## Calculus 8th Edition

$\dfrac{5}{3}$
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, $=\dfrac{5}{3}$