Answer
$$f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{9}x - \frac{1}{{18}};\,\,\,\,\,\,\,\left[ {2,5} \right] \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ then}} \cr
& \frac{1}{9}x - \frac{1}{{18}} \geqslant 0 \cr
& 2x - 1 \geqslant 0 \cr
& x \geqslant \frac{1}{2} \cr
& \left[ {2,5} \right]{\text{ is on the interval }}\left[ {\frac{1}{2},\infty } \right).{\text{ then }}f\left( x \right) \geqslant 0{\text{ for the interval }}\left[ {2,5} \right] \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr
& \int_2^5 {\left( {\frac{1}{9}x - \frac{1}{{18}}} \right)} dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{1}{{18}}{x^2} - \frac{1}{{18}}x} \right)_2^5 \cr
& = \left( {\frac{1}{{18}}{{\left( 5 \right)}^2} - \frac{1}{{18}}\left( 5 \right)} \right) - \left( {\frac{1}{{18}}{{\left( 2 \right)}^2} - \frac{1}{{18}}\left( 2 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{10}}{9} - \frac{1}{9} \cr
& = 1 \cr
& \cr
& {\text{The conditions 1 and 2 are verified}}{\text{, then }} \cr
& f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr} $$
