Thread: [SOLVED] Derive c.d.f. from p.d.f

1. [SOLVED] Derive c.d.f. from p.d.f

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.

2. Originally Posted by Sashikala
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.
$F(x) = 0$ for $x \leq 0$.

$F(x) = \int_{0}^x \frac{x}{9} \, dx$ for $0 < x < 3$.

$F(x) = \int_{0}^3 \frac{x}{9} \, dx + \int_3^x \frac{6-x}{9}$ for $3 \leq x \leq 6$.

$F(x) = 1$ for $x > 6$.

3. Thanks

Thank you Mr. Fantastic