Find k value to get the function pdf
(b) $kx^2(4-x)^3$, for $0 < x < 4$. $0$ otherwise
Can this be done with gamma/beta distribution?
$$\int_{0}^{4} kx^2(4-x)^3dx$$
The definition is:
$$ k\int_0^1 x^{\alpha-1}(1-x)^{\beta-1} = k\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} = B(\alpha,\beta) $$
Doesn't work for $0$ to $4$
If I let $x = 4t$, then
$$\int_{0}^{4} 64t^2(4(1 - t))^3dt = \int_{0}^{4} 64t^24^3(1 - t)^3dt = 4096k \int_{0}^{4}t^2(1-t)^3dt$$
To be in beta form where $\alpha = 3$ and $\beta = 4$
Yes, using a change of variables $x\gets 4t$ works so
$$\int_0^4 k x^2(4-x)^3~\mathsf d x= \int_0^1 k(4t)^2(4-4t)^3\cdot 4\mathsf d t\\= 4^3k\int_0^1 t^2(1-t)^3~\mathsf d t$$