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

As has already been explained at least twice, the best functions follow this pattern: 
a continuous function $f$ with $f(0)=f(1)=0$
that approximates $y = h(x) = 1-\frac\pi8+\sqrt{x-x^2}$ for $0<x<1.$
I propose a family of functions for $n$ a positive integer,
$$f_n(x) = \sqrt{x-x^2} + \left(1-\frac\pi8\right)g_n(x),$$
where
$$g_n(x) = \left(1 + \frac{1}{2n}\right)\left(1-(1-2x)^{2n}\right).$$
Since
$$\int_0^1 1-(1-2x)^{2n}\; dx = \frac{2n}{2n+1},$$
we have $\int_0^1 g_n(x)\;dx = 1,$ and therefore $\int_0^1 f_n(x)\; dx = 1.$
The path integral is more difficult to compute than the area integral, but
$1-(1-2x)^{2n}$ takes on its maximum value, $1$, at $x=\frac12$.
So if we set $h_n(x) = h(x) + \frac{1}{2n}\left(1-\frac\pi8\right)$
we ensure that $f(x) \leq h_n(x)$ for $0 \leq x \leq 1.$
I claim the path length is less than the length of the bounding curve
consisting of the graph of $h_n(x)$ from $x=0$ to $x=1$ 
and the two segments joining $(0,0)$ to $(0,h_n(0))$ and $(1,h_n(1))$ to $(1,0)$.
The length of that bounding path is
$$2+\frac\pi4 + \frac1n\left(1-\frac\pi8\right) < 2+\frac\pi4 + \frac{0.607301}{n}.$$
Therefore if we pick, say $n = 1000000,$ the resulting path exceeds the
theoretical minimum by less than $6.074 \times 10^{-7},$
which is less than one part in $4.5 \times 10^6.$
To within the accuracy possible in any visual graph I could present here,
the graph of $f_n(x)$ for large $n$ is the same as 
the graph of every other near-theoretical-minimum solution:


Alternatively, stealing an idea from robjohn, we have
$$\int_0^1 (x-x^2)^{1/n} = B\left(1+\frac1n, 1+\frac1n\right)
 = \frac{\Gamma\left(1+\frac1n\right)^2}{\Gamma\left(2+\frac2n\right)},$$
so we can set
$$g_n(x) =
 \frac{\Gamma\left(2+\frac2n\right)}{\Gamma\left(1+\frac1n\right)^2}(x-x^2)^{1/n}$$
and proceed as before.
This $n$th-root approach seems to converge faster than my $2n$th-power approach.
