ I = \int_{0}^{\pi} \sin[x] \, dx \]
The integral of \(\sin[x]\) can be computed as:
\[ I = -\cos[x] \bigg|_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

**Hint for Step 1:** Remember that the integral of \(\sin[x]\) is \(-\cos[x]\) and evaluate it at the limits.

### Step 2: Understand the sum involving the greatest integer function
The problem states:
\[ I = \sum_{\epsilon=0}^{n} a_{\epsilon} \sin(\epsilon) \]
We need to determine the coefficients \(a_{\epsilon}\) based on the intervals defined by the greatest integer function.

### Step 3: Break down the integral into intervals
The integral from \(0\) to \(\pi\) can be broken down into intervals based on the greatest integer function:
- From \(0\) to \(1\): \(\sin(0)\)
- From \(1\) to \(2\): \(\sin(1)\)
- From \(2\) to \(3\): \(\sin(2)\)
- From \(3\) to \(\pi\): \(\sin(3)\) (and the remaining part up to \(\pi\))

Thus, we can express the integral as:
\[ I = \int_{0}^{1} \sin(x) \, dx + \int_{1}^{2} \sin(x) \, dx + \int_{2}^{3} \sin(x) \, dx + \int_{3}^{\pi} \sin(x) \, dx \]

### Step 4: Calculate each integral
- \(\int_{0}^{1} \sin(x) \, dx = 1 - \cos(1)\)
- \(\int_{1}^{2} \sin(x) \, dx = \cos(1) - \cos(2)\)
- \(\int_{2}^{3} \sin(x) \, dx = \cos(2) - \cos(3)\)
- \(\int_{3}^{\pi} \sin(x) \, dx = \cos(3) - (-1) = \cos(3) + 1\)

Now, summing these gives:
\[ I = (1 - \cos(1)) + (\cos(1) - \cos(2)) + (\cos(2) - \cos(3)) + (\cos(3) + 1) \]
This simplifies to:
\[ I = 2 \]

### Step 5: Identify coefficients \(a_{\epsilon}\)
From the intervals, we can deduce:
- \(a_0 = 1\) (from \(0\) to \(1\))
- \(a_1 = 1\) (from \(1\) to \(2\))
- \(a_2 = 1\) (from \(2\) to \(3\))
- \(a_3 = \pi - 3\) (from \(3\) to \(\pi\))

### Step 6: Calculate \(a_n + n\)
For \(n = 3\):
\[ a_3 = \pi - 3 \]
Thus,
\[ a_n + n = (\pi - 3) + 3 = \pi \]

### Step 7: Find \(\frac{a_n + n}{\pi}\)
Finally, we compute:
\[ \frac{a_n + n}{\pi} = \frac{\pi}{\pi} = 1 \]

### Final Answer
The value of \(\frac{a_n + n}{\pi}\) is equal to:
\[ \boxed{1} \]