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. 



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



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?

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

We wish to minimize
$$
\int_0^1\sqrt{f'(x)^2+1}\,\mathrm{d}x\tag{1}
$$
while keeping
$$
\int_0^1f(x)\,\mathrm{d}x=1\tag{2}
$$
This means that we wish to find an $f$ so that the variation of length is $0$
$$
\int_0^1\frac{f'(x)\,\delta f'(x)}{\sqrt{f'(x)^2+1}}\,\mathrm{d}x=0\tag{3}
$$
which, after integration by parts, noting that $\delta f(0)=\delta f(1)=0$, becomes
$$
\int_0^1\frac{f''(x)\,\delta f(x)}{\sqrt{f'(x)^2+1}^{\,3}}\,\mathrm{d}x=0\tag{4}
$$
for all variations of $f$, $\delta f$, so that the variation of area is $0$
$$
\int_0^11\,\delta f(x)\,\mathrm{d}x=0\tag{5}
$$
This means that $\frac{f''(x)}{\sqrt{f'(x)^2+1}^{\,3}}$ is perpendicular to all $\delta f$ that $1$ is. This is so only when there is a $\lambda$ so that
$$
\frac{f''(x)}{\sqrt{f'(x)^2+1}^{\,3}}=\lambda\tag{6}
$$
However, $(6)$ just says that the curvature of the graph of $f$ is $\lambda$. That is, the graph of $f$ is an arc of a circle.



Find the Length of the Arc

Since the length of the chord of the circle we want is $1$, we have
$$
2r\sin\left(\frac\theta2\right)=1\tag{7}
$$
Since the area cut off by this chord is $1$, we have
$$
r^2\left[\frac\theta2-\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)\right]=1\tag{8}
$$
Square $(7)$ to get
$$
2r^2(1-\cos(\theta))=1\tag{9}
$$
and rewrite $(8)$ to get
$$
\frac12r^2(\theta-\sin(\theta))=1\tag{10}
$$
Solve  $4(1-\cos(\theta))=\theta-\sin(\theta)$ to get
$$
\theta=4.3760724130128873845\tag{11}
$$
and then $(7)$ gives
$$
r=0.61313651252231835636\tag{12}
$$
This would lead to a minimum length of
$$
L=r\theta=2.6831297778598481320\tag{13}
$$



Problem

Unfortunately, since $\theta\gt\pi$, the minimizing curve is an arc that cannot be represented by a function.



The minimizing curve that is closest to the graph of a function is the curve that joins $(0,0)$ and $(1,0)$ to the endpoints of
$$
y=1-\frac\pi8+\sqrt{x-x^2}\tag{14}
$$



which has a length of
$$
2+\frac\pi4=2.7853981633974483096\tag{15}
$$
However, this curve is not the graph of a function.



A Sequence of Approximations

$$
f_n(x)=\frac1{c_n}\left(1-\frac\pi8+\sqrt{x-x^2}\right)\left(x-x^2\right)^{1/n}\tag{16}
$$
where
$$
c_n=\left(1-\frac\pi8\right)\frac{\Gamma\left(1+\frac1n\right)^2}{\Gamma\left(2+\frac2n\right)}+\frac{\Gamma\left(\frac32+\frac1n\right)^2}{\Gamma\left(3+\frac2n\right)}\tag{17}
$$
As $n\to\infty$, the length of $f_n$ approaches $2+\frac\pi4$.

At $n=100$, we get a length of $L=2.7857313936$, less than $\frac1{3000}$ above the minimum:



At $n=1000$, we get a length of $L=2.7854017568$, less than $\frac1{250000}$ above the minimum.
The most recent overkill contribution - which is at least somewhat interesting, though not the optimal - is the following behemoth: $f(x)=-0.00010156(b^{x-1/2}+b^{-(x-1/2)})+1.120357$ where $b=121716670.4$ You can think of this as a hyperbolic cosine with a different base. This has arc length $2.8788364$. The above answer does beat polynomials of the form $p(x)=a(x-1/2)^n+b$. Such polynomials can do no better than 2.8940115.