Answer
$\dfrac{5}{3}$
Work Step by Step
Integrate the integral first with respect to $z$, then $x$, and then $y$
$ \int_{1}^2 \int_{0}^{2z} \int_{0}^{\ln x} xe^{-y} dy dx dz= \int_{1}^2 \int_{0}^{2z} [-xe^{-y}]_{0}^{\ln x} dx dz=\int_{1}^2 \int_{0}^{2z}(x-1) dx dy$
This implies that $\int_1^2[-x+\dfrac{x^2}{2}]_0^{2z} dz= -\int^{1}_{2}[2z-2z^2] dz$
Hence, $\int_{1}^2 \int_{0}^{2z} \int_{0}^{\ln x} xe^{-y} dy dx dz=- [z^2-(\dfrac{2}{3}) z^3]_{1}^2=\dfrac{5}{3}$
