Given the Probability Density Function: $f(x)=kx(2-x), 0\leq x\leq 1$ Prove that $k=\frac 3 2$
Looks like it should be a Beta Distribution, but all examples of a beta distribution use the format: $f(x)=kx^{(a-1)}(1-x)^{(b-1)}, 0\leq x\leq 1$ where $k=\frac {(a-b-1)!}{(a-1)!(b-1)!}$
$f(x)=\begin{cases}kx(2-x) &: 0\leq x\leq 1\\ 0 &: \text{elsewhere}\end{cases}$
Since $f(x)$ is a probability density function then: $\int\limits_{-\infty}^\infty f(x)\;\mathrm{d}x = 1$
So for the given function you will need to find $k$ such that: $\int\limits_0^1 kx(2-x)\;\mathrm{d}x = 1$
$$\therefore f(x) = \frac 32 x(2-x), 0\leq x\leq 1$$
Mean: $\displaystyle \bar X = \operatorname{E}[X] = \int\limits_{-\infty}^\infty x f(x)\;\mathrm{d}x = \frac 32 \int_0^1 2x^2-x^3\;\mathrm{d}x$
Variance: $\displaystyle \operatorname{Var}[X] = \operatorname{E}[X^2]-\operatorname{E}[X]^2 = \int\limits_{-\infty}^\infty x^2 f(x)\;\mathrm{d}x - (\bar X)^2 = \frac 32 \int_0^1 2x^3-x^4\;\mathrm{d}x-(\bar X)^2$