Answer
$$f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{3}{{98}}{x^2};\,\,\,\,\,\,\,\left[ {3,5} \right] \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ then}} \cr
& \frac{3}{{98}}{x^2}{\text{ is positive for all real number }}x,{\text{ then }}f\left( x \right) \geqslant 0{\text{ for the given interval}} \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr
& \int_3^5 {\left( {\frac{3}{{98}}{x^2}} \right)} dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{{3{x^3}}}{{98\left( 3 \right)}}} \right)_3^5 \cr
& = \frac{1}{{98}}\left( {{{\left( 5 \right)}^3} - {{\left( 3 \right)}^3}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{{98}}\left( {125 - 27} \right) \cr
& = \frac{1}{{98}}\left( {98} \right) \cr
& = 1 \cr
& \cr
& {\text{The conditions are verified}}{\text{, so }} \cr
& f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr} $$
