Is variance the area under the curve of the distribution of a population? I am trying to understand what variance is, I already know the "official" definition
"Variance is the average squared deviations from the mean"
But I am trying to give it a visual entity.
I get to this conclusion through this logic.
Given this formula that represent variance :

If an Xi is way far from the mean, then it will get a big square but given his little proportion of Pi, it become very tiny square. Otherwise, when an element closer to the mean it will get a little square but the big proportion of Pi, turn it into big square.
So in my head it turns into something like this:

Is this idea an approximation to the reality or it his a flagged conclusion?

Based in what you have comment me, and what I have watched in this video that I could look for thanks to you. I've cleared my mind a little more.
https://www.youtube.com/watch?v=ISaVvSO_3Sg
There are some details that I maybe get a little confused and couldn't express myself properly, misusing tags etc
The area of a probability it will always by 1 cause it refers to the total of a proportion that it cat reach to 100% (or 1) right? From what @AdamO have told me in the comments
Let me bring this hipottetical cases to understand a little bit more
Let supposed that, I have a a population of 40 elephants, with the same age, with the exactly same weight, nor even a nanogram  lower or heigher.
If I have to use this formula

It will be no variance since all have the same data right?
Lets suppose now that I have the same set elephants, but now we add 20 more elephants to weight 100 kg less than the original, and another 20 that are 100 kg heavier than the original.
If we keep adding more elephants, with more weight and less weight, keeping the balance and keeping the same mean.
It will start change the variance.
I wouldn't know exactly the figure or the shape of the distribution, but I would know, the final size of the combination of all of them. Isn't that somehow, the area of the distribution of the population? The only case that it didn't exist is in the hipotetical case of a stack of elements with all have the same value therefore, stacks of elements that all are the mean
Thank you very much! @Accumulation @User1865345
 A: The area is the width times the height, but the variance is the width squared times the height. In statistics, we speak of "moments". The n-th moment is the integral of x the n-power times the function. The 0-th moment is $\int x^0 f(x)dx$, or just $\int f(x) dx$. For a PDF, that should be 1 no matter what. The 1st moment is  $\int x^1f(x) dx$, which is the mean. The second moment is $\int x^2f(x) dx$, which for a centered distribution, is the variance. You can get a visual by imagining the function rotated about a vertical line that goes through the mean. The volume that results will be proportional to the variance; if you have probability mass far from the mean, then it will rotate through a larger circle and contribute more to the volume.
A: While both the comments are correct, whuber ♦'s articulated it in a more comprehensive sense that can be conjured up as a "visual entity".
Variances are defined for those random variables which are residing in the space $\mathcal L^2$ where $\mathbb EX^2<\infty.$ It is a measure of spreadness.
For an intuitive take, the concept of moment of inertia comes handy.  $\rm [I]$ expounded it distinctly (emphasis mine):




Variance provides a 'quadratic' measure of how widely the distribution of
an RV is spread about its mean. The 'moment of inertia' idea makes this precise.
Suppose that $X$ is 'continuous'. Imagine that we make a very thin sheet of metal the shape of the area under the graph of $f_x,$ of unit mass per unit area. The variance of $X$ measures how hard it is to spin this metal sheet about a vertical axis through the mean — see Figure $\rm E(i).$ One precise form of this statement is that if the sheet is spinning, at $1$ complete revolution per second, then its total kinetic energy is a certain constant $(2\pi^2)$ times $\operatorname{Var}(X).$ Experience shows that variance provides a much more useful measure of
spread than, for example, $\mathbb E (|X — \mu_X|).$


Reference:
$\rm [I]$ Weighing the Odds: A Course in Probability and Statistics, David Williams, Cambridge University Press, $2001,$ sec. $3.5,$ pp. $67-68.$
