Answer
$$f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{27}}\left( {2x + 4} \right);\,\,\,\,\,\,\,\left[ {1,4} \right] \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ then}} \cr
& \frac{1}{{27}}\left( {2x + 4} \right) \geqslant 0 \cr
& 2x + 4 \geqslant 0 \cr
& x \geqslant - 2 \cr
& \left[ {1,4} \right]{\text{ is on the interval }}\left[ { - 2,\infty } \right).{\text{ then }}f\left( x \right) \geqslant 0{\text{ for the interval }}\left[ {1,4} \right] \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr
& \int_1^4 {\frac{1}{{27}}\left( {2x + 4} \right)} dx\mathop = \limits^? 1 \cr
& {\text{integrating}} \cr
& = \frac{1}{{27}}\left( {{x^2} + 4x} \right)_1^4 \cr
& = \frac{1}{{27}}\left( {{{\left( 4 \right)}^2} + 4\left( 4 \right)} \right) - \frac{1}{{27}}\left( {{{\left( 1 \right)}^2} + 4\left( 1 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{32}}{{27}} - \frac{5}{{27}} \cr
& = 1 \cr
& \cr
& {\text{The condition 1 and 2 are verified}}{\text{, then }} \cr
& f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr} $$
