Integrating over a tetrahedron Let $S$ be the tetrahedron in $\mathbb{R}^3$ having vertices $(0,0,0), (0,1,2), (1,2,3), (-1,1,1)$. Calculate $\int_S f$ where $f(x,y,z) = x + 2y - z$. 
Before I show you guys what I have tried, please no solutions. Just small hints. Now, I have been trying to set up the integral by looking at $x$ being bounded between certain planes, etc. I ended up with $$\int_0^{x+2} \int_{\frac{z}{2} - \frac{x}{2}}^2 \int_{2y - z}^{3z - 4y} f\:\: \mathrm{d}x\mathrm{d}y\mathrm{d}z.$$ But this doesn't seem correct. The question came with a hint: To find a linear diffeomorphism $g$ to use as a change of variables, but I have been unable to find such a mapping between $S$ and the unit cube.

I am reading "Analysis on Manifolds" by James R. Munkres.
This exercise is Exercise 6 on p.151 in section 17 in this book.
I solved this exercise.
Viktor Vaughn's answer is very beautiful.
But my answer is not beautiful.
Let $A:=(0,0,0),B=(1,2,3),C=(0,1,2),D=(-1,1,1)$.
The equation of the plane passing through $A,B,C$ is $p_1(x,y,z)=-x+2y-z=0$.
The equation of the plane passing through $A,B,D$ is $p_2(x,y,z)=-x-4y+3z=0$.
The equation of the plane passing through $A,C,D$ is $p_3(x,y,z)=x+2y-z=0$.
The equation of the plane passing through $B,C,D$ is $p_4(x,y,z)=-x+z=2.$
$p_1(D)>0$.
$p_2(C)>0$.
$p_3(B)>0$.
$p_4(A)<2$.
So, $S=\{(x,y,z)\in\mathbb{R}^3\mid p_1(x,y,z)>0\text{ and }p_2(x,y,z)>0\text{ and }p_3(x,y,z)>0\text{ and }p_4(x,y,z)<2\}.$
Let $$H:=\begin{pmatrix}-1&2&-1\\-1&-4&3\\1&2&-1\end{pmatrix}.$$
Then $\det H=4\neq 0$.
Let $G:=H^{-1}$.
Then, $$G=\begin{pmatrix}-1/2&0&1/2\\1/2&1/2&1\\1/2&1&3/2\end{pmatrix}.$$
Let $g:\mathbb{R}^3\ni(u,v,w)\mapsto ((-1/2)u+(1/2)w,(1/2)u+(1/2)v+w,(1/2)u+v+(3/2)w)\in\mathbb{R}^3.$
Then, $R:=g^{-1}(S)=\{(u,v,w)\in\mathbb{R}^3\mid 0<u\text{ and }0<v\text{ and }0<w\text{ and }u+v+w<2\}.$
$g$ is a diffeomorphism from $R$ to $S$.
$\det Dg=1/4$.
So, by Change of variables theorem, $\int_S x+2y-z=\int_R w\cdot (1/4)=1/6.$
