Answer
$\color{blue}{\dfrac{5}{16}}\quad$ or $\quad \color{blue}{0.3125}$
Work Step by Step
Using Def. 3.4.2,
$\begin{align*}
P\left(\frac{3}{4} \le Y\le 1\right) &= \int_{3/4}^1 \left(\dfrac{2}{3} +\dfrac{2}{3}y\right)\ dy \\
&= \left[ \dfrac{2}{3}y + \dfrac{2}{3}\dfrac{y^2}{2}\right]_{3/4}^1 \\
&= \left[ \dfrac{2}{3}y + \dfrac{1}{3}y^2\right]_{3/4}^1 \\
&= \dfrac{2}{3}(1-3/4) + \dfrac{1}{3}(1^2-(3/4)^2) \\
&= \dfrac{2}{3}\cdot\dfrac{1}{4} + \dfrac{1}{3}(1 - 9/16) \\
&= \dfrac{1}{6} + \dfrac{1}{3}(7/16) \\
&= \dfrac{8}{48} + \dfrac{7}{48} \\
&= \dfrac{15}{48} \\
\color{blue}{P\left(\frac{3}{4} \le Y\le 1\right)} &\color{blue}{= \dfrac{5}{16} \;=\; 0.3125}
\end{align*}$
