1. The problem involves evaluating the triple integral $$\int \int \int xe^{-y} \, dy \, dx \, dz$$ over the given limits. 2. The limits are given as: - For $z$: from 0 to 1 - For $y$: from 0 to $2z$ - For $x$: from 0 to $\ln x$ (this seems inconsistent; assuming $x$ is the variable of integration with limits 0 to 2) 3. Assuming the integral is $$\int_0^1 \int_0^{2z} \int_0^{\ln x} xe^{-y} \, dy \, dx \, dz$$ but since $\ln x$ as an upper limit for $x$ is not valid, we interpret the integral as $$\int_0^1 \int_0^{2z} \int_0^{\ln x} xe^{-y} \, dy \, dx \, dz$$ is likely a misinterpretation. 4. Instead, consider the integral $$\int_0^1 \int_0^{2z} \int_0^{\ln x} xe^{-y} \, dy \, dx \, dz$$ is intended as $$\int_0^1 \int_0^{2z} \int_0^{\ln x} xe^{-y} \, dy \, dx \, dz$$ with $x$ from 0 to 1, $y$ from 0 to $2z$, and $z$ from 0 to 1. 5. To proceed, we integrate with respect to $y$ first: $$\int_0^{2z} xe^{-y} \, dy = x \int_0^{2z} e^{-y} \, dy = x \left[-e^{-y}\right]_0^{2z} = x(1 - e^{-2z})$$ 6. Next, integrate with respect to $x$ from 0 to 1: $$\int_0^1 x(1 - e^{-2z}) \, dx = (1 - e^{-2z}) \int_0^1 x \, dx = (1 - e^{-2z}) \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}(1 - e^{-2z})$$ 7. Finally, integrate with respect to $z$ from 0 to 1: $$\int_0^1 \frac{1}{2}(1 - e^{-2z}) \, dz = \frac{1}{2} \int_0^1 (1 - e^{-2z}) \, dz = \frac{1}{2} \left[z + \frac{e^{-2z}}{2}\right]_0^1 = \frac{1}{2} \left(1 + \frac{e^{-2} - 1}{2}\right) = \frac{1}{2} \left(1 - \frac{1}{2} + \frac{e^{-2}}{2}\right) = \frac{1}{4} (1 + e^{-2})$$ 8. Therefore, the value of the integral is $$\boxed{\frac{1 + e^{-2}}{4}}$$.