# Continuous random variable and probability

Suppose that X is a continuous Random variable with probability density function given by

$$f(x) = x^2 + \frac{2}{3}x + \frac{1}{3} \text{ for } 0 \leq x \leq c$$

What must be the value of c? And why?

• The integral of the probability density is required to be $1$. So take that integral for arbitrary $c$ and set it equal to $1$. – Ian Oct 19 '16 at 14:11

Hint: you want the probability of anything happening equal to one. So therefore $$\int_{0}^{c} x^2 + \frac{2}{3}x + \frac{1}{3} = 1.$$