Is reversing the order of integration possible here. If so where do I make a mistake? I've been looking to solve this problem for a few hours now. But I don't seem to progress to the give solution at all. Maybe you could be of some help to me. The question is as follows:
"Reverse the order of integration and evaluate the resulting iterated integral"
$$\int_{-2}^{4}\int_{\frac{x_2^2}{2}-3}^{x_2+1}(x_1)dx_1dx_2$$
So what I did is this:

$$x_{1}=x_{2}+1 \Rightarrow x_2=x_1-1$$
$$x_1=\frac{(x^2)}{2}-3 \Rightarrow x_2= -\sqrt{2x_1+6}$$
Given that where
$$x_2=4 \Rightarrow x_1 = 4+1 = 5 $$
Then for,
$$x_2=-2 \Rightarrow x_1=\frac{-2^2}{2}-3=-1$$
Which if I'm right gives me the new intervals for the reverse iterated interval. Namely,
$$\int_{-1}^{5}\left(\int_{-\sqrt{2x_1+6}}^{x_1-1}(x_1)dx_2\right)dx_1$$
Then integrated to:
$$\int_{-1}^{5}\left[x_1*(x_1-1)-(x_1*-\sqrt{2x_1+6})\right] dx_1 $$
According to my teachers solution manual the solution should be should be (18/5). Which is the solution to the integral when not reversing the order of integration.

You aren't allowinf for the fact that $x_2$ may be negative.  I think you should break the outer integral up into two cases: $$\begin{align}-2&\leq x_2\leq0\\0&\leq x_2\leq5\end{align} $$
Then consider the sign when you take square roots.
