Intuition behind the extected value formula, for continuous cases

Question

I was thinking about the formula that gives us the expected value of a continuous random variable with pdf $f(x)$, that is
$$E[X]=int_{a}^{b} xf(x)dx$$
How can this, intuitively, gives us the expected value of $X$? I can totally understand it in the discrete case, because it is analogous to just calculating the arithmetic mean.
But here? I am not sure, there isn't even a proper probability in this formula.
The only thought that can give me some idea is that $f(x)dx$ is the probability that $X$ is in the small range of width $dx$. After this thought, and considering the integral as an infinite summation, I can see the analogy with the discrete case.
However I am not certain this is the way to think about it. A more detailed, based on intuition, reasoning would be appreciated. Thanks!

d.k.o. · Accepted Answer

Consider the following Riemann sum with $Delta=(b-a)/n$:
$$
S_n:=sum_{i=1}^n x_i^* f(x_i^*)Delta,
$$
where $x_i^*in [x_{i-1},x_i]$ and $x_i=a+iDelta$. The term $f(x_i^*)Delta$ is, approximately, the probability of $Xin [x_{i-1},x_i]$, and so, $S_n$ can be interpreted as an aprroximate average of $X$. As $n$ increases, this approximation becomes better, and in the limit,
$$
S_nto int_a^b xf(x),dx=mathsf{E}X.
$$

Intuition behind the extected value formula, for continuous cases

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