This is similar to the question posted here but instead of the point being uniformly distributed we choose it based on a beta distribution. Simulations show a simple relationship between the area and $\alpha=\beta$, so I am wondering if there is a closed form here that can be derived.
In the case $\alpha=\beta=1$ we have the special case of the point being uniformly distributed.
In the case of $\alpha=\beta$ I believe the integral involved takes the form:
$$\frac12 \int_0^1 \frac{1}{B(\alpha,\alpha)}x^{\alpha-1}(1-x)^{\alpha-1}x(1-x)dx = \frac12 \int_0^1 \frac{1}{B(\alpha,\alpha)}x^{\alpha}(1-x)^{\alpha}dx$$
In the more complicated case of $\alpha$ and $\beta$ different the integral becomes:
$$\frac12 \int_0^1 \frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}x(1-x)dx = \frac12 \int_0^1 \frac{1}{B(\alpha,\beta)}x^{\alpha}(1-x)^{\beta}dx$$
$$ A(\alpha,\beta)=\int_0^1 \frac{x^{\alpha}(1-x)^{\beta}}{2B(\alpha,\beta)}\,dx=\frac{B(\alpha+1,\beta+1)}{2B(\alpha,\beta)}=\frac{\alpha\beta}{2(\alpha+\beta+1)(\alpha+\beta)}. $$ As expected $A(\alpha,\alpha)\to 1/8$ as $\alpha\to\infty$.