Double integration over a general region $\iint x^2 +2y$ bound by $y=x$ $y=x^3$  $x \geq 0$
this is either a type I or type II since the bounds are already nicely given for a type I, I integrated it as a type I:
Finding the bounds:
$x^3=x \to x^3-x=0 \to x(x^{2}-1)= 0 \to x=0, x=\pm1$ 
Since $-1\lt 0$ my bounds for $x$ are $[0,1]$ and since $x \gt x^3$, $x$ is my upper bound for dy.
$\int_{0}^{1} \int_{x^3}^{x} x^2+2y$ $dydx$
$\int_{x^3}^{x}$ $x^2+y^2$ $\Big\vert_{x}^{x3}$dy $=$ $(2x^2) - (-x^2+x^6)$
$\int_{0}^{1} 3x^2+x^6 dx=$ $x^{3}+\frac{1}{7}x^7 \Big\vert_{0}^{1} \to 1 -\frac{1}{7}$ $=\frac{6}{7}$

Note that $x^3\lt x$ in the interval $(0,1)$.  (A picture always helps in this kind of problem.)
So $y$ travels from $y=x^3$ to $y=x$.
One can see without checking details that the answer $-\frac{4}{21}$ cannot be right. Your integrand is positive in the region, so the answer must be positive.
