Answer
$$k = \frac{3}{{14}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = k{x^{1/2}};\,\,\,\,\,\,\,\,\,\,\,\,\left[ {1,4} \right] \cr
& {\text{The function f is a probability density function of a random variable X }} \cr
& {\text{in the interval }}\left[ {a,b} \right]{\text{ if}}: \cr
& 1{\text{ condition}}{\text{. }}f\left( x \right) \geqslant 0{\text{ for all }}x{\text{ in the interval }}\left[ {a,b} \right].{\text{ then}} \cr
& and \cr
& 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then replacing the interval }}\left[ {1,4} \right]{\text{ and }}f\left( x \right) = k{x^{1/2}} \cr
& \int_1^4 {k{x^{1/2}}} dx = 1 \cr
& {\text{integrating}} \cr
& \left( {\frac{{k{x^{3/2}}}}{{3/2}}} \right)_1^4 = 1 \cr
& \frac{{2k}}{3}\left( {{{\left( 4 \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right) = 1 \cr
& {\text{simplify and solve for }}k \cr
& \frac{{2k}}{3}\left( 7 \right) = 1 \cr
& \frac{{14k}}{3} = 1 \cr
& k = \frac{3}{{14}} \cr
& \cr
& {\text{then }}f\left( x \right) = \frac{{14}}{3}{x^{1/2}} \cr
& {\text{the function is positive for the interval }}\left[ {1,4} \right].{\text{ then the condition 1 it is true}} \cr
& {\text{for }}k = \frac{3}{{14}} \cr} $$
