Prove that `int_(0)^(25)e^(x-[x])dx=25(e-1)`.

Answer

To prove that

\[ \int_{0}^{25} e^{x - [x]} \, dx = 25(e - 1), \]

we will break down the integral using the properties of the greatest integer function \([x]\).

### Step 1: Understanding the function \(e^{x - [x]}\)

The expression \(x - [x]\) represents the fractional part of \(x\), denoted as \(\{x\}\). Thus, we can rewrite the integral as:

\[ \int_{0}^{25} e^{x - [x]} \, dx = \int_{0}^{25} e^{\{x\}} \, dx. \]

### Step 2: Splitting the integral

The function \([x]\) is constant within each interval \([n, n+1)\) for \(n = 0, 1, 2, \ldots, 24\). Therefore, we can split the integral into 25 parts:

\[ \int_{0}^{25} e^{\{x\}} \, dx = \sum_{n=0}^{24} \int_{n}^{n+1} e^{\{x\}} \, dx. \]

### Step 3: Evaluating each integral

Within each interval \([n, n+1)\), we have:

\[ \{x\} = x - n. \]

Thus, we can rewrite the integral over each interval:

\[ \int_{n}^{n+1} e^{\{x\}} \, dx = \int_{n}^{n+1} e^{x - n} \, dx = e^{-n} \int_{n}^{n+1} e^{x} \, dx. \]

Now, we can evaluate the integral:

\[ \int_{n}^{n+1} e^{x} \, dx = [e^{x}]_{n}^{n+1} = e^{n+1} - e^{n}. \]

### Step 4: Combining the results

Now substituting back, we have:

\[ \int_{n}^{n+1} e^{\{x\}} \, dx = e^{-n} (e^{n+1} - e^{n}) = e^{-n} e^{n} (e - 1) = (e - 1). \]

### Step 5: Summing over all intervals

Since there are 25 intervals from \(n = 0\) to \(n = 24\), we can sum the results:

\[ \int_{0}^{25} e^{\{x\}} \, dx = \sum_{n=0}^{24} (e - 1) = 25(e - 1). \]

### Conclusion

Thus, we have shown that:

\[ \int_{0}^{25} e^{x - [x]} \, dx = 25(e - 1). \]