Question: Why does a differential factor appear when going from discrete summation to continuous summation, what is the meaning of the differential factor that appears in the integral? i know its meaning from the point of view that a integral is the area under a curve, but if we are summing a bunch of numbers, why are we multiplying that summation by a differential factor?

Answer: 
I don't know how the notation $"\int \dots\,dx"$ came into use, although it makes some sense and fits well with [Leibniz's notation](https://en.wikipedia.org/wiki/Leibniz%27s_notation) for the derivative $"\frac{d\cdot}{dx}"$.

By your question, I assume you're talking about [Riemann integrals](https://en.wikipedia.org/wiki/Riemann_integral), however let's consider a more advanced and powerful tool: [Lebesgue integration](https://en.wikipedia.org/wiki/Lebesgue_integration).
Keep in mind that if $f$ is properly Riemann integrable, then it's also Lebesgue integrable.

The Lebesgue integral of a function $f$ over a set $E$ looks like this

$$\int\_E f\,d\mu$$

where $\mu$ is a [measure](https://en.wikipedia.org/wiki/Measure_(mathematics)), a function that *weighs* sets. Here are a couple of examples:

* $\lambda$, Lebesgue measure, is the usual one: given an interval $I=(a,b)$ then $\lambda(I)=b-a$ aka the length of the interval
* P, a probability measure, where the measure $P(A)$ is the likelihood that the event represented by the set $A$ will occur. Think about a die, if $A=\{1,2\}$ then $P(A)=\frac{1}{3}$.

Let's go back to Lebesgue integrals: the function $f$ is integrated with respect to the measure indicated by the symbol $d\mu$.

Notice that I haven't said nothing about the domain of the function $f$, not yet.

If $f\colon\mathbb{R}\to\mathbb{R}$ and $\mu\equiv\lambda$, then you get the usual integral:
$$
\int\_\mathbb{R}f\,d\lambda\equiv\int\_\mathbb{R}f(x)\,d(x)
$$

Now consider the sequence $a\_n$, which is a function defined on natural numbers, $a\_n\colon\mathbb{N}\to\mathbb{R}$. The default choice for a measure defined on $\mathbb{N}$ is the so called [counting measure](https://en.wikipedia.org/wiki/Counting_measure):
$$
\mu\_c(A)=
\begin{cases}
|A|\qquad \text{if A is finite}\\
\infty\qquad \text{if A is not finite}
\end{cases}
$$

The counting measure *counts* how many elements are in each set. Of course, if $A$ is a singleton then $\mu\_c(A)=1$.
The Lebesgue integral of a sequence $a\_n$ is $\int\_\mathbb{N}a\_n\,d\mu\_c$, using $\sigma$-additivity:
$$
\int\_\mathbb{N}a\_n\,d\mu\_c=\sum\_{n\in\mathbb{N}}\int\_{\{n\}} a\_n\,d\mu\_c=\sum\_{n\in\mathbb{N}} a\_n\mu\_c(\{n\})=\sum\_{n\in\mathbb{N}}a\_n
$$

As you can see, even when we are "summing a bunch of numbers" there's a "differential factor" $d\mu\_c$. It just happens to hide away because it's equal to 1.

I hope you find this answer useful. I too used to struggle with understanding the relation between integrals and discrete summations. Lebesgue integration theory answered almost all of my doubts.
If you sum a finite "bunch" of numbers well, you are just doing a sum. 

The problem comes when the bunch is infinite. Then you have a series. And you know under which conditions the series can result in a finite number: first terms can be finite, but those in the tail shall become smaller and smaller. We started and know this since 2000 years with the exhaustion method.  

Then for instance, to find the area of the circle you can start from that of the inscribed equilateral triangle, add that of the triangles built on its sides, etc. That is the sum of infinite decreasing terms.  

But you can do that dividing the circumference into small segments of length $\Delta x$ and sum the triangles with vertex in the center: a Riemann sum. Differently from before it is the sum of many terms all comparably small (and the study of this comparison is what led to calculus).  

You can consider it as well as the sum of many finite terms, weighted by $\Delta x/ 2 \pi$ : and here is where $dx$ pops out.