Line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. There is a line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. Find m.
My solution:


When I compute my answer, I get $1-\frac{1}{\sqrt[3]{2}}$ (as seen in Wolfram below)
Solution in Wolfram:

https://www.wolframalpha.com/input/?i=%281-x%29%5E2%28-2x%2B2%29%3D1

Solution in Geogebra:

https://www.geogebra.org/calculator/kshvr8mw

My question: is there a way to find this solution without a GDC/Geogebra? In other words, a method which would lead to a simpler equation in the end?
Thank you.

This is my solution.
First, we recognize that the quadratic has roots at $(0,0)$ and $(1,0)$. So, to first solve for the area under the quadratic:
$\int_0^1 x-x^2.dx=\left[\frac{x}{2}-\frac{x^3}{3}\right]_0^1$
$=\frac{1}{6}$
So, half of this area is $\frac{1}{12}$
Let $x=a$ be the abscissa of the intersection point of line $y=mx$ and curve $y=x(1-x)$ different from $0$.
Then:
$ma=a(1-a)$
$m=(1-a)$
So, the equation of the straight line is $y=(1-a)x$
$\int_0^a(x-x^2)-(1-a)x.dx=\frac{1}{12}$
$\int_0^a-x^2+ax.dx=\frac{1}{12}$
$\left[-\frac{x^3}{3}+\frac{ax^2}{2}\right]_0^a=\frac{1}{12}$
$-\frac{a^3}{3}+\frac{a^3}{2}=\frac{1}{12}$
$\frac{a^3}{6}=\frac{1}{12}$
$a^3=\frac{1}{2}$
$a=\sqrt[3]{\frac{1}{2}}$
As earlier determined, the equation of the straight line is $y=(1-a)x$, so $y=(1-\sqrt[3]{\frac{1}{2}})x$.
