Setting up a double integral domain How would I set up the folowing double integral
$4\displaystyle \iint x^{1/2}dA$$
where the domain is the closed region bounded by $y=x$ and $y=x^2-3x$
I know the domain is  curve shaped so I would do dx cut vertically
So would my integral be
$$\int_0^4\int_{x^2-3x}^x dydx$$

The way it would be set up is
$\int_0^4\int_{y=x^2-3x}^{y=x}(x)^{1/2}dydx$
taking the first integral would give
$(x)^{3/2}-x^{5/2}+3x^{3/2}$
$\int_0^4 4x^{3/2}-x^{5/2}dx$
$\frac{8}{5}x^{5/2}-\frac{2}{7}x^{7/2}$
=$\frac{8}{5}(4)^{5/2}-\frac{2}{7}(4)^{7/2}-0$
