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