I assume [tex]f(x)=0[/tex] otherwise. If [tex]f(x)[/tex] is indeed a proper PDF, then its integral over the support of [tex]X[/tex] is 1.
[tex]\displaystyle \int_{-\infty}^\infty f(x) \, dx = k \int_0^2 (2x + 3x^2) \, dx = 1[/tex]
Compute the integral.
[tex]\displaystyle \int_0^2 (2x + 3x^2) \, dx = (x^2 + x^3)\bigg|_{x=0}^{x=2} = (2^2 + 2^3) - (0^2 + 0^3) = 12[/tex]
Then
[tex]12k = 1 \implies \boxed{k=\dfrac1{12}}[/tex]
