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 \( \int_{0}^{100} e^{x - [x]} \, dx \), we will follow these steps:

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

The expression \( [x] \) denotes the greatest integer function (also known as the floor function), which gives the largest integer less than or equal to \( x \). Therefore, \( x - [x] \) represents the fractional part of \( x \), which is periodic with a period of 1. Specifically, for \( x \) in the interval \( [n, n+1) \) (where \( n \) is an integer), \( [x] = n \) and thus \( x - [x] = x - n \).

### Step 2: Break the integral into intervals

Since \( e^{x - [x]} \) 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:

\[ \int_{0}^{100} e^{x - [x]} \, dx = \sum_{n=0}^{99} \int_{n}^{n+1} e^{x - n} \, dx \]

### Step 3: Simplify the integral

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

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

Thus, we can rewrite the integral as:

\[ \int_{n}^{n+1} e^{x - n} \, dx = \int_{n}^{n+1} e^{(x - n)} \, dx = \int_{0}^{1} e^{u} \, du \]

where \( u = x - n \), and \( du = dx \).

### Step 4: Evaluate the integral

Now we compute the integral \( \int_{0}^{1} e^{u} \, du \):

\[ \int e^{u} \, du = e^{u} + C \]

Evaluating from 0 to 1:

\[ \int_{0}^{1} e^{u} \, du = e^{1} - e^{0} = e - 1 \]

### Step 5: Combine results

Since there are 100 intervals from 0 to 100, we multiply the result of the integral over one interval by 100:

\[ \int_{0}^{100} e^{x - [x]} \, dx = 100 \cdot (e - 1) \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{0}^{100} e^{x - [x]} \, dx = 100(e - 1) \]

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