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