The value of `int_(0)^(1000)e^(x-[x])dx`, is ([.] denotes the greatest integer function) :

Options

A. 1000 e
B. 1000 (e - 1)
C. 1001 (e - 1)
D. None of these

Answer

To solve the integral \( I = \int_{0}^{1000} e^{x - [x]} \, dx \), where \([x]\) denotes the greatest integer function, we can break the integral into segments based on the behavior of the greatest integer function.

### Step 1: Understanding the Greatest Integer Function
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). For any integer \(n\), in the interval \(n \leq x < n+1\), we have \([x] = n\).

### Step 2: Breaking the Integral
We can break the integral from \(0\) to \(1000\) into \(1000\) segments:
\[ I = \sum_{n=0}^{999} \int_{n}^{n+1} e^{x - n} \, dx \]
This is because in each interval \([n, n+1)\), \([x] = n\).

### Step 3: Simplifying the Integral
Now we simplify the integral:
\[ I = \sum_{n=0}^{999} \int_{n}^{n+1} e^{x - n} \, dx = \sum_{n=0}^{999} e^{-n} \int_{n}^{n+1} e^{x} \, dx \]
The integral \(\int_{n}^{n+1} e^{x} \, dx\) can be computed as follows:
\[ \int e^{x} \, dx = e^{x} + C \]
Thus,
\[ \int_{n}^{n+1} e^{x} \, dx = e^{n+1} - e^{n} = e^{n}(e - 1) \]

### Step 4: Substituting Back
Now substituting back into the sum:
\[ I = \sum_{n=0}^{999} e^{-n} (e^{n}(e - 1)) = (e - 1) \sum_{n=0}^{999} 1 = (e - 1) \cdot 1000 \]

### Step 5: Final Calculation
Thus, we have:
\[ I = 1000(e - 1) \]

### Step 6: Applying the Greatest Integer Function
Since the question asks for the value of \(I\) in terms of the greatest integer function, we need to find:
\[ \lfloor I \rfloor = \lfloor 1000(e - 1) \rfloor \]
Using \(e \approx 2.718\), we calculate:
\[ e - 1 \approx 1.718 \implies 1000(e - 1) \approx 1718 \]
Thus,
\[ \lfloor 1000(e - 1) \rfloor = 1718 \]

### Final Answer
The value of \( \int_{0}^{1000} e^{x - [x]} \, dx \) is:
\[ \lfloor I \rfloor = 1718 \]