Area under the parabola is always 1. But why? I have the following matlab function:
function y=par(x,s)
%parabolic density function with basis = 2s
y=0*x;
ind=-s<x & x< s;
y(-s<x & x< s)=(3/(4*(s^3)))*(s*s-x(ind).*x(ind));

This function implements
$$y_s(x) = \begin{cases}
\frac{3}{4s^3}(s^2-x^2) & \text{if } -s<x<s, \\
0 & \text{otherwise.}
\end{cases}$$
The area under the parabola is always one. I tested it with the following matlab procedure:
s = 0.5:0.1:100.5;
for i=1:length(s)
    q =  quad(@(x)par(x,s(i)),-s(i),s(i));
end

q is always 1 here.
btw.
quad is a matlab function to evaluate the integral. Here it calculates $q = \int_{-s}^s y_s(x)\,dx$.
Can somebody explain to me why it is always one ?
 A: The integral of the function is $1$, which is the area under the parabola:
$\displaystyle\int_{-s}^s\frac{3}{4s^3}(s^2-x^2)dx= \frac{3}{4s^3}\left(\int_{-s}^{s}s^2dx-\int_{-s}^{s}x^2dx \right) =\frac{3}{4s^3}\left(2ss^2-2\int_{0}^{s}x^2dx \right)= \\
=\displaystyle\frac{3}{4s^3}\left(2s^3-2\frac{s^3}{3}\right)=1$
A: It may help to rewrite your integrand as
$$ y_s(x) = \frac{3}{4s^3}(s^2 - x^2) = \frac 1s \frac 34 \left( 1 - \left( \frac xs \right)^2 \right). $$
Now, let's first consider the case $s = 1$, which simplifies the integrand to
$$ y_1(x) = \frac 34 (1 - x^2), $$
which is to be integrated from $-1$ to $1$.  From high school calculus, I'm sure you'll remember that
$$ \int_0^1 x^2 \,dx = \frac 13 $$
which implies that
$$ \int_0^1 (1 - x^2) \,dx = 1 - \frac 13 = \frac 23 $$
and thus, since $(-x)^2 = x^2$,
$$ \int_{-1}^1 (1 - x^2) \,dx = 2 \cdot \frac 23 = \frac 43. $$
Multiply that by $\frac 34$, and you get $1$.
Now, what about $s \ne 1$?  The rewritten integrand has a factor of $\frac 1s$ in front of it, so the height of the parabola is divided by a factor of $s$.  On the other hand, the argument $x$ is also divided by $x$, which effectively multiplies the width of the parabola (and, not coincidentally, the range over which it is integrated) by $s$.  Those two effects cancel out, leaving the area unchanged.
In fact, we can generalize this result: for any function $f$ integrable on $[a,b]$ and any non-zero constant $s$,
$$\begin{aligned}
\int_a^b f(x) dx 
&= \frac 1s \int_{sa}^{sb} f\left(\frac xs\right) \,dx \\
&= \int_{sa}^{sb} \frac 1s f\left(\frac xs\right) \,dx .
\end{aligned}$$ 
