Question: Is there a general solution to $\int e^{F(x)}dx$, where $F'(x)=f(x)$? What should one do if one has an integral of this form, without having any real constraints on $F(x)$?

Answer: 
The process is that the function in the lower limit needs to be less than the function in the upper limit over the interval of integration. In your case, $y^2 \leq y+2$ for all $y$ in $[-1,2]$.

As for how to determine *that* inequality, you could do it by algebra, but the quickest way is probably with a graph. You know that $x=y^2$ is a parabola with its vertex at the origin, and which opens to the right. The line $y=x-2$ passes through $(0,-2)$ and $(2,0)$.

[![enter image description here](https://i.stack.imgur.com/kHfz0.png)](https://i.stack.imgur.com/kHfz0.png)

A few more comments:

* You may think that you can just try it with the functions in one order, and if you get a negative number, just drop the minus sign and make it positive. This may work, but only if the graphs don't cross each other in the interval over which you're integrating.
* In multivariable calculus, you really need graphing skills. Trying to do it all by just pattern matching and algebra doesn't help you gain any intuition.
* You asked which “equation” should be in the lower and upper limits of integration, but you mean which *function*. For instance $y=x+2$ is an equation, but when you set up the integral it had to be manipulated into $x=y-2$, because you needed a function of $y$.
You should use $x=y^2$ for the lower bound and $x=y+2$ for the upper because integrals are direction dependent. Values of the function are added when moving in the direction x increased. This is for each fixed y just the fact that $\int\_a^b f(x)dx = - \int\_b^a f(x)dx$ Convention makes this positive if $a<b$ and $f(x)>0$.