Minimizing arclength with a set of parameters

TheStealthTarget
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i am trying make a function who's integral is equal to 1 from 0 to 1 and goes threw points (0,0) and (1,0) with the minimal arc length, is tehre anyway to do this?

i have tried a couple things, which iwll get me xome question, i have made a triangle, and i have made many trapazoids, the trapazoids have a small arc length then the triangle, but i am sure there is a curved line that is even smaller, how would you find this curved line.
 
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What arc lengths do you get for the triangle (2 sides), square (3 sides), trapezoid (still 3 sides), whatever (4 sides), etc.? And what do you get for an ellipse? And what do you get for the function A sin(x) appropriately scaled? Do you see any trends that you can use to prove that <whatever> has minimum arc length?
 
for a triangle i get an arc length of 2sqrt4.25 = 4.123105626
for a square (or trapazoid with sides that have a slope of 10000000000 or negligible amount) i get 3

i am redoing the trapazoid with a slope of 4, i realized a made a mistake.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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