Answer
$38.4$
Work Step by Step
The equation of the line passing through $(0,2)$ and $(4,0)$ is given by: $y-0=(x-4) \times \dfrac{0-2}{4-0} \implies y=\dfrac{-1}{2}x+2$
The iterated integral can be calculated as:
$\iint_{D} x^2y d A=\int_0^4 \int_{\dfrac{-1}{2}x+2}^2 x^2y dy dx\\=\int_0^4 \dfrac{x^2y^2}{2}|_{\frac{-1}{2}x+2}^2 dx \\=\int_0^4 (x^3-\dfrac{x^4}{8}) \ dx \\=[\dfrac{x^4}{4}-\dfrac{x^5}{40}]_0^4 \\=38.4$
