The length X of a side of a square is uniformly distributed between 1 and 6. Find the probability density function of the area of the square. Then find the mean and variance of the area of the square.

\begin{align} The \; p.d.f \; of \; X \; is \\ \\ f(x) &= \frac{1}{5} \\ \\ and \; X \; is \; a \; random \; variable, so \\ \\ 1 &= \int_{1}^{6} \frac{1}{5}dx \quad \enclose{circle}{\color{black}{1}} \\ \\ Let \; A \; be \; the \; area \; of \; the \; square. \; Then \\ \\ x^2 &= a \\ \\ x &= a^{\frac{1}{2}} \\ \\ differentiating \; gives \\ dx &= \frac{1}{2}a^{- \frac{1}{2}} da \quad \enclose{circle}{\color{black}{2}} \\ \\ and \; the \; limits \; [1,6] \; for \; X \\ become \; [1,36] \; for A. \\ Then \; substituting \; \enclose{circle}{\color{black}{2}} \; into \; \enclose{circle}{\color{black}{1}} \; gives \\ \\ 1 &= \int_{1}^{36} \left(\frac{1}{5} \right) \frac{1}{2}a^{- \frac{1}{2}} da \\ \\ so \; the \; p.d.f. \; of \; A \; is \\ \\ g(a) &= \frac{1}{10}a^{- \frac{1}{2}} \;, \; 1 \leqslant \; a \; \leqslant \; 36 \\ \\ \\ To \; find \; the \; mean, \; we \; know \; that \\ \\ E(X) &= \int_{1}^{36}a \frac{1}{10}a^{- \frac{1}{2}} \\ \\ &= \frac{1}{10} \int_{1}^{36}a^{\frac{1}{2}} \\ \\ &= \frac{1}{10} \left[ \frac{2}{3}a^{\frac{3}{2}} \right]_{1}^{36} \\ \\ &= \frac{2}{30} \left(36^{\frac{3}{2}}- 1^{\frac{3}{2}} \right) \\ \\ &= \frac{2}{30}(216-1) \\ \\ &= 14 \frac{1}{3} \; square \; units \\ \\ \\ To \; find \; the \; variance, \; we \; know \; that \\ \\ Var(X) &= E(X^2)- E^2(X) \\ \\ Using \; the \; p.d.f. \; of \; A \;, \\ \\ E(X^2)&= \int_{1}^{36} a^2 \frac{1}{10}a^{- \frac{1}{2}} \\ \\ &= \frac{1}{10} \int_{1}^{36}a^{\frac{3}{2}} \\ \\ &= \frac{1}{10} \left[ \frac{2}{5}a^{\frac{5}{2}} \right]_{1}^{36} \\ \\ &= \frac{2}{50} \left(36^{\frac{5}{2}}- 1^{\frac{5}{2}} \right) \\ \\ &= \frac{2}{50}(7776-1) \\ \\ &= 311 \\ \\ So \quad Var(X) & = 311- \left(14\frac{1}{3}\right)^2 \\ \\ &= 311-205.\dot{4} \\ \\ &= 105.6 \quad (to \; 2 \; d.p.) \end{align}