Expected Area of Triangle from larger segment Take a stick of length one and pick a point from it randomly. This will obviously divide the stick into two pieces. Pick another point from inside the larger interval and divide the larger interval into two pieces. If we assume this is a case where the three pieces form a triangle, what is the expected area of the triangle formed?
Thanks for any help or suggestions!
Note: Question was originally on Overflow but I thought it fit this sit better. 
 A: Let $a$ be the length of the smaller piece from the first break, and $b$ be the length of the leftmost piece from the second break. A triangle is formed if $\frac12-a < b < \frac12$.
By Heron's formula, the area of that triangle is
$$
   \sqrt{\frac12\left(\frac12 - a\right)\left(\frac12-b\right)\left(\frac12-(1-a-b)\right)}
$$
and so for a fixed $a$, the average value of the area is
$$
    \frac1{\frac12-(\frac12-a)}\int_{\frac12-a}^{\frac12}\sqrt{\frac12\left(\frac12 - a\right)\left(\frac12-b\right)\left(\frac12-(1-a-b)\right)}\,\text{d}b.
$$
This looks messy, but before even integrating, we can simplify this into
$$
   \frac1a \sqrt{\tfrac12(\tfrac12-a)} \int_{\frac12-a}^{\frac12} \sqrt{\left(\frac12-b\right)\left(\frac12-(1-a-b)\right)}\,\text{d}b
$$
and after substituting $t = a + b - \frac12$, into
$$
   \frac1a \sqrt{\tfrac12(\tfrac12-a)} \int_0^a \sqrt{t(a-t)}\,\text{d}t.
$$
The region between the $t$-axis and the graph of $\sqrt{t(a-t)} = \sqrt{(\frac a2)^2 - (\frac a2-t)^2}$ is a semicircle with radius $\frac a2$ centered at $(\frac a2, 0)$, so its area is $\frac12 \pi (\frac a2)^2$, and the expression above simplifies to $\frac\pi 8 a \sqrt{\frac12(\frac12-a)}$.
Now we integrate with respect to $a$ (and divide by $\frac12$) to take the average: our final answer will be
$$
    2\int_0^{1/2}\frac\pi 8 a \sqrt{\tfrac12(\tfrac12-a)}\,\text{d}a.
$$
A substitution of $u = \frac12-a$ reduces this to integrating $u^{1/2}$ and $u^{3/2}$, and the whole thing simplifies to $\frac{\pi}{120}$.
