Answer
$\color{blue}{\frac{5}{252}}$
Work Step by Step
$\begin{align*}
E(Y) &= \int_{\mathbb{R}} y \cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y)\ ] \\
&= \int_0^1 y\cdot 5y^4\ dy \qquad \left[\ \text{since}\ f_Y(y)= 5y^4,\ 0\le y \le 1\ \right] \\
&= \int_0^1 5y^5\ dy \\
&= 5\left(\frac{y^6}{6}\ \right\vert_0^1 \\
&= \frac{5}{6}(1^6-0^6) \\
E(Y)&= \frac{5}{6} \\
\mu_Y &= \frac{5}{6}
\end{align*}$
$\begin{align*}
E(Y^2) &= \int_{\mathbb{R}} y^2 \cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ] \\
&= \int_0^1 y^2\cdot 5y^4\ dy \qquad \left[\ \text{since}\ f_Y(y)= 5y^4,\ 0\le y \le 1\ \right] \\
&= \int_0^1 5y^6\ dy \\
&= 5\left(\frac{y^7}{7}\ \right\vert_0^1 \\
&= \frac{5}{7}(1^7-0^7) \\
E(Y^2)&= \frac{5}{7}
\end{align*}$
By Theorem 3.6.1, since $\mu_Y = \dfrac{5}{6}$ exists and $E(Y^2) = \dfrac{5}{7}$ is finite,
$\begin{align*}
\text{Var}(Y) &= E(Y^2) - \mu_Y^2 \\
&= \frac{5}{7} - \left(\frac{5}{6}\right)^2 \\
&= \frac{5}{7} - \frac{25}{36} \\
&= \frac{180-175}{252} \\
\color{blue}{\text{Var}(Y)}\ &\color{blue}{= \frac{5}{252}}
\end{align*}$
