Answer
$\color{blue}{\frac{3}{80}}$
Work Step by Step
$\begin{align*}
E(Y) &= \int_{\mathbb{R}} y\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y)\ ] \\
&= \int_0^1 y\cdot(3(1-y)^2)\ dy \qquad [\ \text{since}\ f_Y(y)=3(1-y)^2,\ y\in[0,1]\ ] \\
&= 3\int_0^1(y-2y^2+y^3)\ dy \\
&= 3\left( \frac{y^2}{2} - 2\frac{y^3}{3} + \frac{y^4}{4}\ \right\vert_0^1 \\
&= 3\left[\left( \frac{1^2}{2} - 2\frac{1^3}{3} + \frac{1^4}{4}\ \right) - \left( \frac{0^2}{2} - 2\frac{0^3}{3} + \frac{0^4}{4}\ \right)\right] \\
&= 3\left[\left(\frac{1}{2}-\frac{2}{3} + \frac{1}{4}\right) - (0-0+0)\right] \\
&= 3\left(\frac{6-8+3}{12}\right) \\
E(Y) &= \frac{1}{4} \\
\mu &= \frac{1}{4}
\end{align*}$
$\begin{align*}
E(Y^2) &= \int_{\mathbb{R}} y^2\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ] \\
&= \int_0^1 y^2\cdot(3(1-y)^2)\ dy \qquad [\ \text{since}\ f_Y(y)=3(1-y)^2,\ y\in[0,1]\ ] \\
&= 3\int_0^1(y^2-2y^3+y^4)\ dy \\
&= 3\left( \frac{y^3}{3} - 2\frac{y^4}{4} + \frac{y^5}{5}\ \right\vert_0^1 \\
&= 3\left[\left( \frac{1^3}{3} - \frac{1^4}{2} + \frac{1^5}{5}\ \right) - \left( \frac{0^3}{3} - \frac{0^4}{2} + \frac{0^5}{5}\ \right)\right] \\
&= 3\left[\left(\frac{1}{3}-\frac{1}{2} + \frac{1}{5}\right) - (0-0+0)\right] \\
&= 3\left(\frac{10-15+6}{30}\right) \\
E(Y^2) &= \frac{1}{10} \;\lt\; \infty
\end{align*}$
By Theorem 3.6.1, since $\mu= \dfrac{1}{4}$ exists and $E(Y^2) = \dfrac{1}{10}$ is finite,
$\begin{align*}
\text{Var}(Y) &= E(Y^2) - \mu^2 \\
&= \frac{1}{10} - \left(\frac{1}{4}\right)^2 \\
&= \frac{1}{10} - \frac{1}{16} \\
&= \frac{8-5}{80} \\
\color{blue}{\text{Var}(Y)}\ &\color{blue}{= \frac{3}{80}}
\end{align*}$
