1. **Stating the problem:** We want to find the function $$F(x) = \int_{0}^{x} \lambda e^{-\lambda t} dt$$ where $$\lambda > 0$$ is a constant. 2. **Formula and rules:** This is a definite integral of an exponential function. The integral of $$e^{at}$$ with respect to $$t$$ is $$\frac{1}{a} e^{at}$$ plus a constant. 3. **Intermediate work:** \[ F(x) = \int_0^x \lambda e^{-\lambda t} dt = \lambda \int_0^x e^{-\lambda t} dt \] Calculate the integral inside: \[ \int_0^x e^{-\lambda t} dt = \left[-\frac{1}{\lambda} e^{-\lambda t} \right]_0^x = -\frac{1}{\lambda} e^{-\lambda x} + \frac{1}{\lambda} e^{0} = \frac{1}{\lambda} (1 - e^{-\lambda x}) \] Multiply by $$\lambda$$: \[ F(x) = \lambda \times \frac{1}{\lambda} (1 - e^{-\lambda x}) = 1 - e^{-\lambda x} \] 4. **Explanation:** The integral represents the cumulative distribution function (CDF) of an exponential distribution with rate $$\lambda$$. It starts at 0 when $$x=0$$ and approaches 1 as $$x \to \infty$$. **Final answer:** $$F(x) = 1 - e^{-\lambda x}$$