Solution
= Solution
The <probability density function> is normalized because $\int_1^\infty3x^{-4}\,dx=1$. Its <expected value> and second moment are
$$
\mathbb E[X]=3\int_1^\infty x^{-3}\,dx=\frac32,\qquad
\mathbb E[X^2]=3\int_1^\infty x^{-2}\,dx=3.
$$
Both <improper integrals> converge, so the <variance> is
$$
\boxed{\mathbb E[X]=\frac32,\qquad \operatorname{Var}(X)=3-\left(\frac32\right)^2=\frac34.}
$$