Orthogonality of Functions in Function Space The dot product of two functions, $f$ and $g$ in $C[a,b]$ is given by
$$f \cdot g = \int_a^b p(x)f(x)g(x)dx$$
where $p(x)$ is the weight function. If $p(x) = 1$ then,
$$f \cdot g = \int_a^b f(x)g(x)dx$$
The dot product for vectors is the sum of the multiplication of like components. Functions that are continuous over an interval can have infinite many values because there are infinitely many values you can put into the function. That is why an integral is used in the above definition above of the dot product of two functions. 
My question involves the $dx$. I know it tells you what variable you are integrating over, but in Calc I, $dx$ had the purpose of giving a small length to a rectangle so that you could sum up all the areas of the small rectangles. If the dot product of $f$ and $g$ was defined as 
$$f \cdot g = \int_a^bf(x)g(x)$$
without the differential, it would make sense to me, but you need a differential in all integrals. So why is the dot product the area under the function $q(x) = f(x)g(x)$? This question comes up a lot, especially in physics, since my professors will say we need to sum up something, but they just tack on the $d$ whatever. 
 A: When they taught you the "geometrical meaning" of the symbol $dx$ it was merely for pedagogical and mnemotechnical purposes, as you say, it's there to tell you over which variable you're integrating and some other things. But in the end it's just notation and it's there to remind you what the integral symbol stands for. 
I think your problem is of other nature. You're trying to compare the "usual" dot product with this new dot product. Even though they are similar, they are two different products in different spaces and represent different things, and that's where I think the confusion comes from. For some mysterious reasons this dot products work great in different physical problems, but in the end they are just dot products (a bilinear function such that ...). There's nothing more to it.
Using a physicist's answer: With the units of the usual dot product you'll end up with a scalar with the units of the product of the units (huh..). In the case of the integral dot product you'll end up with an extra unit that comes from the integration (actually whenever you use any of this dot products you always take into account this).
