This question follows on from my previous question I asked at Computing Coefficients of a Quadratic Equation given definite Integral
This time however I have two free variables.
For the quadratic function $$-ax^2 + bx$$ an upside down parabola with one root at $x_1=0$, is there a way to compute a and b such that the definite integral of $f(x)$ between the roots ($x_1, x_2: f(x_1) \land f(x_2)= 0$) equals $1?$ And furthermore bound such that the height of the parabola does not exceed a given value $V?$ and limits on the rate of change f'(x), A.
Starting with a quadratic with a roots at $x = 0$ and $x = x_1$, we get a general form $f(x) = a x (x-x_1)$.
Your integral condition requires that $1 = \int_0^{x_1} a x (x-x_1) \mathrm{d}x \ = \frac{-a x_1^3}{6}$
The parabola reaches a maxima at $x = \frac{x_1}{2}$, with the value $-\frac{a x_1^2}{4}$. Thus, $x_1 = \frac 6 {4 v}$ and $a = - \frac{64}{36} V^3$, giving a maximum derivative of $A = \frac 8 3 V^2$.
On the interval $[0,x_1]$, the absolute value of the derivative has a maxima at $x = 0$ and $x = x_1$, with value $-a x_1$ Thus, $x_1 = \sqrt{\frac{A}{6}}$ and $a = - \sqrt{6 A}$, giving a maximum value of $V = \frac{A^{\frac{2}{3}}}{4 \sqrt{ 6}}$
Depending on your choice of $V$ and $A$, you may not satisfy both conditions, in which case you'll have to choose the stricter of the two. Checking the sign of $\frac 8 3 V^2- A$ is a good test for that...