Observe the following probability density function for a continuous random variable X
$$f (x) = \begin{cases}
k\sqrt x (1-x) &\text{ for }x\in(0,1)\\
0 &\text{ otherwise}
\end{cases}
$$
Find the value of $k$ which makes $f$ a density function.
My thoughts, is it the integral from $0$ to $1$ of $f(x)$?
Best Answer
We must choose $k$ so that $\int_0^1 k(1-x)\sqrt{x}\,dx=1$. So we need to evaluate the integral.
For the evaluation, let $x=u^2$. Then $dx=2u\,du$, and we find $k$ such that $$\int_0^1 2k(u^2-u^4)\,du=1.$$