On the volume of a non-right tetrahedron and the integral of a linear function over it I've solved quite a bit of these questions with tetrahedrons with a 90 degree angle, but I am not sure how to approach a non-right angle tetrahedron. I am unsure how to find the triple integral's bounds. Calculation itself is no problem. 
$ \int_B (x + y + 2z) dV$
where B is the tetrahedron with vertices
$(0, 0, 0)$, $(0, 2, 0)$, $(1, 3, 1)$ and $(1, 1, 0)$.
Is the way of finding the bounds same for non-right angle tetrahedrons? 
 A: Actually the integration of a linear function on a tetrahedron with a vertex at the origin does not require to find the volume of such tetrahedron, but we may solve both problems at once. $\mathbb{R}^3$ is spanned by $v_1=(0,2,0)^T$, $v_2=(1,3,1)^T$ and $v_3=(1,1,0)^T$ and the wanted volume is just 
$$ \frac{1}{6}\left|\det M\right|=\frac{1}{6}\left|\det\begin{pmatrix}0 & 1 & 1 \\ 2 & 3 & 1 \\ 0 &  1 & 0\end{pmatrix}\right|=\frac{1}{3} $$
by a Laplace expansion along the first column. Each point of the given tetrahedron $T$ is a linear combination of $v_1,v_2,v_3$ with non negative coefficients, whose sum does not exceed $1$. If we enforce the substitution
$$(x,y,z)^T = M (X,Y,Z)^T$$
we have that
$$\iiint_{T}(x+y+2z)\,d\mu = 2\iiint_{\substack{X,Y,Z\in[0,1]\\X+Y+Z\leq 1}}2X+6Y+2Z\,d\mu $$
and the integration problem over a generic tetrahedron boils down to the integration problem over a trirectangular tetrahedron. By the linearity of the integral and symmetry, this further simplifies into just computing
$$ \iiint_{\substack{X,Y,Z\in[0,1]\\X+Y+Z\leq 1}}X\,d\mu = \int_{0}^{1}X\cdot\frac{1}{2}(1-X)^2\,dX=\frac{1}{24}.$$
It follows that the value of the wanted integral is $\frac{2(2+6+2)}{24}=\color{red}{\frac{5}{6}}$.
