Answer
$\dfrac{5}{3}$
Work Step by Step
Need to integrate the given integral first with respect to $y$, then $x$, and then $z$
$ \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$
or, $=\int_{1}^2 \int_{0}^{2z}[-1+x]dx dy$
or, $=\int_1^2[-x+\dfrac{x^2}{2}]_0^{2z} dz$
or, $= -\int^{1}_{2}[2z-2z^2] dz$
or, $=- [z^2-\dfrac{2}{3}z^3]_{1}^2$
or, $=[-4+\dfrac{16}{3}+1-\dfrac{2}{3}]$
or, $=\dfrac{5}{3}$
