Answer
$C=\dfrac{3}{32}$and $P(3 \leq X \leq 4)=\dfrac{5}{32}$
Work Step by Step
We are given that $p(x)=Cx(4-x)$
$\int_0^{4}Cx(4-x) dx =\dfrac{4Cx^2}{2}-\dfrac{Cx^3}{3}]_0^4\\=\dfrac{32 C}{3}$
The condition will be satisfied when
$\int_0^{4}Cx(4-x) dx=\dfrac{32C}{3}=1$ or, $C=\dfrac{3}{32}$
We will find
$P(3 \leq X \leq 4)=\int_3^4 \dfrac{3}{32} x(4-x) dx \\=\dfrac{3}{32}[2x^2-\dfrac{x^3}{3}]_3^4\\=\dfrac{5}{32}$
