What is a simple, integral-free definition of the area of a complicated plane figure that is digestible at or near the level of introductory calculus? What is an integral-free definition of the area of a complicated plane figure at/near the level of introductory calculus?
 A: Here is an axiomatic approach: (Note: by $(a,b)$ I mean any of the intervals $[a,b], ]a,b[, [a,b[$ or ]a,b]).
A simple region is a subset $S$ of $\mathbb{R}^2$ that is a union of finitely many rectangles $S=\bigcup_{i=1}^n (a_i,b_i)\times (c_i,d_i)$ such that $[a_i,b_i]\cap [a_j,b_j]$ is either a point or empty for $i\neq j$, and similarly for the intervals $[c_i,d_i]$.
Now suppose that there is a set $\mathcal{M}$ of subsets of $\mathbb{R}^2$ and a function $a:\mathcal{M}\to \mathbb{R}$ such that:


*

*For every $A\in \mathcal{M}$ we have $a(A)\geq 0$.

*If $A,B\in\mathcal{M}$ then $A\cup B\in\mathcal{M}$ and $A\cap B\in\mathcal{M}$ and we have
$$
a(A\cup B) = a(A) + a(B) - a(A\cap B).
$$

*If $A,B\in\mathcal{M}$ and $A\subseteq B$, then $B\setminus A\in \mathcal{M}$ and
$$
a(B\setminus A) = a(B)-a(A).
$$

*If $A\in \mathcal{M}$ and $T:\mathbb{R}^2\to\mathbb{R}^2$ is a displacement (i.e. rigid motion), then $T(A)\in\mathcal{M}$ and
$$
a(T(A)) = a(A).
$$

*For $a<b$ and $c<d$ we have that $(a,b)\times(c,d)\in\mathcal{M}$ and
$$
a((a,b)\times (c,d))=(b-a)(d-c).
$$

*Let $A\subseteq \mathbb{R}^2$. If there exists a unique $c\in\mathbb{R}$ such that
$$
a(S)\leq c\leq a(S')
$$
for every pair of simple regions $S,S'$ such that S\subseteq A\subseteq S', then $A\in\mathcal{M}$ and $a(A)=c$.


Then we can define a subset $A$ of $\mathbb{R}^2$ to be "measurable" if it belongs to $\mathcal{M}$, and define the area of $A$ as $a(A)$.
May be the most complicated part to introduce is that of a rigid motion, because it needs some amount of linear algebra, but it can be treated in an intuitive way, if the aim is to present it at the level of introductory calculus.
Each one of the axioms can be interpreted geometrically. You can also note that some of this axioms are redundant, but for an introductory level it does not matter (I think).
A: For "figures" that are finite in extent, I propose the following: 
Take a large square $S$ of "area"  $U$ (computed by the width x height rule) that completely encloses your figure. Choose a point uniformly randomly in $S$, and if it's inside your figure, produce the value $1$; otherwise produce $0$. The expected value for this process, multiplied by $U$, is called the area of the figure. 
To make this computational, generate many many random points, say $N$ of them; suppose that $P$ end up inside the figure. Then the area is approximately $\frac{P}{N} U$, and as $N$ increases, the approximation gets better and better. 
The very sharpest students might reasonably ask, "How do you know that there is an expected value?" You answer by complimenting them on their insight. :) 
It's at least clear that the expected value (if it exists) is larger than that of any collection of rectangles whose union is inside the shape, and less than that of any collection of rectangles whose union contains the shape, and this can be used to estimate the expected value in some very useful practical cases. 
This presentation has the advantage that it sets students up for the practical ways we compute areas of shapes for which antiderivatives are impractical, and it ties into the classic chemist's trick: to compute the area under a curve, you weigh the (rectangular) paper on which the curve is plotted and note its area; then you use scissors to cut out the area under the curve and weigh just that. The ratio of the small to the large weight is the ratio of the small region's area to the area of the rectangle. 
