f(x) = {1/9x, 0<x<3
1/9(6-x), 3≤ x≤6
0 otherwise
Given that f(x) is a p.d.f., How to find the c.d.f for it? Pls help me on this.
$\displaystyle F(x) = 0$ for $\displaystyle x \leq 0$.
$\displaystyle F(x) = \int_{0}^x \frac{x}{9} \, dx$ for $\displaystyle 0 < x < 3$.
$\displaystyle F(x) = \int_{0}^3 \frac{x}{9} \, dx + \int_3^x \frac{6-x}{9}$ for $\displaystyle 3 \leq x \leq 6$.
$\displaystyle F(x) = 1$ for $\displaystyle x > 6$.