Determine whether the function is a probability density function over the given interval.
-$$\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 3 } \mathrm { x } , 6 \leq \mathrm { x } \leq 8$$

Question

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For a function to be a probability density function (PDF) over a given interval, it must satisfy two conditions: the integral over the interval must equal 1, and the function must be non-negative over that interval. While $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 3 } \mathrm { x }$ is non-negative over the interval $6 \leq \mathrm { x } \leq 8$, integrating $\frac { 1 } { 3 } \mathrm { x }$ from 6 to 8 does not yield 1. The integral is $\frac { 1 } { 6 } \mathrm { x }^2 $ evaluated from 6 to 8, which equals $\frac { 1 } { 6 } (64 - 36) = \frac { 28 } { 6 } = \frac { 14 } { 3 }$, not 1.

Determine Whether the Function Is a Probability Density Function Over f(x)=13x,6≤x≤8\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 3 } \mathrm { x } , 6 \leq \mathrm { x } \leq 8f(x)=31​x,6≤x≤8

Determine Whether the Function Is a Probability Density Function Over $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 3 } \mathrm { x } , 6 \leq \mathrm { x } \leq 8$