Where do I fail with this integral (coordinates on a simplex)? Let us regard the standard simplex in two dimensions
$S = \{ x_1 + x_2 \leq 1, x_1 \geq 0, x_2 \geq 0 \}$
Where does the following calculation fail:
$ \int_S x_1^2 = \int_0^1 x_1^2 \int_0^{1-x_1} dx_2 dx_1
= \int x_1^2 ( 1 - x_1 ) dx_1 = \int_0^1 x_1^2 - x_1^3 dx_1
= \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$
while
$ \int_S x_1^2 = \int_0^1 \int_0^{1-x_2} x_1^2 dx_1 dx_2
= \int_0^1 \dfrac{(1-x_2)^3}{3} dx_2 = \frac{1}{3}\sum^3_{i=0}\int_0^1 x_2^i dx_2 = \frac{1}{3}\sum^3_{i=0} \dfrac{1}{i+1} dx_2$
$ = \frac{1}{3}( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} ) = \frac{7}{36}$
I suppose the error is the error is in the first equality signs (except I have done some very dumb mistake that I am blind to.

There is an oversight in the second calculation.
You wanted
$$\int \frac{(1-x_2)^3}{3}\,dx_2,$$
and decided to expand $(1-x_2)^3$ and integrate term by term.
Note that by the Binomial Theorem, or otherwise,
$$(1+t)^3=1+3t+3t^2+t^3,$$
so $(1-x_2)^3=1-3x_2 +3x_2^2-x_2^3$. 
Expanding is in any case an inefficient way to evaluate the integral. Instead, make the substitution $u=1-x_2$. Then our integral becomes
$$\int_1^0 -\frac{u^3}{3}\,du.$$
