To find the minimum value of given integral.

Question

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 $$\int _{0}^2 |f'(x)|dx .$$

Answer 1

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.

Answer 2

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.

Answer 3

Interesting question!

Note: $$\int_0^2 |f'(x)|dx=\int_0^2 |2ax+b|dx=\int_0^2 |2ax+1-2a|dx=$$ $$\begin{cases} 2 (min), \ \ if \ \ a\in [-\frac{1}{2},\frac{1}{2}] \ \ \\ -2a-\frac{1}{2a} >2, \ \ if \ \ a\in (-\infty, -\frac{1}{2}) \\ 2a+\frac{1}{2a}>2, \ \ if \ \ a\in (\frac{1}{2}, +\infty) \end{cases}$$