Answer
$$f\left( x \right){\text{ }}is{\text{ }}not{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^3}}}{{81}};\,\,\,\,\,\,\,\left[ {0,3} \right] \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ then}} \cr
& \frac{{{x^3}}}{{81}} \geqslant 0 \cr
& {x^3} \geqslant 0 \cr
& x \geqslant 0 \cr
& \left[ {0,3} \right]{\text{ is on the interval }}\left[ {0,\infty } \right).{\text{ then }}f\left( x \right) \geqslant 0{\text{ for the interval }}\left[ {0,3} \right] \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr
& \int_0^3 {\left( {\frac{{{x^3}}}{{81}}} \right)} dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{{{x^4}}}{{4\left( {81} \right)}}} \right)_3^3 \cr
& = \frac{1}{{324}}\left( {{x^4}} \right)_0^3 \cr
& = \frac{1}{{324}}\left( {{{\left( 3 \right)}^4} - {{\left( 0 \right)}^4}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{81}}{{324}} \cr
& = \frac{1}{4} \cr
& \int_a^b {f\left( x \right)} dx \ne 1 \cr
& \cr
& {\text{Condition 2 is not satisfied}}{\text{, so }} \cr
& f\left( x \right){\text{ }}is{\text{ }}not{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr} $$
