Answer
$\color{blue}{F_Y(y) = \begin{cases} 0, & y \lt 0 \\ y^4, & 0\le y \lt 1 \\ 1, & y \ge 1 \end{cases} \\
P(0 \le Y \le \frac{1}{2}) = F_Y(\frac{1}{2}) -F_Y(0) = \dfrac{1}{16}}$
Work Step by Step
$\underline{\text{Find}\ F_Y(y)}$
$\begin{align*}
F_Y(y) &= P(Y\le y) \\
&= \int^y_{-\infty} f_Y(u)\ du \\
&= \begin{cases} \displaystyle \int^y_{-\infty} 0 \ du, & y \lt 0 \\ \displaystyle\int^0_{-\infty} 0\ du + \int^y_0 4u^3\ du, & 0\le y \lt 1 \\ \displaystyle\int^0_{-\infty} 0\ du + \int^1_0 4u^3\ du + \int^y_1 0\ du, & y \ge 1 \end{cases} \\
&= \begin{cases} 0, & y \lt 0 \\ 0 + \left( 4\dfrac{u^4}{4}\right]_0^y, & 0\le y \lt 1 \\ 0 + 1 + 0, & y \ge 1 \end{cases} \\
&= \begin{cases} 0, & y \lt 0 \\ \left( y^4 - 0^4\right), & 0\le y \lt 1 \\ 1, & y \ge 1 \end{cases} \\
\color{blue}{F_Y(y)} &\color{blue}{= \begin{cases} 0, & y \lt 0 \\ y^4, & 0\le y \lt 1 \\ 1, & y \ge 1 \end{cases}} \\
\end{align*}$
$\underline{\text{Using}\ F_Y(y), \text{find}\ P(0\le Y\le \frac{1}{2})}$
$\begin{align*}
P(0\le Y\le {\scriptsize\frac{1}{2}}) &= F_Y(1/2) - F(0) \\
&= (1/2)^4 - 0 \\
\color{blue}{P(0\le Y\le {\scriptsize\frac{1}{2}})} &\color{blue}{= 1/16}
\end{align*}$
