Answer
$=\frac{1}{18}x^{2}-\frac{1}{18}x-\frac{1}{9}$
Work Step by Step
We are given $f(x)=\frac{1}{9}x-\frac{1}{18}$
for $2\leq x \leq 5$
The cumulative distribution function is given by
$F(x)=P(X\leq x)=\int^{x}_{2}(\frac{1}{9}t-\frac{1}{18})dt$
$=\frac{1}{18}t^{2}-\frac{1}{18}t|^{x}_{2}$
$=\frac{1}{18}x^{2}-\frac{1}{18}x-\frac{1}{9}$
