|(-1/6)a^3| = |[1/2][(1/6)a^3 - 200a + 8000/3]|
a^3 + 1200a - 16000 = 0 <--- I cheated with Mathematica
And here's the interesting part: there's only ONE real positive solution!
a = 11.92
Let's try it:
∫ [x^2 - 11.92x] from 0 to 11.92 = -282.278 (pretend its positive though)
∫ [x^2 - 11.92x] from 11.92 to 20 = 564.94
564.94 / 282.278 = 2.001
Done!
Edit; I also found the result for a general endpoint other that 20. It's hard to type out, but it's here:
http://www.wolframalpha.com/input/?i=a+%…
Where "b" is the second point. You'll note that subbing in 20:
http://www.wolframalpha.com/input/?i=b^2…
Gives the result I already stated (you can see the exact form now though).
Edit 2: I one-upped myself again. It turns out that the relationship between b and a is linear. By differentiating the general function above, we obtain:
http://www.wolframalpha.com/input/?i=1%2…
Which is just a number, approximately equal to 0.59607 as you can see. Thus, all you have to do, given some endpoint b, to determine the a such that that area from 0 to a is half the area from a to b is to multiply b by 0.59607 (for this particular function)!
20 x 0.59607 = 11.92