Answer
$\dfrac{e^{8}-1}{4}$
Work Step by Step
Consider:
$I= \int_0^{4} \int_0^{\sqrt {4-y}} \dfrac{x e^{2y}}{4-y} dx dy$
or, $=\int_0^{4} [ \dfrac{x^2 e^{2y}}{4-y} ]_0^{\sqrt {4-y}}dx dy$
or, $= \int_0^{4} \dfrac{(\sqrt {4-y})^2 e^{2y} }{2 (4-y)} dy$
or, $=\int_{0}^{4} \dfrac{e^{2y}}{2} dy$
or, $[\dfrac{e^{2y}}{4}]_0^4=\dfrac{e^{8}-1}{4}$
