Determine the value of k that makes the function below a pdf

Determine the value of k that makes the function below a pdf for a continuous random variable.

Solution

f(x) must be normalized, so

Integral [f(x) dx] = 1

Integral [k(x^2+2) dx]|(1,2) = 1

k Integral [(x^2+2) dx]|(1,2) = 1

k [x^3/3 + 2x]|(1,2) = 1

k[(2^3/3 +2*2) - (1^3/3 + 2*1)] = 1

k(4.333333333) = 1

k = 0.230769231 or 3/13 [ANSWER]