Answer
$$\dfrac{e^{8}-1}{4}$$
Work Step by Step
$$\int_{R} dA= \int_0^{4} \int_0^{\sqrt {4-y}} \dfrac{x e^{2y}}{4-y} \space dx \space dy \\=\int_0^{4} [ \dfrac{x^2 e^{2y}}{4-y} ]_0^{\sqrt {4-y}} \space dx \space dy\\= \int_0^{4} \dfrac{(\sqrt {4-y})^2 e^{2y} dy}{2 (4-y)} \\=\int_{0}^{4} \dfrac{e^{2y }dy}{2} \\ =\dfrac{1}{4}[e^{2y}]_0^4\\=\dfrac{1}{4}(e^{8}-1) \\=\dfrac{e^{8}-1}{4}$$
