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

Options

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

Explanation

To solve the integral \( I = \int_{0}^{100} e^{x - [x]} \, dx \), where \([x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Understanding the Function
The expression \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). Therefore, we can rewrite the integral as:
\[ I = \int_{0}^{100} e^{\{x\}} \, dx \]

### Step 2: Periodicity of the Function
The fractional part function \( \{x\} \) is periodic with a period of 1. This means:
\[ \{x + 1\} = \{x\} \]
Thus, the integral from 0 to 100 can be broken down into 100 intervals of length 1:
\[ I = \sum_{n=0}^{99} \int_{n}^{n+1} e^{\{x\}} \, dx \]

### Step 3: Evaluating the Integral Over One Period
For \( n \leq x < n+1 \), we have \( \{x\} = x - n \). Therefore, we can rewrite the integral over one period:
\[ \int_{n}^{n+1} e^{\{x\}} \, dx = \int_{n}^{n+1} e^{x - n} \, dx = e^{-n} \int_{n}^{n+1} e^{x} \, dx \]

### Step 4: Calculating the Integral
Now, we evaluate the integral:
\[ \int_{n}^{n+1} e^{x} \, dx = [e^{x}]_{n}^{n+1} = e^{n+1} - e^{n} \]
Thus,
\[ \int_{n}^{n+1} e^{\{x\}} \, dx = e^{-n} (e^{n+1} - e^{n}) = e^{-n} e^{n} (e - 1) = (e - 1) \]

### Step 5: Summing Over All Periods
Since there are 100 such intervals from 0 to 100:
\[ I = \sum_{n=0}^{99} (e - 1) = 100(e - 1) \]

### Final Answer
Thus, the value of the integral is:
\[ I = 100(e - 1) \]