Finding $c$ when $X$ has pdf $f(x) = cx(1 - x)$ for $0<x<1$

Question

Let $X$ be a random variable having density function as $f(x) = cx(1 - x)$, for $0 < x < 1$.

Find the value of $c$.

After solving this question I got the answer as $6$, whereas the answer given in the book is $1/2.$ Can anyone please verify.

Answer 1

If the PDF is $f(x) = cx(1-x),$ for $0 < x < 1,$ then we must have $c = 6.$ It is easy to see that $\int_0^1 x(1-x)\,dx = 1/6,$ so that $c = 6.$

This is a member of the beta family of distributions. The general family density function is $$f(x) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},$$ for $0 < x < 1$ with parameters $\alpha, \beta > 0.$

Specifically in this question, we have parameters $\alpha = \beta = 2.$ For positive integers $k$, we have $\Gamma(k) = (k-1)!,$ so that $c = \frac{3!}{1!\cdot 1!} = 6.$

Answer 2

A probability density function satisfies the following properties:

1. $$f(x) \geq 0 \quad \forall x$$

2. $$ \int_{-\infty}^{\infty} f(x) dx = 1$$