`int_(-1)^(41//2)e^(2x-[2x])dx`, where `[*]` denotes the greatest integer function.

Answer

To solve the integral

\[ I = \int_{-1}^{\frac{41}{2}} e^{2x - [2x]} \, dx, \]

where \([2x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Rewrite the integrand
Using the property of the greatest integer function, we can express \( [2x] \) as:

\[ [2x] = 2x - \{2x\}, \]

where \(\{2x\}\) is the fractional part of \(2x\). Thus, we can rewrite the integrand:

\[ e^{2x - [2x]} = e^{2x - (2x - \{2x\})} = e^{\{2x\}}. \]

So, the integral becomes:

\[ I = \int_{-1}^{\frac{41}{2}} e^{\{2x\}} \, dx. \]

### Step 2: Change of variable
To simplify the integral, we can perform a substitution. Let:

\[ t = 2x \implies dx = \frac{dt}{2}. \]

We also need to change the limits of integration. When \(x = -1\), \(t = 2(-1) = -2\), and when \(x = \frac{41}{2}\), \(t = 2 \cdot \frac{41}{2} = 41\). Thus, the integral becomes:

\[ I = \int_{-2}^{41} e^{\{t\}} \cdot \frac{dt}{2} = \frac{1}{2} \int_{-2}^{41} e^{\{t\}} \, dt. \]

### Step 3: Evaluate the integral
The function \(\{t\}\) has a period of 1, meaning it repeats every integer. Therefore, we can split the integral into segments from \(-2\) to \(41\):

\[ \int_{-2}^{41} e^{\{t\}} \, dt = \int_{-2}^{-1} e^{\{t\}} \, dt + \int_{-1}^{0} e^{\{t\}} \, dt + \sum_{n=0}^{40} \int_{n}^{n+1} e^{\{t\}} \, dt. \]

### Step 4: Calculate each segment
1. **From \(-2\) to \(-1\)**:
- Here, \(\{t\} = t + 2\) (since \(t\) is in the range \([-2, -1)\)):
\[ \int_{-2}^{-1} e^{\{t\}} \, dt = \int_{-2}^{-1} e^{t + 2} \, dt = e^2 \int_{-2}^{-1} e^t \, dt = e^2 \left[e^t\right]_{-2}^{-1} = e^2 \left(e^{-1} - e^{-2}\right) = e^2 \left(\frac{1}{e} - \frac{1}{e^2}\right) = e^2 \left(\frac{e-1}{e^2}\right) = e - 1. \]

2. **From \(-1\) to \(0\)**:
- Here, \(\{t\} = t + 1\):
\[ \int_{-1}^{0} e^{\{t\}} \, dt = \int_{-1}^{0} e^{t + 1} \, dt = e \int_{-1}^{0} e^t \, dt = e \left[e^t\right]_{-1}^{0} = e \left(1 - \frac{1}{e}\right) = e - 1. \]

3. **From \(0\) to \(41\)**:
- Each integral from \(n\) to \(n+1\) where \(n\) is an integer contributes the same value:
\[ \int_{n}^{n+1} e^{\{t\}} \, dt = \int_{0}^{1} e^{t} \, dt = [e^t]_{0}^{1} = e - 1. \]
- There are \(41\) such intervals (from \(0\) to \(40\)), thus:
\[ \sum_{n=0}^{40} \int_{n}^{n+1} e^{\{t\}} \, dt = 41(e - 1). \]

### Step 5: Combine all parts
Now, we can combine all parts:

\[ \int_{-2}^{41} e^{\{t\}} \, dt = (e - 1) + (e - 1) + 41(e - 1) = 2(e - 1) + 41(e - 1) = 43(e - 1). \]

### Final Step: Multiply by \(\frac{1}{2}\)
Finally, we multiply by \(\frac{1}{2}\):

\[ I = \frac{1}{2} \cdot 43(e - 1) = \frac{43}{2}(e - 1). \]

Thus, the final answer is:

\[ \boxed{\frac{43}{2}(e - 1)}. \]