Find the value of $k$ which minimises $F(k)= \int_{0}^{4} |x(4-x)-k|dx$ 
QUESTION:
Find the value of $k$ which minimises $$F(k)=\int_{0}^{4} |x(4-x)-k|dx$$


MY APPROACH: Clearly $4x-x^2$ is a downward concave parabola with the roots $0$ and $4$. And $y=k$ is a line parallel to the $x$ axis. Now, the modulus of the area bounded by these two curves can be divided into two parts -
A- the area under the parabola and above the line.
B- the area above the parabola and under the line.
Now, the line intersects the parabola at $4x-x^2=k$. Call the two roots of this equation $\alpha_k$ and $\beta_k$ (obviously, $\alpha$ and $\beta$ are functions of $k$). Without loss of generality, assume that $\alpha_k \le \beta_k$ (equality is achieved when the line is tangential to the parabola, at $(2,4)$ )
$$\therefore F(k)= \underbrace{ \int_{0}^{\alpha_k} \big( k - x(4-x) \big) dx}_{m}  + \int_{\alpha_k}^{\beta_k} \big( x(4-x) - k \big) dx + \underbrace{\int_{\beta_k}^{4} \big( k - x(4-x) \big)}_{n} dx$$
Now, due to symmetry of the problem, $m$ and $n$ must have the same value. Hence, we can write,
$$F(k)= 2 \int_{0}^{\alpha_k}\big( k - x(4-x) \big) dx +  \int_{\alpha_k}^{\beta_k} \big( x(4-x) - k \big) dx $$
And the rest of the problem can be solved by Leibniz rule of differentiating under the integral sign. But the problem is $$\alpha_k=\frac{4-\sqrt{16-4k}}{2}$$ and $$\beta_k= \frac{4 + \sqrt{16-4k}}{2}$$ which obviously doesn't look very nice..
I am stuck here..
Am I even going in the right direction ?
Your help to complete this sum and/or any alternate (hopefully simpler) solution is much appreciated..
Thank you so much ..
(I attach a graph below, for better clarity on whatever I have said above.. Note that: the green line is variable, that's $y=k$)

 A: Refer to the following picture.

Notice that $F(k)$ is the area of highlighted region. If we increase $k$ by an infinitesimal $\epsilon$, left and right regions’ area increase by $x_{1}\epsilon$ and $(4-x_{2})\epsilon$ respectively, while middle region’s area decrease by $(x_{2}-x_{1})\epsilon$.
In other word, $\frac{d}{dk}F(k)=4-2(x_{2}-x_{1})$. Notice that $x_{1}=1$ , $x_{2}=3$ yields $\frac{d}{dk}F(k)=0$.
Substitute either $1$ or $3$ to $x(4-x)$ to obtain $k=3$
A: Note that \begin{align}F(k)&=\int_0^4|x(4-x)-k|\,dx=2\int_0^2|x(4-x)-k|\,dx\\&=2\int_0^\alpha (x^2-4x+k)\,dx-2\int_\alpha^2(x^2-4x+k)\,dx\end{align} where $\alpha=2-\sqrt{4-k}$ is the smaller root of the quadratic.
As $\int_0^x(t^2-4t+k)\,dt=x^3/3-2x^2+kx$ we have $$F(k)=4\left(\frac{\alpha^3}3-2\alpha^2+k\alpha\right)-2\left(\frac{2^3}3-2\cdot2^2+2k\right)=\frac43\alpha(\alpha^2-6\alpha+3k)+\frac{32}3-4k.$$ As $\alpha^2-4\alpha+k=0$, we can simplify to get $$\frac43\alpha(\alpha^2-6\alpha+3k)=\frac43\alpha(4\alpha-k-6\alpha+3k)=\frac83(k\alpha-\alpha^2)=\frac83((k-4)\alpha+k)$$ so that $F(k)=\frac43(2(k-4)\alpha-k+8)$. Hence it suffices to minimise $$G(k):=\frac34\left(F(k)-\frac{32}3\right)=2(k-4)(2-\sqrt{4-k})-k=3k+2(4-k)^{3/2}-16.$$ Now $G'(k)=3-3(4-k)^{1/2}$ and setting to zero yields $k=3$. This is a minimum as $G''(3)>0$.
A: First see this proposition

Now divide your function into two monotonic intervals. First from 0 to 2 and the second from 2 to 4. In both the cases, the midpoints of abscissa yield the same ordinate f(3)=f(1)=3
So k=3 is your required answer
