# Chapter 11 - Probability and Calculus - 11.1 Continuous Probability Models - 11.1 Exercises - Page 575: 2

$$f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function$$

#### Work Step by Step

\eqalign{ & f\left( x \right) = \frac{1}{3}x - \frac{1}{6};\,\,\,\,\,\,\,\left[ {3,4} \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}{3}x - \frac{1}{6} \geqslant 0 \cr & 2x - 1 \geqslant 0 \cr & x \geqslant \frac{1}{2} \cr & \left[ {3,4} \right]{\text{ is on the interval }}\left[ {\frac{1}{2},\infty } \right).{\text{ then }}f\left( x \right) \geqslant 0{\text{ for the interval }}\left[ {3,4} \right] \cr & \cr & 2{\text{ condition}}:\int_a^b {f\left( x \right)} dx = 1.{\text{ then}} \cr & \int_3^4 {\left( {\frac{1}{3}x - \frac{1}{6}} \right)} dx \cr & {\text{integrating}} \cr & = \left( {\frac{1}{{3\left( 2 \right)}}{x^2} - \frac{1}{6}x} \right)_3^4 = \left( {\frac{1}{6}{x^2} - \frac{1}{6}x} \right)_3^4 \cr & = \frac{1}{6}\left( {{x^2} - x} \right)_3^4 \cr & = \frac{1}{6}\left( {{{\left( 4 \right)}^2} - \left( 4 \right)} \right) - \frac{1}{6}\left( {{{\left( 3 \right)}^2} - \left( 3 \right)} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{6}\left( {12} \right) - \frac{1}{6}\left( 6 \right) \cr & = 2 - 1 \cr & = 1 \cr & \cr & {\text{The conditions are both verified}}{\text{, so }} \cr & f\left( x \right){\text{ }}is{\text{ }}a{\text{ }}probability{\text{ }}density{\text{ }}function \cr}

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