Understanding Integration This is a question coming from a Math newb.
When I learn something I try to connect it to what I already know, so please bear with me.
I know how to calculate integrals, but I never really understood what they mean, until I read this and felt like it was a really good explanation. So I tried to test the hypothesis with some examples.
When you have a static function like 
$$f(x) = 2$$
And you make a definite integration, it's ok:
$$\int_0^32\space dx = 2  \cdot 3- 2 \cdot 0 = 6$$
That's basically like saying 3 $\cdot$ y, with y behaving according to $y = 2$. But as soon as I try to test what he said with a variable, e.g.: $$f(x) = 2x$$
$$\int_0^3 2x\space dx= \frac{2 \cdot 3^2}{2} - \frac{2 \cdot 0^2}{2} = 9$$
However I expected it to be like $3 \cdot y$, with $y$ behaving according to $y = 2$.
In the end I thought, what I was actually thinking of was $\sum$, but I still don't really get what integration (a/o differentiation) means.
I know when I calculate the area under the function curve, but what does that mean?
Let's think only in positive integer values. I have an unlimited amount of apple baskets. The first one will have two apples, the second one will have four apples, the one after that 6, etc.
So that function would be:
$$\operatorname{How many apples do I have after basket}(x) = 2x$$
Then what would the value of the definite integral between 0 and 3 (which is $9$) imply?
 A: I think that your confusion stems from a misunderstanding of how summations relate to integrals. Think of an integral as the limit of a Riemann Sum, or $Lim_{\Delta x\rightarrow 0} \Sigma f(x)\Delta x$. Let's think for a second about how we derive this.
We know an integral is the area under a curve. But how can we calculate this without calculus? Well, we can start with an approximation. Take the function $f(x) = x^3$, for example, on the interval 0 to 3. One way to begin approximating this is to use a series of boxes that fit under the curve, since we can easily calculate the area of a box. Starting with 4 boxes, our approximation would, graphically, look something like this:

And we could get are approximate integral by taking the sum of the areas of the rectangles. Using the simple $area = h * w$ formula and the graph, we can see that the height of any one rectangle is the function value at the rectangle's leftmost point, and the width of a rectangle is a constant .5. This is nice, and all, but there is a lot of area under the curve that the rectangles don't 'cover'. One way to 'cover' more area, and get a better approximation, is to use slimmer rectangles:

Now, there is much less area that is not 'covered' by the rectangles under the curve, which will result in a better approximation. So if error decreases as we make the rectangles slimmer, what happens if we make the rectangles infinitely slim? The error would go to 0, and we would get our perfect amount of area under the curve. Abusing notation for a second, we can think of this as $Lim_{width\rightarrow 0} \Sigma height * width$, or, a sum of infinitely slim rectangles under the curve. Since height is the function value at the leftmost point of the rectangle, and width is just a distance along the x axis, we can substitute our variables and get the integral formula we began with,  $Lim_{\Delta x\rightarrow 0} \Sigma f(x)\Delta x$
The reason that your how many apples question fails is that your apple 'function' is not continuous. Imagine graphing your apple function, which is defined only on the positive integers. It would only be a series of points at x=1, 2, 3, etc... and not a smooth curve. What the integral is calculating is the area under the smooth curve defined on all real numbers within your bounds. 
A: Your intuition is very true. You are adding apples up. For a non-wild function  $\int_a^b f(x)dx$ means adding up (infinitely many) values $f(x)$. But why does it become finite? Because each value $f(x)$ only occupies the single point $x$ on the interval $[a,b]$. So, it is adding up very many values but each value being multiplied by an almost zero segment. Thus, it is the interplay of these that allows for the final result to become finite number.
A second interpretation: Imagine a function that is $1$ from $x=0$ to $x=2$, is $5$ from $x=2$ to $x=7$. What is a fair average for this function? Is it $\frac{1+5}{2}$ because those are the only values $f$ takes on? I don't think it will be fair. Because $f$ tries "hard" to be $5$ on a longer interval. If you drive at $10$ mph for $30$ minutes and then continue at $70$ mph for 3 hours, will your average speed be $50$?! Of course not. So, to define a nice average, you must take into account the bandwidth, the segment, the portion where $f$ takes on a certain value. In integrating $f(x)$ over $[a,b]$ you add up values of $f$ taking into account "how much $x$-space" that value of $f$ occupies.
Note to mathematicians: This way of looking at $\int f$ is closer to Lebesgue integral than to Riemann. One wonders why the latter was invented first!
