The value of `int_(0)^(100) e^(x- [x])dx` is

Options

A. 100e
B. `100 (e-1)`
C. `100 (e+1)`
D. none

Answer

To solve the integral \( \int_{0}^{100} e^{x - [x]} \, dx \), we can use the periodic property of the function involved. Here’s a step-by-step solution:

### Step 1: Understand the function
The expression \( e^{x - [x]} \) can be rewritten as \( e^{\{x\}} \), where \( \{x\} \) is the fractional part of \( x \). The fractional part function \( \{x\} = x - [x] \) is periodic with a period of 1.

### Step 2: Use the periodic property
Since \( e^{\{x\}} \) is periodic with period 1, we can express the integral from 0 to 100 as a sum of integrals over each interval of length 1:
\[ \int_{0}^{100} e^{x - [x]} \, dx = \int_{0}^{100} e^{\{x\}} \, dx = 100 \int_{0}^{1} e^{\{x\}} \, dx \]
This is because there are 100 intervals of length 1 from 0 to 100.

### Step 3: Calculate the integral from 0 to 1
Now we need to compute \( \int_{0}^{1} e^{\{x\}} \, dx \). Since \( \{x\} = x \) for \( x \) in the interval [0, 1), we have:
\[ \int_{0}^{1} e^{\{x\}} \, dx = \int_{0}^{1} e^{x} \, dx \]

### Step 4: Evaluate the integral
The integral \( \int e^{x} \, dx \) is \( e^{x} \). Therefore:
\[ \int_{0}^{1} e^{x} \, dx = \left[ e^{x} \right]_{0}^{1} = e^{1} - e^{0} = e - 1 \]

### Step 5: Combine the results
Now substituting back into our earlier expression:
\[ \int_{0}^{100} e^{x - [x]} \, dx = 100 \int_{0}^{1} e^{x} \, dx = 100 (e - 1) \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{100} e^{x - [x]} \, dx \) is:
\[ \boxed{100(e - 1)} \]