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 down the integral into segments based on the behavior of the greatest integer function.

### Step 1: Understand the Greatest Integer Function
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). For example:
- If \(0 \leq x < 1\), then \([x] = 0\).
- If \(1 \leq x < 2\), then \([x] = 1\).
- If \(2 \leq x < 3\), then \([x] = 2\).
- This pattern continues up to \(1000\).

### Step 2: Break the Integral into Intervals
We can express the integral as a sum of integrals over each interval where \([x]\) is constant:
\[ I = \sum_{n=0}^{999} \int_{n}^{n+1} e^{x - n} \, dx \]
This is because \([x] = n\) for \(x \in [n, n+1)\).

### Step 3: Evaluate Each Integral
Now, we evaluate each integral:
\[ \int_{n}^{n+1} e^{x - n} \, dx = \int_{n}^{n+1} e^{x} e^{-n} \, dx = e^{-n} \int_{n}^{n+1} e^{x} \, dx \]
The integral of \(e^{x}\) is:
\[ \int e^{x} \, dx = e^{x} + C \]
Thus,
\[ \int_{n}^{n+1} e^{x} \, dx = e^{n+1} - e^{n} = e^{n}(e - 1) \]
So,
\[ \int_{n}^{n+1} e^{x - n} \, dx = e^{-n} \cdot e^{n}(e - 1) = (e - 1) \]

### Step 4: Sum the Integrals
Now, we sum over all intervals from \(n = 0\) to \(n = 999\):
\[ I = \sum_{n=0}^{999} (e - 1) = 1000(e - 1) \]

### Step 5: Final Result
Thus, the value of the integral is:
\[ I = 1000(e - 1) \]