Question

Determine the constant \(c\) so that \(f(x)=c x(3-x)^{4}, 0

Short answer

Therefore, the constant \(c\) that makes \(f(x) = c x(3-x)^{4}\) a valid pdf is \(c = \frac{2}{9}\).

Step-by-step solution

Setting Up the Integral

A pdf must satisfy the property that its integral over its full domain is 1. So we need to solve the equation \(\int_{0}^{3} f(x) dx = 1\). In this case, the integral to solve is \(\int_{0}^{3} c x(3-x)^{4} dx = 1\).

Substitution

Use substitution to simplify the integral. Let \(u = 3 - x\), then \(du = -dx\) and \(x = 3 - u\). The limits of the integral change to 3 for the lower limit and 0 for the upper limit. The integral to solve becomes \(-c \int_{3}^{0} u^{4}(3 - u)du\).

Switch the Limits of Integration and Simplify

Switching the limits of integration will get rid of the negative sign in front of the integral. This results in \(c \int_{0}^{3} u^{4}(3-u) du \). Expand the integrand \(u^{4}(3-u) = 3u^{4} - u^{5}\). So the integral to solve becomes \(c \int_{0}^{3} (3u^{4} - u^{5}) du\).

Evaluate the Integral

The integral can be solved using the power rule for integration: \(\int_{a}^{b} x^n dx = \frac{1}{n+1}x^{n+1} |_{a}^{b}\). Thus, the integral \(c \int_{0}^{3} (3u^{4} - u^{5}) du = c(\frac{3}{5}u^{5} - \frac{1}{6}u^{6}) |_{0}^{3} = c(45 - 81/2) = c(9/2)\). We set this equal to 1 and solve for \(c\).

Solve For \(c\)

\(\frac{9}{2}c = 1 \Rightarrow c = \frac{2}{9}\).