`int_(-1)^(2) e^(-x)dx`

Answer

To solve the integral \( \int_{-1}^{2} e^{-x} \, dx \), we will follow these steps:

### Step-by-Step Solution:

1. **Set up the integral**:
We start with the integral:
\[ I = \int_{-1}^{2} e^{-x} \, dx \]

2. **Substitution**:
We will use the substitution \( t = -x \). Then, differentiating both sides gives:
\[ dt = -dx \quad \Rightarrow \quad dx = -dt \]

3. **Change the limits of integration**:
When \( x = -1 \):
\[ t = -(-1) = 1 \]
When \( x = 2 \):
\[ t = -2 \]
Thus, the limits change from \( x: -1 \to 2 \) to \( t: 1 \to -2 \).

4. **Rewrite the integral**:
Substituting \( e^{-x} \) and \( dx \) into the integral, we have:
\[ I = \int_{1}^{-2} e^{t} (-dt) = -\int_{1}^{-2} e^{t} \, dt \]

5. **Reversing the limits**:
Changing the limits of integration changes the sign:
\[ I = \int_{-2}^{1} e^{t} \, dt \]

6. **Integrate**:
The integral of \( e^{t} \) is \( e^{t} \):
\[ I = \left[ e^{t} \right]_{-2}^{1} \]

7. **Evaluate the limits**:
Now we evaluate the definite integral:
\[ I = e^{1} - e^{-2} \]

8. **Final answer**:
Thus, the final result is:
\[ I = e - e^{-2} \]