## Calculus with Applications (10th Edition)

$$k = \frac{2}{9}$$
\eqalign{ & f\left( x \right) = kx;\,\,\,\,\,\,\,\,\,\,\,\,\left[ {0,3} \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}}: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[ {0,3} \right]{\text{ and }}f\left( x \right) = kx \cr & \int_0^3 {kx} dx = 1 \cr & {\text{Integrating}} \cr & \left( {\frac{{k{x^2}}}{2}} \right)_0^3 = 1 \cr & \frac{k}{2}\left( {{{\left( 3 \right)}^2} - {{\left( 0 \right)}^2}} \right) = 1 \cr & {\text{Simplify and solve for }}k \cr & \frac{{9k}}{2} = 1 \cr & k = \frac{2}{9} \cr & {\text{then }}f\left( x \right) = \frac{2}{9}x \cr & {\text{The function is positive for the interval }}\left[ {0,3} \right].{\text{ then the condition 1 it is true}} \cr & {\text{for }}k = \frac{2}{9} \cr}