Answer
$\color{blue}{\frac{9}{4} = 2.25\ \text{years}}$
Work Step by Step
$\begin{align*}
E(Y) &= \int_\mathbb{R} y\cdot f_Y(y)\ dy \\
&= \int_{-\infty}^\infty y\cdot f_Y(y)\ dy \\
&= \int_0^3 y\cdot\left(\frac{1}{9}y^2\right)\ dy \qquad \text{since}\ f_Y(y)= \frac{1}{9}y^2,\ 0\le y\le 3 \\
&= \frac{1}{9}\int_0^3 y^3\ dy \\
&= \frac{1}{3^2}\left[ \frac{y^4}{4}\ \right\vert_0^3 \\
&= \frac{1}{(3^2)(4)}\left( 3^4 - 0^4\right) \\
&= \frac{3^4}{(3^2)(4)} \\
\color{blue}{E(Y)}\ &\color{blue}{= \frac{9}{4}}
\end{align*}$
The average time that such a patient spends in remission is about $\color{blue}{\dfrac{9}{4} = 2.25\ \rm years}$.
