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?
$$
\begin{array}{c|ll}
\hline 
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\ 
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\ 
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\ 
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\ 
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\ 
\text{6} & -4x\ln x & {} & {} & 3.21360 \\ 
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\ 
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\ 
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\ 
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\ 
\end{array}$$

$p_0 = (0, 0), p_i = (i/n, y_n), p_n = (1,0)$
$A = \frac{1}{2} \cdot \sum_{i=1}^n{(y_i + y_{i-1})} \cdot \frac{1}{n} = 1$
$L = \sum{|p_{i+1} - p_i|}$
$\frac{\partial A}{\partial y_i} = \frac{\partial }{\partial y_i} \frac{1}{2n} \cdot (y_i + y_{i-1} + y_{i+1} + y_i) = \frac{1}{n}$
$\frac{\partial L}{\partial y_i} = \frac{\partial }{\partial y_i} \bigg[\sqrt{\frac{1}{n^2} + (y_i - y_{i-1})^2} + \sqrt{\frac{1}{n^2} + (y_{i+1} - y_i)^2}\bigg] = $
$ = \frac{y_i - y_{i-1}}{\sqrt{\frac{1}{n^2} + (y_i - y_{i-1})^2}} + \frac{y_i - y_{i+1}}{\sqrt{\frac{1}{n^2} + (y_{i+1} - y_i)^2}} = \lambda / n$
Lagrange multiplier
Does anyone know how make this a differential equation? $n \rightarrow \infty, 1/n \rightarrow dx$
$\frac{\delta_i}{\sqrt{(dx)^2 + (\delta_i)^2}} - \frac{\Delta_i}{\sqrt{(dx)^2 + (\Delta_i)^2}} = \lambda \cdot dx$
$\frac{\delta_i^2}{{(dx)^2 + (\delta_i)^2}} = \lambda^2 \cdot (dx)^2 + \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}} + 2 \frac{\Delta_i}{\sqrt{(dx)^2 + (\Delta_i)^2}} \lambda \cdot dx $
$\bigg[\frac{\delta_i^2}{{(dx)^2 + (\delta_i)^2}} - \lambda^2 \cdot (dx)^2 - \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}}\bigg]^2 = 4 \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}} \lambda^2 \cdot (dx)^2 $
