Universal parabolic constant

Take ${\textstyle y={\frac {x^{2}}{4f}}}$ as the equation of the parabola. The focal parameter is ${\displaystyle p=2f}$ and the semilatus rectum is ${\displaystyle \ell =2f}$.

$${\displaystyle {\begin{aligned}P&:={\frac {1}{p}}\int _{-\ell }^{\ell }{\sqrt {1+\left(y'(x)\right)^{2}}}\,dx\\&={\frac {1}{2f}}\int _{-2f}^{2f}{\sqrt {1+{\frac {x^{2}}{4f^{2}}}}}\,dx\\&=\int _{-1}^{1}{\sqrt {1+t^{2}}}\,dt&(x=2ft)\\&=\operatorname {arsinh} (1)+{\sqrt {2}}\\&=\ln(1+{\sqrt {2}})+{\sqrt {2}}.\end{aligned}}}$$

P is a transcendental number.

Proof. Suppose that P is algebraic. Then ${\displaystyle \!\ P-{\sqrt {2}}=\ln(1+{\sqrt {2}})}$ must also be algebraic. However, by the Lindemann–Weierstrass theorem, ${\displaystyle \!\ e^{\ln(1+{\sqrt {2}})}=1+{\sqrt {2}}}$ would be transcendental, which is not the case. Hence P is transcendental.

Since P is transcendental, it is also irrational.

The average distance from a point randomly selected in the unit square to its center is[5]

${\displaystyle d_{\text{avg}}={P \over 6}.}$

Proof.

${\displaystyle {\begin{aligned}d_{\text{avg}}&:=8\int _{0}^{1 \over 2}\int _{0}^{x}{\sqrt {x^{2}+y^{2}}}\,dy\,dx\\&=8\int _{0}^{1 \over 2}{1 \over 2}x^{2}(\ln(1+{\sqrt {2}})+{\sqrt {2}})\,dx\\&=4P\int _{0}^{1 \over 2}x^{2}\,dx\\&={P \over 6}\end{aligned}}}$

There is also an interesting geometrical reason why this constant appears in unit squares. The average distance between a center of a unit square and a point on the square's boundary is ${\displaystyle {P \over 4}}$. If we uniformly sample every point on the perimeter of the square, take line segments (drawn from the center) corresponding to each point, add them together by joining each line segment next to the other, scaling them down, the curve obtained is a parabola. [6]