What is the value of `int_(0)^(1)(x-1)e^(-x) dx`?

Options

A. `0`
B. `e`
C. `1/e`
D. `(-1)/e`

Explanation

To solve the integral \( \int_{0}^{1} (x-1)e^{-x} \, dx \), we will use integration by parts. The integration by parts formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = x - 1 \) (which is algebraic)
- \( dv = e^{-x} \, dx \) (which is exponential)

### Step 2: Differentiate \( u \) and integrate \( dv \)

Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = dx \]
- Integrate \( dv \):
\[ v = \int e^{-x} \, dx = -e^{-x} \]

### Step 3: Apply the integration by parts formula

Now we can apply the integration by parts formula:
\[ \int (x-1)e^{-x} \, dx = (x-1)(-e^{-x}) \bigg|_{0}^{1} - \int (-e^{-x}) \, dx \]

### Step 4: Evaluate \( (x-1)(-e^{-x}) \) at the limits

Calculating the first term:
\[ (x-1)(-e^{-x}) \bigg|_{0}^{1} = \left[ (1-1)(-e^{-1}) - (0-1)(-e^{0}) \right] = [0 - 1] = -1 \]

### Step 5: Evaluate the remaining integral

Now we need to calculate the remaining integral:
\[ -\int (-e^{-x}) \, dx = \int e^{-x} \, dx = -e^{-x} \bigg|_{0}^{1} \]
Calculating this gives:
\[ -e^{-x} \bigg|_{0}^{1} = -\left( -e^{-1} + e^{0} \right) = e^{-1} - 1 \]

### Step 6: Combine the results

Now we combine both parts:
\[ \int_{0}^{1} (x-1)e^{-x} \, dx = -1 + (e^{-1} - 1) = e^{-1} - 2 \]

### Final Answer

Thus, the value of the integral is:
\[ \int_{0}^{1} (x-1)e^{-x} \, dx = e^{-1} - 2 \]