Answer
$=\frac{1}{6}x^{2}-\frac{1}{6}x-1$
Work Step by Step
We are given $f(x)=\frac{1}{3}x-\frac{1}{6}$
for $3\leq x \leq 4$
The cumulative distribution function is given by
$F(x)=P(X\leq x)=\int^{x}_{3}(\frac{1}{3}t-\frac{1}{6})dt$
$=\frac{1}{6}t^{2}-\frac{1}{6}t|^{x}_{3}$
$=\frac{1}{6}x^{2}-\frac{1}{6}x-1$
