Is this line of reasoning correct/valid? I'm only in the second month of my first calculus course, so I'm not sure how much sense this question will make. I'll give it a try anyways though.

Let's say you have the sum of an infinite series like this
$$\lim_{n\to \infty}\sum_{x=1}^n f(x,n)$$  
Now imagine that each whole number value for $f(x,n)$ creates an area with width $1$ and length $f(x,n)$ (meaning the area will just equal $f(x,n)$). By dividing by the limit of $n$ (or shrinking the width), I would imagine you could make it so that the combined width of all of the terms in the series would instead effectively equal 1. My reason for thinking this is that there are $n$ terms in the series, each with width $\frac{1}{n}$. Thus $\frac{1}{n} \times{n}=1$.
I should note, that when I say $f(x,n)$, I really want $x$ to be the variable, and $n$ to be a constant where $n \to \infty$. Weather one should multiply or divide by $n^{c}$ to get meaningful results, probably depends on the function. Now...
correct me if I'm wrong, but I think that for very specific functions, one could create a whole new type of function by doing this. One where integer-ial changes of $x$ For $f(x,n)$, are equal to infinitesimal changes for this new function.
Lets call this new function $g(x)$, for example. Then the $n^{th}$ term of $f(x,n)$ for $x$, would be equal to $g(1)$. Similarly, the $\frac{1}{2}n^{th}$ term would be equal to $g(\frac{1}{2})$.
I'm wondering if my thinking is correct. Even partial answers would be appreciated.
 A: Your question is a great one, because by answering it, you will get introduced to the delightful concept of integral. Keep asking questions, even if they sound crazy to you.
First I will show you an example of what you are proposing. You start with a small $n$ and you get a rough graph of $f(x,n)$, like

If you increase $n$ you will see $f(x,n)$ is getting a lot smoother.

In this case, I was using $f(x,n)=|sin(\frac{13 \pi}{n}x)|$. This means that $f(x,n)$ is a discretization of $sin(13 \pi t)$. For a wide-enough class of functions (i.e. continuous by parts functions) $h(t)$, whith $t\in[0,1]$ you can apply the following reasoning to such discretizations by setting $f(x,n)=h(\frac{x}{n})$ where $x\in \{1,...,n\}$. I will add that if you want to find the nearest value of $h(t)$ in the discretization $f(x,n)$ for some t, you can use $h(t)\approx g(t,n)= f(\lfloor tn \rfloor,n)$, ($\lfloor \cdot \rfloor$ is the floor function.) In the limit you will obtain 
$$h(t)=\lim_{n\rightarrow\infty} g(t,n)$$
Which means that, as you were saying, in the limit, integer changes in $x$ will cause infinitesimal changes in $h(\frac{x}{n})$.
Now let's attack the sums you were talking about. To make it clearer, let's redefine the discretization by setting $\Delta t =\frac{1}{n}$. Also,  as you were hypothesizing, I will need to multiply each sum by $\Delta t$. Let's write all that down:
$$I_{\Delta t}=\sum_{x=1}^{\frac{1}{\Delta t}}f(x,\frac{1}{\Delta t})\Delta t=\sum_{x=1}^{\frac{1}{\Delta t}}h(x\Delta t)\Delta t$$
And taking the limit
$$I=\lim_{\Delta t\rightarrow 0}I_{\Delta t}$$ You will see nothing less than
$$I=\int_{0}^{1}h(t)dt$$ 
Which is the definite integral of $h(t)$ in $[0,1]$. Although it is not the most general definition of the integral, as you will eventually learn, it is still interesting, don't you think? Also, it's very easy to extend this formulation for $t$ in any interval $[a,b]$. 
