The answer is the convex polygon inscribed in a semicircle of diameter $A_0A_n$, that is, $\angle A_0A_iA_n = 90^\circ$ for all $i = 1, 2, \dots, n-1$.

First of all, the polygon must be convex because if it isn't the case one can obtain a polygon of greater area by reflecting the sides of the polygon inside of its convex hull.

Consider an arbitrary vertex $A_i$ and fix the polygons $A_0A_1 \dots A_i$, $A_iA_{i+1} \dots A_n$; in particular, $A_0A_i$ and $A_iA_n$ are fixed. Now the area of $A_0A_1A_2 \dots A_n$ is the sum of the areas of the fixed polygons $A_0A_1 \dots A_i$, $A_iA_{i+1} \dots A_n$ and the triangle $A_0A_iA_n$, which attains its maximum if and only if $\angle A_0A_iA_n = 90^\circ$, since the area of the triangle is $\frac{1}{2}A_iA_0 \cdot A_iA_n \sin \angle A_0A_iA_n$ and $A_iA_0$ and $A_iA_n$ are both fixed.

It remains to show that such polygon is unique. Indeed, if $A_0A_n = 2r$ and $O$ is the midpoint of $A_0A_n$ and center of the semicircle, then $\angle A_iOA_{i+1} = 2 \arcsin \frac{x_i+x_{i+1}}{2r}$. But $\sum_{i=1}^n 2 \arcsin \frac{x_i+x_{i+1}}{2r} = \pi$ and $\arcsin$ is increasing in $[0, 1]$, so $r$ is unique and hence the polygon is unique.