<s>[INST] Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:


*

*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;

*$f(0)=0$ and $f(1)=0$;

*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$. 


Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above. 

$\mathbf{\color{red}{\text{Contest results:}}}$ 
$$
\begin{array}{c|ll}
\hline 
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\ 
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\ 
\text{3} & \text{mickep} & {} & {} & 2.81108 \\ 
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\ 
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline 
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\ 
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\ 
\end{array}$$

Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline 
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\ 
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\ 
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\ 
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\ 
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\ 
\text{6} & -4x\ln x & {} & {} & 3.21360 \\ 
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\ 
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\ 
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\ 
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\ 
\end{array}$$
 [/INST] An easy approach is to simply construct an ellipse with its upper half satisfying the above conditions.
An ellipse is defined via $2$ numbers $a$ and $b$ which are each the half of the major and minor axis of the ellipse.

Then all points $(x,y)$ which suffice the following equation are on the ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Or to get the upper half of the eclipse as a function:
$$  y = b \, \sqrt{\left(1 - \frac{x^2}{a^2}\right) } $$
The area $A$ of the complete ellipse is given via $A = \pi\,a\,b$ and therefore our first condition translates to:
$$ \frac{1}{2}\,\pi\,a\,b = 1$$
Also, as we want that $f(0) = 0 = f(1)$, we have:
$$ 2\,a = 1 $$
That already gives us
$$ a = \frac{1}{2} \\ b = \frac{4}{\pi} $$
and therefore an ellipse with the correct size. However, this results in an ellipse which intersects the $x$-Axis at $x_1=-0.5$ and $x_2=0.5$. To meet our conditions, we move the ellipse $0.5$ to the right and get:
$$  y = \frac{4}{\pi} \, \sqrt{1 - 4\,(x-0.5)^2 }\tag{$\dagger$} $$
Now we simply let Wolfram Alpha do the computation for the arc length. The result is 
$$ 2.919463, $$
and the graph in $(\dagger)$ appears below:

</s>