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 \( I = \int_0^{\pi} \sin[x] \, dx \) and relate it to the summation provided in the question.

### Step 1: Break down the integral
We know that the greatest integer function \( [x] \) takes integer values. Therefore, we can break the integral into segments based on the intervals defined by integers:

\[ I = \int_0^{\pi} \sin[x] \, dx = \int_0^1 \sin[0] \, dx + \int_1^2 \sin[1] \, dx + \int_2^3 \sin[2] \, dx + \int_3^{\pi} \sin[3] \, dx \]

### Step 2: Evaluate each segment
1. From \( 0 \) to \( 1 \):
\[ \int_0^1 \sin[0] \, dx = \int_0^1 0 \, dx = 0 \]

2. From \( 1 \) to \( 2 \):
\[ \int_1^2 \sin[1] \, dx = \sin(1) \cdot (2 - 1) = \sin(1) \]

3. From \( 2 \) to \( 3 \):
\[ \int_2^3 \sin[2] \, dx = \sin(2) \cdot (3 - 2) = \sin(2) \]

4. From \( 3 \) to \( \pi \):
\[ \int_3^{\pi} \sin[3] \, dx = \sin(3) \cdot (\pi - 3) \]

### Step 3: Combine the results
Now, we can combine all these results to find \( I \):
\[ I = 0 + \sin(1) + \sin(2) + \sin(3)(\pi - 3) \]

### Step 4: Relate to the summation
According to the problem, we have:
\[ I = \sum_{\epsilon=0}^{n} a_{\epsilon} \sin \epsilon \]
From the integral evaluation, we can identify the coefficients:
- \( a_0 = 0 \)
- \( a_1 = 1 \)
- \( a_2 = 1 \)
- \( a_3 = \pi - 3 \)

### Step 5: Determine \( n \)
The maximum value of \( n \) in the summation corresponds to the highest integer value used, which is \( n = 3 \).

### Step 6: Calculate \( a_n + n \)
Now we need to find \( a_n + n \):
\[ a_3 + n = (\pi - 3) + 3 = \pi \]

### Step 7: Find the final value
Finally, we need to calculate:
\[ \frac{a_n + n}{\pi} = \frac{\pi}{\pi} = 1 \]

### Final Answer
Thus, the value of \( \frac{a_n + n}{\pi} \) is equal to \( 1 \).

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