To solve the integral \( I = \int_0^{10} [x] e^{[x]} e^{-(x-1)} \, dx \), where \([x]\) is the greatest integer function (GIF), we can break down the integral into intervals where the value of \([x]\) is constant.
### Step 1: Identify the intervals
The greatest integer function \([x]\) takes constant integer values in the intervals:
- \([0, 1)\) where \([x] = 0\)
- \([1, 2)\) where \([x] = 1\)
- \([2, 3)\) where \([x] = 2\)
- \([3, 4)\) where \([x] = 3\)
- \([4, 5)\) where \([x] = 4\)
- \([5, 6)\) where \([x] = 5\)
- \([6, 7)\) where \([x] = 6\)
- \([7, 8)\) where \([x] = 7\)
- \([8, 9)\) where \([x] = 8\)
- \([9, 10)\) where \([x] = 9\)
### Step 2: Break the integral into parts
We can express the integral as a sum of integrals over each interval:
\[
I = \sum_{n=0}^{9} \int_n^{n+1} [x] e^{[x]} e^{-(x-1)} \, dx
\]
This simplifies to:
\[
I = \sum_{n=0}^{9} \int_n^{n+1} n e^n e^{-(x-1)} \, dx
\]
### Step 3: Evaluate each integral
For each interval \([n, n+1)\):
\[
I_n = n e^n \int_n^{n+1} e^{-(x-1)} \, dx
\]
The integral can be computed as:
\[
\int e^{-(x-1)} \, dx = -e^{-(x-1)}
\]
Evaluating from \(n\) to \(n+1\):
\[
I_n = n e^n \left[-e^{-(x-1)}\right]_{n}^{n+1} = n e^n \left[-e^{-(n+1-1)} + e^{-(n-1)}\right]
\]
This simplifies to:
\[
I_n = n e^n \left[-e^{-n} + e^{-(n-1)}\right] = n e^n \left[-\frac{1}{e^n} + \frac{e}{e^n}\right] = n e^n \left[\frac{e - 1}{e^n}\right] = n (e - 1)
\]
### Step 4: Sum over all intervals
Now we sum \(I_n\) from \(n=0\) to \(n=9\):
\[
I = \sum_{n=0}^{9} n (e - 1) = (e - 1) \sum_{n=0}^{9} n
\]
The sum of the first 9 natural numbers is:
\[
\sum_{n=0}^{9} n = \frac{9 \cdot (9 + 1)}{2} = \frac{9 \cdot 10}{2} = 45
\]
Thus,
\[
I = (e - 1) \cdot 45
\]
### Final Result
The value of the integral is:
\[
I = 45(e - 1)
\]
