Answer
$\dfrac{23}{5}$
Work Step by Step
Need to integrate the given integral first with respect to $z$, then $x$, and then $y$
$ \int_{0}^1 \int_{y}^{2y} \int_{0}^{x+y} 6xy dz dx dy= \int_{0}^1 \int_{y}^{2y} |6xyz|_{0}^{x+y} dx dy$
or, $=\int_{0}^1\int_y^{2y} 6xy(x+y) dx dy$
or, $=\int_0^1 \int_y^{2y} [6x^2y+6xy^2]dx dy$
or, $=(6) \int^{0}_{1}[\dfrac{1}{3} 8y^4+\dfrac{4y^4}{2}] dy$
or, $=(6) [\dfrac{8y^5}{15}+\dfrac{2y^5}{5}]^{0}_{1}$
or, $=[\dfrac{23y^5}{5}]_0^1$
or, $=\dfrac{23}{5}$
