How do I scale my parabolas so that their integrals over [0,1] are always the same?

Question

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.

Answer 1

$$\int_{0}^{1} 1-N\left(x-\frac{1}{2}\right)^2 \mathrm{d}x=1-\frac{N}{12}$$ 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 $$\int_0^1 M*\left(1-N\left(x-\frac{1}{2}\right)^2 \right)\mathrm{d}x=C$$

Answer 2

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 $N$.

Answer 3

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 $1$).