Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^3}y{\rm{d}}A = \frac{1}{{5376}}$
Work Step by Step
Let ${\cal D}$ denote the region between the curves $y = {x^2}$ and $y = x\left( {1 - x} \right)$.
We need to find the boundaries of ${\cal D}$, so we solve for the intersection of the curves:
${x^2} = x\left( {1 - x} \right)$, ${\ \ \ \ \ }$ ${x^2} = x - {x^2}$
$2{x^2} - x = 0$
$x\left( {2x - 1} \right) = 0$
So, the boundaries of ${\cal D}$ is at $x=0$ and $x = \frac{1}{2}$.
We describe ${\cal D}$ as a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le \frac{1}{2},{x^2} \le y \le x\left( {1 - x} \right)} \right\}$
Evaluate the double integral of $f\left( {x,y} \right) = {x^3}y$ over ${\cal D}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^3}y{\rm{d}}A = \mathop \smallint \limits_{x = 0}^{1/2} \mathop \smallint \limits_{y = {x^2}}^{x\left( {1 - x} \right)} {x^3}y{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^{1/2} {x^3}\left( {{y^2}|_{{x^2}}^{x\left( {1 - x} \right)}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^{1/2} {x^3}\left( {{x^2}\left( {1 - 2x + {x^2}} \right) - {x^4}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^{1/2} \left( {{x^5} - 2{x^6}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{1}{6}{x^6} - \frac{2}{7}{x^7}} \right)|_0^{1/2}$
$ = \frac{1}{2}\left( {\frac{1}{6}\cdot\frac{1}{{64}} - \frac{2}{7}\cdot\frac{1}{{128}}} \right) = \frac{1}{{5376}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^3}y{\rm{d}}A = \frac{1}{{5376}}$.
