Arc length contest! Minimize the arc length of f(x)f(x) when given three conditions.

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.