The two curves meet in the first quadrant when
[tex]x^2 = 2-x \implies x^2 + x - 2 = (x + 2) (x - 1) = 0 \implies x=1[/tex]
Then the integral in question is
[tex]\displaystyle \iint_R x \, dA = \int_0^1 \int_{x^2}^{2-x} x \, dy \, dx \\\\ ~~~~~~~~ = \int_0^1 x (2-x-x^2) \, dx \\\\ ~~~~~~~~ = \int_0^1 (2x - x^2 - x^3) \, dx = 1 - \frac13 - \frac14 = \boxed{\frac5{12}}[/tex]
