Thread: [SOLVED] p.d.f; how do I find the c.d.f?

1. [SOLVED] p.d.f; how do I find the c.d.f?

I have a p.d.f given by

$f(x)=\frac{3}{32}(4-x^2)$ for $-2\leq x\leq 2$
$f(x)=0$ otherwise.

The question asks me to find the c.d.f. In the textbook, it says I do this by integrating the function, which gives me:

$\frac{3}{8}x-\frac{1}{32x^3}$
The answer is $\frac{3}{8}x-\frac{1}{32x^3}+\frac{1}{2}$

but I can't spot my mistake - so where has the half come from?

2. Originally Posted by Quacky
I have a p.d.f given by

$f(x)=\frac{3}{32}(4-x^2)$ for $-2\leq x\leq 2$
$f(x)=0$ otherwise.

The question asks me to find the c.d.f. In the textbook, it says I do this by integrating the function, which gives me:

$\frac{3}{8}x-\frac{1}{32x^3}$
The answer is $\frac{3}{8}x-\frac{1}{32x^3}+\frac{1}{2}$

but I can't spot my mistake - so where has the half come from?
For x < -2, F(x) = 0.

For $-2 \leq x \leq 2$, $F(x) = \left[ \frac{3}{8} u - \frac{1}{32} u^3 \right]_{u=-2}^{u=x} = ....$

For x > 2, F(x) = 1.

3. Thankyou for the explanation