Distribution of values of a function over a certain range I don't have that much of knowledge about statistics and probability theory and I need to have the distribution of values of a function in a certain range $(a_1,a_2)$.
Consider an arbitrary function $y=f\left( x \right)$ :

I want to have a sense of the distribution of $y$s over a specific range. The closest I can think of is a made up formula:
$$\delta(y)=\sum \left|f' \right|_{f(x)=y}$$
Meaning that the $\delta(y)$ is the sum of derivatives of $f$ at the points where it reaches $y$. 
Or maybe:
$$\delta(y)=\frac{d}{dy}\int^{a_1}_{a_2} Ramp(f(x)-y)dx$$
If the range is big enough I expect this distribution function to:


*

*have a maxima at the average

*be zero below the global minima and above the global maxima

*have a zero derivative at global minima/maxima and average

*to be differentiable 


I would appreciate if you could help me know if this distribution function has a name? I need some keywords which I can use to read about it. and if there is any way to analytically calculate it?
P.S.1 @Moderators, I have no clue what tags should I put here. please help me with that.
 A: This whole inquiry can go in a number of directions, and I'm just gonna address the most basic and standard one.
Your first proposal involving $|f'|$ is on the right track to formalize the following intuitive notion:

"when $x$ is uniformly sampled with points of constant density, how dense would the corresponding points along $y$ be around $y=f(x)$?"

Now, there's is no well-defined uniform distribution for $x$ over the entire real line. For a bounded domain of length $L$ (for example $x = -L/2$ to $L/2$), what you are looking for is (adopting your notation of $\delta$ meaning the distribution)
$$ \delta_Y(y) = \frac1{L}\dfrac{1}{\bigl| df(x) / dx\bigr|_{x=f^{-1}(y)}} \qquad \text{where $\displaystyle \frac1{L}$ comes from} \qquad \delta_X(x) = \frac1{L}$$
This is the standard transformation of random variables, where $X$ is uniform. That is, $$\delta_Y(y) = \delta_X\bigl(\, x(y)\, \bigr) \cdot | J |$$ where $J$ is the Jacobian of the inverse mapping $x = h(y) = f^{-1}(y)$. In the case of single variable (univariate), the Jacobian is just the derivative $h'$, which is just $1/f'$.
(btw, actually your second proposal of "differentiate the integral" is also sort of on the right track "from the other direction", albeit with a slightly wrong guess on the cumulative function) 
I think I don't need to explain to you how this works intuitively, because you already derived it yourself. 
Let me just remind you that, while it is conceptually straightforward, technically one has to split the mapping $f(x)$ into segments of one-to-one mapping so that the inverse exists for each segment. The result would be the sum of the segments (or the countable union, if infinitely many cuts).
