Confused about derivation of center of mass formulae So, not sure if I'm really dumb and not getting something obvious or if there is more complexity here...Let's do 2 objects first. Say C.O.M. is x°. 
Then $x° = \dfrac{x_1M_1 +x_2M_2}{M_1+M_2} $
I get the "idea" of this formula–you are in a sense averaging the value of the masses. But is there some blatantly obvious reason for why this is true? It can easily be derived from:
$(x°-x_1)M_1 = (x_2-x°)M_2$ 
but I don't see why the initial statement ($x° = \dfrac{x_1M_1 +x_2M_2}{M_1+M_2} $) is a priori true. Is there a geometric intuition? 
If we are doing a continuous non-constant function:
$\int_{a}^{x°}f(x)(x°-x)dx= \int_{x°}^{b}f(x)(x-x°)dx $
It can be derived that this is equal (if I did it right) to $\dfrac{\int_{a}^{b}xf(x)dx}{\int_{a}^{b}f(x)dx} $
But again, is there a more obvious geometric intuition? Thanks
 A: Notice the definition of center of mass $x_c$ for two particles rearranges to 
$$x_c = \Big(\frac{m_1}{m_1+m_2}\Big)x_1+\Big(\frac{m_2}{m_1+m_2}\Big)x_2$$
Let's call the ratio $(\frac{m_1}{m_1+m_2})$ "$W_1$". It can can be interpreted as the percentage of mass that object 1 contributes to the system, generally $(\frac{mass of object 1}{total mass})$. To convert to units of percentage simply multiply the resulting decimal by 100%. Notice that $W_1 \leq 1$, that is object 1 (any object) cannot contribute more that 100% of the mass of the system.
In general a weighted average (arithmetic mean) $x_{av}$ of $n$ components is
$$x_{av} = w_1x_1 + w_2x_2 + ... + w_nx_n = \sum_{i=1}^{n} w_ix_i$$
where $w_i$ is the weight (in the statistical sense) of component $x_i$. For convenience we often wish for the weight factors to be normalized, that is, 
$$\sum_{i=1}^{n} W_i = 1$$
Notice here I wrote $W_i$, the normalized weight factors, instead of $w_i$, which are not necessarily normalized. 
For your physics application of this math, the mass $m_i$ of object $i$ corresponds to the "crude" weight factor $w_i$. To normalize weight factors (so that none is greater than 100%), we divide the crude factor by the total weight of all components:
$$W_i = \frac{w_i}{\sum_{i=1}^{n} w_i}$$
or, for us, the object mass by the total mass of the system:
$$W_i = \frac{m_i}{\sum_{i=1}^{n} m_i} = \frac{m_i}{M} $$
(where "$M$" is a common abbreviation for the total system mass that is, $M = \sum_{i=1}^{n} m_i$.) This is why our weight factors for calculating center of mass consist of the mass of the object divided by the total mass of the system. For two objects ($n=2$), the weight factor $W_i$ for object $i$ is of course 
$$W_i = \frac{m_i}{\sum_{i=1}^{2} m_i} = \frac{m_i}{m_1+m_2} $$
Thus, 
$$x_c = \Big(\frac{m_1}{m_1+m_2}\Big)x_1+\Big(\frac{m_2}{m_1+m_2}\Big)x_2 \\
= \sum_{i=1}^{2} \Big(\frac{m_i}{m_1+m_2}\Big) x_i \\
= \frac{\sum_{i=1}^{2} m_ix_i}{\sum_{i=1}^{2} m_i} \\
= \frac{1}{M} \sum_{i=1}^{2}x_im_i$$
I've written a few forms of $x_c$ above so you can mull over how this expression is constructed. For the general center of mass of a system of $n$ objects, simply replace $2$ with $n$ in the final expression.
Let's explicitly juxtapose our center of mass formula with the weighted average formula to drive the point home:
$$x_{av} = \sum_{i=1}^{n} W_ix_i \\
\\
x_c= \sum_{i=1}^{n} \Big(\frac{m_i}{M}\Big) x_i$$
Thus, the center of mass is the weighted average of position with respect to mass.
$$\\ \\$$
P.S. If the system has a continuous distribution of mass rather than $n$ discrete chunks (as for any real object that is not being approximated as a dimensionless point particle), instead of 
$$x_c = \frac{1}{M} \sum_{i=1}^{n}x_im_i$$
We can use 
$$x_c = \frac{1}{M} \int xdm$$
which arises when we take the limit as $n \to \infty$. (Note this form is not the most practical to calculate so we usually make the substitution $dm = \rho dV$ where $\rho$ is the density of the object, which may vary with position.)
A: From the point of view of mathematics, there is no further explanation. The formula 
$$(x°-x_1)M_1 = (x_2-x°)M_2$$ 
has to be considered as an axiom. The motivation comes from physics, and is related to the following image of masses balancing each other:

