1. The problem is to find the function $$F(x) = \int_{0}^{x} \lambda e^{-\lambda t} dt$$ where $$\lambda$$ is a constant. 2. We use the formula for the integral of an exponential function: $$\int e^{at} dt = \frac{1}{a} e^{at} + C$$. 3. Applying this to our integral, we have: $$F(x) = \lambda \int_0^x e^{-\lambda t} dt = \lambda \left[ \frac{e^{-\lambda t}}{-\lambda} \right]_0^x$$ 4. Simplify the expression inside the brackets: $$= \lambda \left( \frac{e^{-\lambda x} - e^{0}}{-\lambda} \right) = \lambda \left( \frac{e^{-\lambda x} - 1}{-\lambda} \right)$$ 5. Cancel $$\lambda$$: $$= - (e^{-\lambda x} - 1) = 1 - e^{-\lambda x}$$ 6. Therefore, the final answer is: $$F(x) = 1 - e^{-\lambda x}$$ This function represents the cumulative distribution function (CDF) of an exponential distribution with rate parameter $$\lambda$$.