I have a p.d.f given by

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

$\displaystyle 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:

$\displaystyle \frac{3}{8}x-\frac{1}{32x^3}$

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

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