Can the sum of $f(x)$ where $a<x<b$ be approximated to the area underneath?

Question

How to find the sum of a function within a given range? For example $\sum f(x)$ when $x$ varies from say $0$ to $1$?

My line of reasoning,

1. Divide the interval $(0,1)$ into small equal parts of length $dx$ and assume (approximate) that the function has a constant value $f(x_c)$ in that interval $dx$ ($x_0$ is either at the beginning of $dx$ or at the end of $dx$).

2. Now the sum of the function in this interval $dx$ is just $f(x_c) + f(x_c) +.... f(x_c)$ ((i.e.) add $f(x_c)$, $dx$ times) which means the sum of $f(x)$ in the interval $dx$ is approximated to $(f(x_0) \times dx)$ that is, Sum $\approx$ area of the rectangle with height $f(x_0)$ and breadth $dx$.

3. Move on to the next interval $dx$ and do the same and finally add all these small areas.

So in conclusion the objective, of finding the sum of $f(x)$ in the interval $0 <x< 1$ is approximately equal to the area underneath it?

Is this correct or have I gone wrong somewhere?

Edit:

Context: I'm learning continuous probability distributions and am trying to reason out why area underneath = the probability to be in between the given range (as a sum of all probabilities).

I understand there's no meaning in the phrase "Sum of f(x)" but that's exactly what we are trying to get around with continuous probability distributions aren't we?

Answer 1

Your intuition about summing up infinitesimal probabilities is correct. The values of the probability density function $f_X$ of a continuous random variable $X$ are not actual probabilities, though - the density function is by definition the derivative of the cumulative distribution function. So your observation follows from the fundamental theorem of calculus. That is, $$P(a<X<b)=\int_a^b f_X(x) dx.$$ Remember that $f_X(x)$ is not the same thing as $P(X=x)$. It's better to think of it as $$f_X(x)=\lim_{h\to 0}\frac{P(x<X<x+h)}h.$$