Chapter 3 Random Variables - 3.4 Continuous Random Variables - Questions - Page 136: 1

$\color{blue}{1/16}$ or $\color{blue}{0.0625}$

Using Def. 3.4.2, $\begin{align*} P\left(0\le X\le \frac{1}{2}\right) &= \int_0^{1/2} 4y^3\ dy \\ &= \left[ 4\cdot \dfrac{y^4}{4} \right]_0^{1/2} \\ &= \left[ y^4\right]_0^{1/2} \\ &= (1/2)^4 - 0^4 \\ &= 1/2^4 - 0 \\ \color{blue}{P\left(0\le X\le \frac{1}{2}\right)} &\color{blue}{= 1/16 \;=\; 0.0625} \end{align*}$