If `int_0^25e^(x-[x])dx=k(e-1)`, then the value of k is equal to

Options

A. 12
B. 25
C. 23
D. 24

Answer

To solve the integral \( \int_0^{25} e^{x - [x]} \, dx = k(e - 1) \), we will follow these steps:

### Step 1: Understand the integrand
The expression \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). This function has a period of 1, meaning it repeats every integer.

### Step 2: Rewrite the integral
Since the function \( e^{x - [x]} = e^{\{x\}} \) is periodic with a period of 1, we can break the integral from 0 to 25 into 25 intervals of length 1:
\[ \int_0^{25} e^{x - [x]} \, dx = \sum_{n=0}^{24} \int_n^{n+1} e^{x - [x]} \, dx \]
This simplifies to:
\[ \int_0^{25} e^{x - [x]} \, dx = \sum_{n=0}^{24} \int_0^1 e^{x} \, dx \]

### Step 3: Evaluate the integral over one period
Now we need to compute \( \int_0^1 e^{x} \, dx \):
\[ \int_0^1 e^{x} \, dx = [e^{x}]_0^1 = e^1 - e^0 = e - 1 \]

### Step 4: Sum the contributions from each interval
Since there are 25 intervals from 0 to 25, we have:
\[ \int_0^{25} e^{x - [x]} \, dx = 25 \int_0^1 e^{x} \, dx = 25(e - 1) \]

### Step 5: Set the equation equal to the given expression
We know from the problem statement that:
\[ \int_0^{25} e^{x - [x]} \, dx = k(e - 1) \]
Thus, we can equate:
\[ 25(e - 1) = k(e - 1) \]

### Step 6: Solve for \( k \)
Assuming \( e - 1 \neq 0 \) (which is true since \( e \approx 2.718 \)), we can divide both sides by \( e - 1 \):
\[ k = 25 \]

### Conclusion
The value of \( k \) is \( \boxed{25} \).