Suppose that $X$ is a random variable that has a p.d.f. given by the formulas $$p_X(x)=\begin{cases}a⋅(1−x^3),&0≤x≤1\\0,&\textrm{otherwise}\end{cases}$$ where $a>0$ is a constant. Find a formula for the c.d.f. $F_X(x)$ when $0≤x≤1$.
Using the theorems I know from a PDF, I know a is $4/3$ (this way the integral from $0$ to $1$ equals $1$).
Now from my understanding, in order to get the CDF, we need to integrate the PDF. I thought I should integrate from $1$ to x which would give us the CDF but this was wrong. Any hints on achieving the CDF of this function from $0$ to $1$.
Yes, this is correct. $a$ should satisfy: $$ \int_0^1a(1-x^3)dx=1. $$ So, $a=4/3$. Then the CDF of $X$ is $$ \int_0^y\frac{4}{3}(1-x^3)dx=\frac{1}{3}y(4-y^3) $$ for $0\le y\le 1$.