For $f(x)$ to be a probability density function, the integral of $f(x)$ over its domain should be equal to 1. Since $f(x)$ is non-zero only over the interval $[0,3]$, we have:
$$\int_{0}^{3} f(x) dx = 1$$
Substituting $f(x)$, we get:
$$\int_{0}^{3} k(3x-x^{2}) dx
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