To find the minimum value of given integral. Consider the polynomial $f(x)=ax^2+bx+c$. If $f(0)=0,f(2)=2$, then find the minimum value of $$\int_{0}^2  |f'(x)| dx.$$
My try: 
$f(0)=0$
$\implies c=0$ and $f(2)=2 \implies 2a+b=1$ 
Also $f'(x) =2ax+b\implies f'(1)=2a+b =1\implies f'(1)=1$
Now to how to calculate the minimum value of $$\int _{0}^2 |f'(x)|dx .$$
 A: Use the integral identity $$\int_{0}^2  |f'(x)| dx \geqslant |\int_{0}^2  f'(x) dx|.$$ Do some algebra and voila! You are done.
A: Idea:
From the facts given it is easy to see that $f(x)=kx(x-2)+x$ for some $k \in \mathbb{R}$.
\begin{align*}
\int_0^2|f'(x)| \, dx & = \int_0^2|2k(x-1)+1| \, dx\\
 & = \int_0^2|2kx+(1-2k)| \, dx
\end{align*}
If we look at the function $y=2kx+(1-2k)$, which is a straight line, then the integral represents the area of the absolute value of this function. If this line intersects the $x-$axis in between $(0,2)$, then the area will be the sum of two triangles (because of the absolute value) lying above the $x-$axis. 
So to minimize the area, the intersection point of this line with the $x-$axis must be one of the end-points (either at $x=0$ or $x=2$). This happens when $k=1/2$ (or $k=-1/2$).
Hopefully you can take it from here.   
A: Interesting question! 
Note:
$$\int_0^2 |f'(x)|dx=\int_0^2 |2ax+b|dx=\int_0^2 |2ax+1-2a|dx=$$
$$\begin{cases} 2 (min), \ \ if \ \ a\in [-\frac{1}{2},\frac{1}{2}] \ \ \\ -2a-\frac{1}{2a} >2, \ \ if \ \ a\in (-\infty, -\frac{1}{2}) \\ 2a+\frac{1}{2a}>2, \ \ if \ \ a\in (\frac{1}{2}, +\infty) \end{cases}$$
