I am trying to find a continuous and closed form function $f(x)$ where $x$ is on the closed interval $[0..1]$ such that the following conditions are met for some given value of $c$ that is between 0 and 1.
$f ( 0 ) =0$
$f ( 1 ) =1$
$\int_{0}^{1} \! f ( x ) \, {\rm d}x=c$
$f' ( x ) > 0$ for all $x \in (0..1)$
$\lim _{x\rightarrow 0^+}f' ( x ) =0$
$\lim _{x\rightarrow 1^-}f' ( x ) =0$
$\lim _{x\rightarrow 0^+}f'' ( x ) =0$
$\lim _{x\rightarrow 1^-}f'' ( x ) =0$
$f'' ( c ) =0$
$f'' ( x ) > 0$ for all $x \in (0..c)$
$f'' ( x ) < 0$ for all $x \in (c..1)$
Obviously, $f(x)$ is also dependent on the value of $c$, but no constraints can be placed on $c$ other than that is properly between $0$ and $1$.
This is a difficult problem in terms of the fact that you wish the area under the curve, $c$ to also be the inflection point, $f''(c)=0$. I will give a partial solution and then describe a pathway to a complete solution. We begin with my definition of the superparabola:
$$f(x;p)=(1-x^2)^p$$
A series of curves is shown on the Wikipedia page. If we look at the region $x\in[-1,0]$ we see that we can have asymmetric sigmoidal curves. The area under the curve is given by
$$\text{Area}=\int_{-1}^0 f(x;p)\ dx =\frac{\sqrt{\pi}\ \Gamma(p+1)}{2\ \Gamma(p+3/2)}$$
Since the area is equal to the parameter $c$, then we can determine the necessary value of $p$. Unfortunately, however, we also have the requirement that $f''(c)=0$. So,
$$f''(x;p)=2p(1-x^2)^(p-2)((2p-1)x^2-1)=0$$
This gives $c=1/(2p-1)$ and it's highly unlikely that this is equal to the area. There is one possibility for rectifying this situation. We can generalize the superparabola to superconics, which are given as follows
$$f(x;p)=(1-x^q)^p$$
This gives us the additional parameter $q$ which may allow you to achieve the desired goal of $c$ equal to the area and inflection point. Unfortunately, I can make no promises here. I leave this in your capable hands.