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=\frac1{12}}`