Find a $m$ divides the area under and above x-axis of $y=x (1 - x)$ My questions is, suppose that for specific $m$, $y=mx$ divides the area under and above $x$-axis of $y=x(1- x)$ into $2$ equal areas, so what is $m$? In an interval, I estimated fast with $m \in (\frac{1}{5}, \frac{2}{5}]$, is this right? I plot this equation on graphs, and only estimated it. I didn't calculate integral precisely.

First find the area of $x(1-x)$ from $0$ to $1$ (which is the area above the x axis) which comes out to be $\frac{1}{6}$, now the area between the line $y=mx$ and $y=x(1-x)$ would be $\frac{1}{12}$, so $$\int_0^{1-m}\left[x(1-x)-mx\right]dx=\frac{1}{12}$$
So, $$\frac{(1-m)^2}{2}-\frac{(1-m)^3}{3}-\frac{m(1-m)^2}{2}=\frac{1}{12}$$
$$\frac{m^2-2m+1}{2}+\frac{m^3-3m^2+3m-1}{3}-\frac{m^3-2m^2+m}{2}=\frac{1}{12}$$
$$3m^2-6m+3+2m^3-6m^2+6m-2-3m^3+6m^2-3m=\frac{1}{2}$$
$$-m^3+3m^2-3m+1=\frac{1}{2}$$
$$(1-m)^3=\frac{1}{2}$$
$$m=1-\frac{1}{\sqrt[3]{2}}$$
