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Consider the polynomial $f(x) = ax^2+bx+c$ and $f(0) = 0$ and $f(2) = 2\;,$ Then

Minimum value of $\displaystyle \int_{0}^{2}|f'(x)|dx$

$\bf{My\; Try::}$ From $f(0) =0\;,$ We get $c=0$ and $f(2) = 2\;,$ We get $4a+2b=2\Rightarrow 2a+b=1$

So we get $f'(x)=2ax+b=2ax+(1-2a)$

Now $\displaystyle \int_{0}^{2}|f'(x)|dx = \int_{0}^{2}|2ax+(1-2a)|dx$

Now How can I solve after that, Help Required, Thanks

Community
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juantheron
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    $f'(x)$ changes sign at $x_0 = \frac{1}{2a} - 1$; you'll need to consider the cases where $x_0$ is outside the interval $[0, 2]$ (in which you can remove the absolute value signs, perhaps inserting a negative as needed) and where $x_0$ is in the interval, in which case you'll need to examine $\int_1^{x_0}|f'(x)| dx + \int_{x_0}^2 |f'(x)| dx$, carefully keep track of the sign of $f'(x)$ in each integral. – John Hughes May 24 '16 at 11:10
  • Thanks John, I am getting $\displaystyle \frac{1}{6}\leq a\leq \frac{1}{2}$ – juantheron May 24 '16 at 11:30

1 Answers1

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(The following is not the intended treatment of the problem, but anyway ...)

The integral $V:=\int_0^2 |f'(t)|\>dt$ can be considered as the length of the curve $$\gamma:\quad t\mapsto f(t)\in{\mathbb R}\qquad(0\leq t\leq2)$$ on the number line. The quantity $V$ is also called the total variation of $f$ on $[0,2]$. As $f(0)=0$ and $f(2)=2$ the length of $\gamma$ is at least $2$, but is larger if $f$ is not monotonically increasing on $[0,2]$, i.e., if the curve goes back and forth.

Christian Blatter
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