It looks like the integral is
[tex]\displaystyle \iint_D x^2 (y-x) \,\mathrm dx\,\mathrm dy[/tex]
where D is the set
D = {(x, y) : 0 ≤ x ≤ 1 and x ² ≤ y ≤ √x}
So we have
[tex]\displaystyle \iint_D x^2(y-x)\,\mathrm dx\,\mathrm dy = \int_0^1 \int_{x^2}^{\sqrt x} x^2(y-x)\,\mathrm dy\,\mathrm dx \\\\ = \int_0^1 \left(\frac{x^2y^2}2-x^3y\right)\bigg|_{y=x^2}^{y=\sqrt x} \\\\ = \int_0^1 \left(\frac{x^3}2-x^{7/2}+x^5-\frac{x^6}2\right)\,\mathrm dx \\\\ = \left(\frac{x^4}8 - \frac{2x^{9/2}}9 + \frac{x^6}6 - \frac{x^7}{14}\right)\bigg|_{x=0}^{x=1} = \frac18-\frac29+\frac16-\frac1{14} = \boxed{-\frac{1}{504}}[/tex]
