Find the value of k that bisects the area Find the value of $k$ for which the line $ y= kx$ bisects the area enclosed by the curve $4y=4x-x^2$ and the $x$ - axis.
I have tried to solve this and the solution seems odd... the solution that came out was $k=1 - \sqrt [3]{2}$
The step I took was to find the interval of the curve $4y=4x-x^2$ (which I got 0 to 4) then integrated after subtracting the line from the curve but the solution I got was $4/3$ or something similar to that.
Any help would be appriciated.

Hint:
The abscissa of intersections of $y=0, 4y=x^2-4x$ are the roots of $$x^2-4x=0$$
Total area $$\int_0^4\dfrac{4x-x^2}4dx=?$$
The abscissa of intersections of $y=kx, 4y=x^2-4x$ are the roots of $$x^2-4x=-4kx\iff x=0,x=4-4k$$
$$\int_0^{4-4k}\dfrac{4x-x^2}4dx=?$$
