Proving the Shoelace Formula with Elementary Calculus?

The shoelace formula found here or here tells you how to calculate the area of any polygon given its coordinates.

The second link I mentioned gives a proof of it, but it is a bit beyond my level of comprehension. Could anyone try to simplify the proof (or provide their own) to a level up to and including single variable calculus?

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A simple induction proof can be based on this fact [*]:

Every simple polygon has a diagonal, that is, a segment joining two non-consecutive vertices that lies totally inside the polygon.

Take the polygon and split it along a diagonal. The area of the polygon is the sum of the areas of the two parts. By induction, these areas are given by the formula. When you combine them, the terms corresponding to the diagonal cancel.

Here’s a really good response I found on YahooAnswers a while ago:

The proof in the link is sheer madness.

For this clockwise order to make sense, you need a point $O$ inside the
polygon so that the angles form $(OA_i, OA_{i+1})$ and $(OA_n,OA_1)$ be all positive.

Then the formula is just adding up the areas of the triangles
$OA_iA_{i+1}$ and $OA_nA_1$.

So all you need is area of $ OA_1A_2 = \dfrac {b_2a_1-b_1a_2}{2} $, which is
elementary.

So here’s my summary of the above answer:

  1. I don’t quite agree that the AoPS proof is sheer madness $-$ it is pretty rigorous and I like it, in the long run.
  2. Pick a point $O$ inside the polygon so that we have positive angles.
  3. You want the sum of the areas $\left[\triangle OA_iA_{i+1}\right]+\left[\triangle OA_nA_1\right]$ over all points $A_i$.
  4. Using the area of a triangle, we have $ \left[\triangle OA_1 A_2\right] = \dfrac {b_2a_1 – b_1 a_2}{2} $. Summing over all indices, you get the desired result.

$ \blacksquare $