Answer
$${\text{ }}k = \frac{1}{{21}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = k{x^2};\,\,\,\,\,\,\,\,\,\,\,\,\left[ {1,4} \right] \cr
& {\text{The function f is a probability density function of a random variable X in the interval }} \cr
& \left[ {a,b} \right]{\text{ if}} \cr
& 1{\text{ condition}}:f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ }} \cr
& and \cr
& \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ Then replacing the interval }}\left[ {1,4} \right]{\text{ and }}f\left( x \right) = k{x^2} \cr
& \int_1^4 {k{x^2}} dx = 1 \cr
& {\text{integrating}} \cr
& \left( {\frac{{k{x^3}}}{3}} \right)_1^4 = 1 \cr
& \frac{k}{3}\left( {{{\left( 4 \right)}^3} - {{\left( 1 \right)}^3}} \right) = 1 \cr
& {\text{simplify and solve for }}k \cr
& \frac{k}{3}\left( {64 - 1} \right) = 1 \cr
& \frac{{63k}}{3} = 1 \cr
& 21k = 1 \cr
& k = \frac{1}{{21}} \cr
& \cr
& {\text{then }}f\left( x \right) = \frac{1}{{21}}{x^2} \cr
& {\text{The function is positive for the interval }}\left[ {1,4} \right].{\text{ So condition 1 is true}} \cr
& {\text{for }}k = \frac{1}{{21}} \cr} $$
