The value of `sum_(n=1)^100 int_(n-1)^n e^(x-[x])dx = `

Options

A. `100(e-1)`
B. `100e`
C. `100(e+1)`
D. `100(1-e)`

Answer

To solve the problem, we need to evaluate the expression:

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

where \([x]\) denotes the greatest integer function (or floor function).

### Step 1: Understanding the Integral
The expression \(e^{x - [x]}\) can be simplified. The term \(x - [x]\) represents the fractional part of \(x\), denoted as \(\{x\}\). Thus, we can rewrite the integral as:

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

### Step 2: Evaluating the Integral
The fractional part \(\{x\}\) for \(x\) in the interval \([n-1, n]\) is given by:

- For \(x \in [n-1, n)\), \(\{x\} = x - (n-1) = x - n + 1\).

Thus, we can rewrite the integral:

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

### Step 3: Calculating the Integral
Now we need to compute the integral \(\int_{n-1}^{n} e^{x} \, dx\):

\[ \int e^{x} \, dx = e^{x} + C \]

Evaluating from \(n-1\) to \(n\):

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

### Step 4: Putting it All Together
Now substituting back into our expression:

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

### Step 5: Summing Over n
Now we sum this result from \(n = 1\) to \(n = 100\):

\[ \sum_{n=1}^{100} (e - 1) = 100(e - 1) \]

### Final Answer
Thus, the value of the original expression is:

\[ 100(e - 1) \]