Answer:
[tex]\frac{7}{2}[/tex]
Step-by-step explanation:
We are asked to find the integral:
[tex]\int\int\limits_D4xy\:dA[/tex]
So, let's apply the vertices given in the question.
[tex]\int\limits^1_0\int\limits_{2x}^{3-x}4xy\:dydx=\int\limits^1_0 2xy^2|_{2x}^{3-x}\:dx=\int\limits^1_0 2x[(3-x)^2-(2x)^2]\:dx=\\\\=\int\limits^1_0 2x(9-6x-3x^2)\:dx=6\int\limits^1_0 (3x-2x^2-x^3)\:dx=\\\\=6(\frac{3}{2}x^2-\frac{2}{3}x^3-\frac{1}{4}x^4)|^1_0=6(\frac{3}{2}-\frac{2}{3}-\frac{1}{4})=\frac{7}{2}[/tex]
