If =`int_(0)^(1) x^(n)e^(-x)dx "for" n in N "then" I_(n)-nI_(n-1)=`

Options

A. e
B. `1//e`
C. `-1//e`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( I_n - n I_{n-1} \) where \( I_n = \int_0^1 x^n e^{-x} \, dx \).

### Step 1: Define \( I_n \)
We start with the definition of \( I_n \):
\[ I_n = \int_0^1 x^n e^{-x} \, dx \]

### Step 2: Apply Integration by Parts
We will use integration by parts, where we let:
- \( u = x^n \) (thus \( du = n x^{n-1} \, dx \))
- \( dv = e^{-x} \, dx \) (thus \( v = -e^{-x} \))

Using integration by parts, we have:
\[ I_n = \left[ -x^n e^{-x} \right]_0^1 + \int_0^1 n x^{n-1} e^{-x} \, dx \]

### Step 3: Evaluate the Boundary Terms
Now, we evaluate the boundary terms:
\[ \left[ -x^n e^{-x} \right]_0^1 = -1^n e^{-1} - \lim_{x \to 0} (-x^n e^{-x}) = -e^{-1} - 0 = -\frac{1}{e} \]

### Step 4: Substitute Back into the Integral
Now substituting back into the equation for \( I_n \):
\[ I_n = -\frac{1}{e} + n \int_0^1 x^{n-1} e^{-x} \, dx \]
This integral is \( I_{n-1} \):
\[ I_n = -\frac{1}{e} + n I_{n-1} \]

### Step 5: Rearranging the Equation
Now we can rearrange this equation to find \( I_n - n I_{n-1} \):
\[ I_n - n I_{n-1} = -\frac{1}{e} \]

### Final Result
Thus, we have:
\[ I_n - n I_{n-1} = -\frac{1}{e} \]

### Summary of Steps
1. Define \( I_n \) as the integral.
2. Use integration by parts to express \( I_n \).
3. Evaluate the boundary terms from integration by parts.
4. Substitute back to express \( I_n \) in terms of \( I_{n-1} \).
5. Rearrange to find the desired expression.