Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 8 / c

The function is a probability density function, rather than a cumulative distribution: its integral over both intervals must be one. Thus $2\alpha +2\alpha =1$ and

The density is symmetric about $7/2$, giving expected value $\mathbb{E}[X]=7/2$. Its second moment is $\mathbb{E}[X^2]=\frac{1}{4}({\int }_1^3{x}^{2}dx+{\int }_4^6{x}^{2}dx)=89/6$. ThereforeEquivalently, each equally weighted interval has variance $1/3$ and its centre is $3/2$ from the overall mean, giving $1/3+9/4$ by the law of total variance.