We require
[tex]\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=1[/tex]
Since
[tex]f(x)=\begin{cases}k(9x-x^2)&\text{for }0\le x\le9\\0&\text{otherwise}\end{cases}[/tex]
the integral above reduces to
[tex]\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[/tex]
[tex]\implies k=\dfrac2{243}[/tex]
