What is the value of `int_0^1 (x-1) e^(-x) dx`

Options

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

Answer

To solve the integral \( I = \int_0^1 (x - 1) e^{-x} \, dx \), we can break it down into two parts:

1. **Separate the Integral**:
\[ I = \int_0^1 x e^{-x} \, dx - \int_0^1 e^{-x} \, dx \]

Let's denote:
\[ I_1 = \int_0^1 x e^{-x} \, dx \]
\[ I_2 = \int_0^1 e^{-x} \, dx \]

Therefore, we have:
\[ I = I_1 - I_2 \]

2. **Calculate \( I_1 \)**:
To solve \( I_1 = \int_0^1 x e^{-x} \, dx \), we will use integration by parts. We choose:
- \( u = x \) (which gives \( du = dx \))
- \( dv = e^{-x} \, dx \) (which gives \( v = -e^{-x} \))

By the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ I_1 = \left[ -x e^{-x} \right]_0^1 + \int_0^1 e^{-x} \, dx \]

Evaluating the boundary term:
\[ \left[ -x e^{-x} \right]_0^1 = -1 \cdot e^{-1} - (-0 \cdot e^{0}) = -\frac{1}{e} - 0 = -\frac{1}{e} \]

Thus,
\[ I_1 = -\frac{1}{e} + \int_0^1 e^{-x} \, dx \]

3. **Calculate \( I_2 \)**:
Now, we compute \( I_2 = \int_0^1 e^{-x} \, dx \):
\[ I_2 = \left[ -e^{-x} \right]_0^1 = -e^{-1} - (-e^{0}) = -\frac{1}{e} + 1 \]

4. **Combine the Results**:
Now we can substitute \( I_1 \) and \( I_2 \) back into our original equation:
\[ I = I_1 - I_2 = \left( -\frac{1}{e} + \left( -\frac{1}{e} + 1 \right) \right) \]
Simplifying this gives:
\[ I = -\frac{1}{e} - \frac{1}{e} + 1 = 1 - \frac{2}{e} \]

5. **Final Answer**:
Therefore, the value of the integral is:
\[ I = 1 - \frac{2}{e} \]