Double integration over a general region as a type I $\iint xy$ bound by the curves $y=x^2$ and $y=3x$
The easiest way to integrate this is as a type I integral $dydx$
since $3x \gt x^2$ in the interval $3x$ is my upper bound and my interval is $[0,3]$ by setting $x^2=3x$ and solving.
$$\int_0^3 \int_{x^2}^{3x} xy \, dy\,dx$$
$$\int_0^3 \left(\int_{x^2}^{3x} x\frac{y^{1+1}}{1+1}dy\right)\,dx=$$ $\int_{0}^{3}$ $\Big[\int_{x^2}^{3x}$ $x\frac{y^{2}}{2}dy \Big\vert_{x^2}^{3x}\Big]$ $dx =$$x(\frac{1}{2}{(3x)^2})-x(\frac{1}{2}x^4) \Big\vert_{x^2}^{3x}$ $=$ $\int_{0}^{3} \frac{9x^{3+1}}{3+1} -\frac{x^{5+1}}{2} \div 5+1 dx$ $=$ $\int_{0}^{3} \frac{9x^4}{4}-\frac{x^{6}}{12} \Big\vert_{0}^{3}dx =$ $\frac{729}{4}-\frac{729}{12} = \frac{243}{2}$
Is there an error in my solution?
Checking an integral calculator gave me $\frac{243}{8}$ How is that possible?

We have:
$$
\int_{x^2}^{3x}xydy=\left[\frac{1}{2}xy^2\right]_{x^2}^{3x}=\frac{1}{2}(9x^3-x^5) 
$$
so the double integral becomes:
$$
\frac{1}{2}\int_0^3(9x^3-x^5) dx=\frac{1}{2}\left[ \frac{9}{4}x^4-\frac{1}{6}x^6\right]_0^3=\frac{729}{8}-\frac{729}{12}=\frac{243}{8}
$$
