What is the expected length of the shorter segment when the stick snaps?

[Ackbach]

Here is this week's POTW, a problem submission by Track. Thanks, Track!

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I thought y'all could use some more stochastic love. Requires knowledge of calculus-based probability theory.

Suppose a $12$-inch, uniformly-shaped wooden stick is held securely at both ends, such that the stick does not move. The stick is then subjected to significant pressure until it snaps cleanly. Let $X$ be the distance from the left end of the stick at which the break occurs. The probability density function of $X$ is given by:
$$f_X(x)=\left\{\begin{array}{rl}
\left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right), & 0\le x \le 12 \\
0, & \text{otherwise}
\end{array}\right.$$
What is the expected length of the shorter segment when the stick snaps?

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[Ackbach]

Congratulations to Opalg for his correct solution, which follows:

Spoiler

The function $f_X(x)$ is symmetric about $x=6$. If the break does not occur in the left half of the stick (at $X$, with $X\leqslant6$) then it will occur at $12-X$ and the length of the shorter segment will again be $X$. Putting these two possibilities together, the expected length of the shorter segment must be $$2\int_0^6xf_X(x)\,dx = \int_0^6\frac{x^2}{12}\Bigl(1-\frac x{12}\Bigr)\,dx = \Bigl[\frac{x^3}{36} - \frac{x^4}{12\cdot48}\Bigr]_0^6 = 6 - \frac94 = \frac{15}4.$$ Thus the expected length of the shorter segment is $3\frac34$ inches.