1. **State the problem:** Evaluate the definite integral $$\int_{-1}^{\alpha} e^{-x} \, dx$$ and write the result to four decimal places. 2. **Recall the formula:** The integral of $$e^{-x}$$ with respect to $$x$$ is $$-e^{-x} + C$$. 3. **Apply the definite integral formula:** $$\int_{-1}^{\alpha} e^{-x} \, dx = \left[-e^{-x}\right]_{-1}^{\alpha} = -e^{-\alpha} - \left(-e^{1}\right) = -e^{-\alpha} + e^{1}$$ 4. **Simplify the expression:** $$= e - e^{-\alpha}$$ 5. **Final answer:** The value of the integral is $$e - e^{-\alpha}$$, which can be approximated numerically once $$\alpha$$ is known. Since $$\alpha$$ is a variable, the exact numeric value depends on its value. If you provide a specific value for $$\alpha$$, I can compute the numerical result to four decimal places.