Answer
$\frac{7}{4}$
Work Step by Step
The region $D$ can be defined as
$D = \{(x,y)| 0\leq x \leq 1, 2x \leq y \leq 3-x \}$
Now, we rewrite our integral as
$\int_0^1\int_{2x}^{3-x} 2xy \ dydx$
Integrating with respect to y, we get
$\int_0^1 [xy^2]^{3-x}_{2x} \ dx$
Plugging in the limits, we get
$\int_0^1 x(3-x)^2 - x(2x)^2 \ dx$
Expanding and combinging like terms gets us
$\int_0^1 -3x^3 -6x^2 + 9x \ dx$
Integrating with respect to x, we get
$[-\frac{3}{4}x^4-2x^3+\frac{9}{2}x^2]^1_0$
Plugging in the bounds, we get
$-\frac{3}{4}(1)^4-2(1)^3+\frac{9}{2}(1)^2 - (-\frac{3}{4}(0)^4-2(0)^3+\frac{9}{2}(0)^2) = -\frac{3}{4}-2+\frac{9}{2} = \frac{7}{4}$
