The proportion of assignments completed within a given day is described by the probability density function f (x) = 12(x - x3) for 0 ≤ x ≤ 1. Find the expected proportion of completed assignments.

Check

$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{1}12(x-x^3)dx$

$=[6x^2-3x^4]\begin{matrix} 1\\ 0 \end{matrix}=6-3=3\not=1$

Let $f (x) = 4(x - x^3).$ for $0 ≤ x ≤ 1.$

Check

$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{1}4(x-x^3)dx$

$=[2x^2-x^4]\begin{matrix} 1\\ 0 \end{matrix}=2-1=1$$E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{0}^{1}x(4(x-x^3))dx$

$=[\dfrac{4}{3}x^3-\dfrac{4}{5}x^5]\begin{matrix} 1\\ 0 \end{matrix}=\dfrac{8}{15}$