Question

Let f(x) = k(9x − x2) if 0 ≤ x ≤ 9 and f(x) = 0 if x < 0 or x > 9. (a) for what value of k is f a probability density function?

Answer 1

We require

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=1$

Since

$f(x)=\begin{cases}k(9x-x^2)&\text{for }0\le x\le9\\0&\text{otherwise}\end{cases}$

the integral above reduces to

$\displaystyle\int_0^9 k(9x-x^2)\,\mathrm dx=k\left(\dfrac92x^2-\dfrac13x^3\right)\bigg|_{x=0}^{x=9}=\dfrac{243k}2=1$
$\implies k=\dfrac2{243}$