Linear transformations map lines to lines, this is something you may already know or otherwise can show easily.
Slightly more general than your statement: linear transformations also map parallelograms to parallelograms. Note that a parallelogram is completely determined by three (non-collinear) points (vertices), call them $\vec a$, $\vec b$ and $\vec c$. The parallelogram is then given by the set of all points $\vec x$:
$$\vec x = \vec a + \lambda \bigl( \vec b - \vec a \bigr)+ \mu \bigl( \vec c - \vec a \bigr) \quad \quad \left( 0 \le \lambda, \mu \le 1 \right)$$
A linear transformation $L$ maps these points to:
L(\vec x) & = L\left(\vec a + \lambda \bigl( \vec b - \vec a \bigr)+ \mu \bigl( \vec c - \vec a \bigr)\right) \quad \quad\quad \left( 0 \le \lambda, \mu \le 1 \right)\\
& = L(\vec a) + \lambda \bigl( L(\vec b) - L(\vec a) \bigr)+ \mu \bigl( L(\vec c) - L(\vec a) \bigr)
And the set of all points $L(\vec x)$ is now a parallelogram determined by the points $L(\vec a)$, $L(\vec b)$ and $L(\vec c)$. Note that these vertices are not collinear since $\vec a$, $\vec b$ and $\vec c$ weren't.
I would greatly appreciate it if people could please take the time to explain why this phenomenon occurs and what the benefits of it are to us.
In the context of integration, you can try to use this to map an ugly parallogram to a nice(r) parallogram, at least as a region of integration. If you can map an arbitrary parallogram to a rectangle in the coordinate space, the integral limits will become simple constants.