Integrals as a sum of a series? I'm just getting started with calculus - enough to be comfortable with derivatives, but we haven't been taught integrals yet. However, as far as I can tell, they are all about finding the area under a graph (perhaps not the most rigorous definition). I was thinking, is it possible to express an integral as the limit of the average of y values between an interval multiplied by the delta x in that interval (as x approaches infinity)? More specifically:
$$\lim_{x \to \infty} \left(a\sum_{n=0}^x \left(na\frac{1}{x}\right)^2\right) = \int_{0}^ax^2dx,\quad x\in[0,a], x\in \mathbb{N}$$
The basic idea is that the function takes smaller and smaller intervals of the function $x^2$, hence the $\frac{1}{x}$. Thus, $(na\frac{1}{x})^2$ results in the y value of $x^2$ at the the nth $x$ value in the series $0, \frac1x, \frac2x, ..., \frac{a}x$.
I made a graph to check my results, but I can't tell if it's just a coincidence because of the simplicity of the function I'm integrating ($y=x^2$), so I would appreciate some sort of affirmation or rebuttal to my understanding of integrals.
I apologize if the wording of my question is awkward or hard to understand - since I don't understand integrals, I don't know much of the proper mathematical language behind them, and as such I realize the mathematical notation in my question also probably horrible. 
 A: 
I was just wondering if the intuition behind it was correct (that integrals are the average of a function over an interval multiplied by the interval length).  

Yes. This is related to the Mean Value Theorem. The theorem is defined as such (due to Wikipedia):

More precisely, if a function ${\displaystyle f}$ is continuous on the closed interval  ${\displaystyle [a,b]}$ , where ${\displaystyle a<b}$ , and differentiable on the open interval ${\displaystyle (a,b)}$ , then there exists a point ${\displaystyle c}$ in ${\displaystyle (a,b)}$ such that 
  $$f'(c)  = \frac{f(b) - f(a)}{b-a}$$  

If we let $f'(x)=g(x)$, and apply the Fundamental Theorem of Calculus, we get the following:
$$g(c) = \frac{1}{b-a}\int_a^b g(x) dx$$
Mutiply both sides by $b-a$ and your claim follows!!  
Edit:
I should really stress that you should look up Riemann Sums. They are usually how we define integrals, and are indeed a sum as your intuition gave! Props to you on a correct intuition in two regards :)
Also,  my proof above does require that you know the Fundamental Theorem of Calculus, namely that $\int_a^b f(x) dx = F(b)-F(a)$, where $F(x) $ is the antiderivative of $f(x)$. If you don't want to use this see my addendum below, once you become familiar with Riemann Sums and want to try a proof. I would love to help if you need it! 
Notes and Addendum:
The above formula, as I wrote it, calculates the average value of a function along an interval, and I have thus often heard it referred to as the "Average Mean Value Theorem". I personally found it very enjoyable to prove directly using Riemann sums and the definition of the arithmetic mean once I became familiar with both (basically you average along smaller rational intervals, and note that this becomes the definition of a Riemann Sum and conclude). This works because you can analytically continue the sum over the rationals to all real numbers by continuity, but I digress. While I'm at it, I ought also to mention that the Mean Value Theorem can be used to find a secant line for the function on the interval... Proving this and also proving the "Average MVT" separately really helped me understand the connections between the Integral, Antiderivative, and Derivative, as the first and third show up in the derivations and the simple substitutions $f'(x) =g(x)$ really ties them together, bringing in the antiderivative in the process and ALSO uses the most fundamental theorem of calculus! 
