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

Answer

To solve the integral

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

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 know that:

\[ 2x = [2x] + \{2x\} \]

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

\[ e^{2x - [2x]} = e^{\{2x\}} \]

This simplifies our integral to:

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

### Step 2: Change of variable
Let’s perform a substitution. Set \(t = 2x\). Then, we have:

\[ dt = 2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{2} \]

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

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

### Step 3: Analyze the periodicity of the fractional part
The function \(\{t\}\) is periodic with a period of 1. Therefore, we can break the integral into segments of length 1. The integral from \(-2\) to \(41\) can be calculated as follows:

1. From \(-2\) to \(0\): This covers 2 full periods.
2. From \(0\) to \(41\): This covers 41 full periods.

### Step 4: Calculate the integral over one period
The integral over one period from \(0\) to \(1\) is:

\[ \int_0^1 e^{\{t\}} \, dt = \int_0^1 e^{t} \, dt \]

Calculating this integral gives:

\[ \int_0^1 e^{t} \, dt = [e^{t}]_0^1 = e - 1 \]

### Step 5: Combine the results
Now, we can calculate the total integral:

1. From \(-2\) to \(0\): The integral is \(2 \cdot (e - 1)\).
2. From \(0\) to \(41\): The integral is \(41 \cdot (e - 1)\).

Thus, the total integral is:

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

### Step 6: Final result
Now, substituting back into our expression for the integral, we have:

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

Thus, the final answer is:

\[ \int_{-1}^{\frac{41}{2}} e^{2x - [2x]} \, dx = \frac{43}{2}(e - 1) \]