How do I scale my parabolas so that their integrals over [0,1] are always the same? I need an upside-down parabola that grows from 0 to 1/2, then decreases symmetrically from 1/2 to 1. So I've opted for the following:
$f(x) = 1-2\left(x-\frac{1}{2}\right)^2$
But I also need to be able to adjust the "sharpness" of the parabola, which I've found I can do using a parameter, that I'm calling N:
$g(N,x) = 1-N\left(x-\frac{1}{2}\right)^2$
The higher N is, the sharper my parabola becomes. The trouble is that I'd like all parabolas (for all values of N between 0 and 4) to have the same integral over [0,1], and I can't figure out how to scale it. I have a feeling that I could multiply it by some expression that depends on N, but cannot figure out which one.
$h(N,x) = M\left(1-N\left(x-\frac{1}{2}\right)^2\right)$
I figured out how to scale the integral (I can provide some details if they're at all useful) but not how to scale the original function. What does M have to be to ensure that the integral is always the same for all N in ]0,4[?
Note that I talk about scaling, but adding a constant might also be satisfactory, as long as the parabola is always strictly positive over [0,1], which is also why I'm restricting N to ]0,4[. Ideally, the integral would not just be constant be equal to 1 (which I realise cannot be the case for my values of N without using a scaling factor or an additional constant), but I can live without it.
 A: $$\int_{0}^{1} 1-N\left(x-\frac{1}{2}\right)^2 \mathrm{d}x=1-\frac{N}{12}$$
So:
$$\frac{1}{1-\frac{N}{12}}\int_{0}^{1} 1-N\left(x-\frac{1}{2}\right)^2 \mathrm{d}x=1$$
So for $M=C*\frac{12}{12-N}$ the integral
$$\int_0^1 M*\left(1-N\left(x-\frac{1}{2}\right)^2 \right)\mathrm{d}x=C$$
A: We just integrate the expression with $N$ included:
\begin{align}
\int_0^1 1-N\left(x-\frac12\right)^2\,\mathrm dx &= \int_{-1/2}^{1/2} 1-Nx^2\,\mathrm dx \\
&= x-\frac N3x^3~\Bigg\vert_{x=-1/2}^{1/2} \\
&= 1-\frac N{12} = \frac{12-N}{12}
\end{align}
It follows that if we choose $M = \dfrac{12}{12-N}$, we achieve the desired property of the value of the integral being independent of $N$.
A: A slightly easier version to work with might be the following function:
$$
h_{M,N}(x)=N\left(x-\frac{1}{2}\right)^2+M
$$
This allows you to change the tightness of the parabola and its height.  In this case, we can compute the integral as
$$
\int\left(N\left(x-\frac{1}{2}\right)^2+M\right)dx=\frac{1}{3}N-\frac{1}{2}N+\frac{1}{4}N+M.
$$
To make this constant, you need 
$$
\frac{1}{3}N-\frac{1}{2}N+\frac{1}{4}N+M=C
$$
or that
$$
M=C-\frac{1}{3}N+\frac{1}{2}N-\frac{1}{4}N
$$
for any constant $C$ (including $1$).
