I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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 SolutionThis 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{c = \frac{5}{{8192}}}}\]