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### Section 8.5 : Probability

3. Determine the value of $$c$$ for which the function below will be a probability density function.

$f\left( x \right) = \left\{ {\begin{array}{ll}{c\left( {8{x^3} - {x^4}} \right)}&{{\mbox{if }}0 \le x \le 8}\\0&{{\mbox{otherwise}}}\end{array}} \right.$ Show Solution

This problem is actually easier than it might look like at first glance.

First, in order for the function to be a probability density function we know that the function must be positive or zero for all $$x$$. We can see that for $$0 \le x \le 8$$ we have $$8{x^3} - {x^4} \ge 0$$. Therefore, we need to require that whatever $$c$$ is it must be a positive number.

To find $$c$$ we’ll use the fact that we must also have $$\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} = 1$$. So, let’s compute this integral (with the $$c$$ in the function) and see what we get.

$\int_{{ - \infty }}^{\infty }{{f\left( x \right)\,dx}} = \int_{0}^{8}{{c\left( {8{x^3} - {x^4}} \right)\,dx}} = \left. {c\left( {2{x^4} - \frac{1}{5}{x^5}} \right)} \right|_0^8 = \frac{{8192}}{5}c$

So, we can see that in order for this integral to have a value of 1 (as it must in order for the function to be a probability density function) we must have,

$\require{bbox} \bbox[2pt,border:1px solid black]{{c = \frac{5}{{8192}}}}$