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

The answer is that you will need to have a constant curvature, which is the partial circle solution by robjohn. If you do want the curve within (0,1) then the rectangle + 1/2 circle solution by both rob and xan.
Why is that? it is actually a physics problem. The solution is a shape of a membrane under pressure.
Parametric function:
$$x=
\begin{cases}
 1&  0\leq s\leq h\\
 \frac{1}{2}+\frac{1}{2}\cos \left( 2\left( s-h \right) \right) &  h<s<h+\frac{2}{\pi}\\
 0&  h+\frac{2}{\pi}<s<2h+\frac{2}{\pi}\\
\end{cases}
$$
$$y=
\begin{cases}
 s&  0\leq s\leq h\\
 h+\frac{1}{2}\sin \left( 2\left( s-h \right) \right)&  h<s<h+\frac{2}{\pi}\\
 \left( 2h+\frac{\pi}{2}-s \right)&  h+\frac{\pi}{2}<s<2h+\frac{2}{\pi}\\
\end{cases}
$$
Area:
$$1=\int_{0}^{1}{y}dx$$
$$
1=\int_{h+\frac{\pi}{2}}^h{\left[ h+\frac{1}{2}\sin \left( 2\left( s-h \right) \right) \right]}d\!\left[ \frac{1}{2}+\frac{1}{2}\cos \left( 2\left( s-h \right) \right) \right] 
$$
Take:
$$\theta=2(s-h)$$
$$4=\int_{\pi}^0{\left[ 2h+\sin \left( \theta \right) \right]}d\!\left( \cos \left( \theta \right) \right) =4h+\frac{\pi}{2}$$
$$h=1-{\pi \over 8}$$
$${\rm Length}=s_{\max}=2h+\frac{\pi}{2}=2+\frac{\pi}{4}=2.785398163...$$
