Answer
$\dfrac{1}{3}$
Work Step by Step
We need to express the domain $D$ in the Type-1 as :$
D=\left\{ (x, y) | 0 \leq x\leq 1, \ 0 \leq y \leq x \right\}
$
Thus, $\iint_{D} x dA=\int_{0}^{1} \int_{0}^{x} x dy dx \\= \int_0^1 [xy]_{0}^{x} dx \\ = \int_0^1 x^2 dx \\= [\dfrac{x^3}{3}]_0^1\\=\dfrac{1^3}{3}-0 \\= \dfrac{1}{3}$
Next, we need to express the domain $D$ in the Type-2 as :$
D=\left\{ (x, y) | 0 \leq y \leq 1, \ y \leq x \leq 1 \right\}
$
Thus, $$\iint_{D} x dA=\int_{0}^{1} \int_{y}^{1} x \ dx \ dy \\= \int_0^1 [\dfrac{x^2}{2}]_{y}^{1} dx \\ = \int_0^1 \dfrac{1}{2}-\dfrac{1} {2} y^2 \ dy \\ [\dfrac{y}{2}-\dfrac{y^3}{6}]_0^1 \\= \dfrac{1}{3}$$
