The original question is:
Suppose a 12 inch metal rod inside a factory of some sort is secured at both ends. The rod is then placed under significant pressure until it breaks cleanly. Let $X$ be the distance from the left end where the break occurs, and the pdf of $X$ is given as: $$f_X(x)=\left\{\begin{array}{rl} \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right), & 0\le x \le 12 \\ 0, & \text{otherwise} \end{array}\right.$$ Find the expected length of the shorter segment when the rod breaks.
I found the expected value using: $$E[X]=\dfrac{1}{24} \int_0^{12}\left({x^2-\dfrac{x^3}{12}}\right)dx=6$$ But the answer is given as $3.75$ and I have no idea how they got that. Wouldn't I expect the rod to break exactly in the middle, and therefore, there wouldn't be a shorter or longer segment in the first place? Or is this conditional expectation?
Let $Y = \min\{X,12-X\}$. Observing that $0 \leq X \leq 12$, we have that:
so in other words:
\begin{equation} Y=\left\{ \begin{array}{lr} X & ;\,\,X\leq 6\\ 12-X & ;\,\,X\geq 6 \end{array}\right.\end{equation}
Hence the integral becomes:
$$\mathbb{E}(Y)=\int_0^6x\cdot \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right)\,dx\,+\,\int_6^{12}(12-x)\cdot \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right)\,dx$$