Question

) Consider the following probability density function of the random variable X. f(x) = k(2x + 3x2 ), 0 ≤ x ≤ 2. (i) Determine the value of the constant k.

Answer 1

I assume $f(x)=0$ otherwise. If $f(x)$ is indeed a proper PDF, then its integral over the support of $X$ is 1.

$\displaystyle \int_{-\infty}^\infty f(x) \, dx = k \int_0^2 (2x + 3x^2) \, dx = 1$

Compute the integral.

$\displaystyle \int_0^2 (2x + 3x^2) \, dx = (x^2 + x^3)\bigg|_{x=0}^{x=2} = (2^2 + 2^3) - (0^2 + 0^3) = 12$

Then

$12k = 1 \implies \boxed{k=\dfrac1{12}}$