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I'm trying to find the area between 2 functions. I have no calculator but have equations. I want to determine which function is above the other so I subtract integral in correct order. Is there a way to do that other than finding points?

These are the 2 functions, $y=-x^2$ and $y=-x^3+6x$

Integrate from -2 to 3

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To find the limits of integration, I suppose you looked for the points of intersection. Note that they intersect not only in $x=-2$ and $x=3$, but also in $x=0$.

It is thus possible that the 'upper' function is not the same in the intervals $[-2,0]$ and $[0,3]$; in fact: it will change in $x=0$ in your case.

Now in each of those intervals, you can easily plug in a value for $x$ to check which function is the larger one. Is there a reason why you want to avoid this, or what do you mean by other than finding points? Take easy values, e.g. $x=-1$ and $x=1$.

Alternatively: you can calculate the integrals on the two intervals (separately!) by subtracting one from the other at random. Then simply take the absolute value: if you accidentally subtracted the larger from the smaller (you'll get the opposite value), the absolute value covers for this.

Applying this to $\color{red}{f(x)=-x^2}$ and $\color{blue}{g(x)=-x^3+6x}$; the area enclosed by the graphs is given by: $$\int_{-2}^3 \left| \color{red}{f(x)}-\color{blue}{g(x)} \right| \,\mbox{d}x = \int_{-2}^0 \color{red}{f(x)}-\color{blue}{g(x)} \,\mbox{d}x +\int_{0}^3 \color{blue}{g(x)}-\color{red}{f(x)} \,\mbox{d}x $$ because $\color{red}{f(x)} \ge \color{blue}{g(x)}$ on $[-2,0]$ and $\color{red}{f(x)} \le \color{blue}{g(x)}$ on $[0,3]$.

If you don't want to check this, you can simply do: $$\int_{-2}^3 \left| \color{red}{f(x)}-\color{blue}{g(x)} \right| \,\mbox{d}x = \left|\int_{-2}^0 \color{red}{f(x)}-\color{blue}{g(x)} \,\mbox{d}x\right| +\left|\int_{0}^3 \color{red}{f(x)}-\color{blue}{g(x)} \,\mbox{d}x\right| $$

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    I did that but it takes way too long. – 2017-01-04
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    Plugging in $x=\pm 1$ takes _too long_...?! By the way: I added a paragraph. – 2017-01-04
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    So if I use absolute value the order doesn't matter? That seems better idea. – 2017-01-04
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    Correct: but then you still have to split into two integrals and take the absolute value of both integrals. If you take the absolute value not _of the integral_, but _of the difference of the functions_, then you'll have _written_ it with a single integral but you'll still need to split it into two to calculate it. – 2017-01-04
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You can use WolframAlpha to plot the functions:

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As you can see in one part one figure is above and in other part another.

Hope you can solve it further.