Example 5.4.12.
Consider our previous example 5.4.4. To compute the mean 1 and variance 2 (and hence the standard deviation) for this distribution,
\begin{equation*}
\mu = \int_{-1}^2 x \cdot x^2/3 dx = \int_{-1}^2 x^3/3 dx = \frac{2^4}{12} - \frac{(-1)^4}{12} = \frac{15}{12} = \frac{5}{4}
\end{equation*}
and by using the alternate formulas 5.4.6
\begin{align*}
\sigma^2 & = E[X^2] - \mu^2\\
& = \int_{-1}^2 x^2 \cdot x^2/3 \; dx - \mu^2\\
& = \int_{-1}^2 x^4/3 \; dx - \left ( \frac{5}{4} \right )^2\\
& = \frac{2^5}{15} - \frac{(-1)^5}{15} - \frac{25}{16}\\
& = \frac{33}{15} - \frac{25}{16} = \frac{51}{80}
\end{align*}
which gives
\begin{equation*}
\sigma = \sqrt{\frac{51}{80}} \approx 0.7984.
\end{equation*}
For skewness 3, note that in computing the variance above you also found that
\begin{equation*}
E[X^2] = \frac{11}{5}.
\end{equation*}
So, once again by using the alternate formulas 5.4.6
\begin{equation*}
E[X^3] = \int_{-1}^2 x^3 \cdot x^2/3 dx = \frac{x^6}{18} |_{-1}^2 = \frac{7}{2}
\end{equation*}
and so
\begin{equation*}
\gamma_1 = \frac{\frac{7}{2} - 3 \cdot \frac{5}{4} \cdot \frac{11}{5} + 2 \cdot (\frac{5}{4})^3}{\sqrt{\frac{51}{80}}^3}
\end{equation*}
For kurtosis 4, you can reuse \(E[X^3] = \frac{7}{2}\) and \(E[X^2] = \frac{11}{5}\) and the alternate formulas 5.4.6 to determine
\begin{equation*}
E[(X-\mu)^4] = E[X^4] - 4 \mu \cdot E[X^3] + 6 \mu^2 \cdot E[X^2] - 3 \mu^4
\end{equation*}
which is the numerator for the kurtosis.