Originally Posted by

**mr fantastic** $\displaystyle \int^{x}_{1} \frac{3}{4}(1-u^2) \ du = \frac{3}{4}x - \frac{1}{4}x^3 + \frac{1}{2}$ is correct ...... But the pdf and hence integral is up in the air I think ....

Well I don't see any reason for us to worry about the algebra so I let Maxima do the work:

Code:

(%i5) integrate((3/4)*(1-z^2), z, -1, x);
(%o5) (3*(2/3-(x^3-3*x)/3))/4
(%i6) radcan(%);
(%o6) -(x^3-3*x-2)/4
(%i7)

and as this is $\displaystyle 1$ at $\displaystyle x=1$, we have that $\displaystyle (3/4)(1-x^2)$ for $\displaystyle x \in (-1,1)$ and zero otherwise is a density, but the $\displaystyle (3/4)x^2 (1-x^2)$ of the original post is not (it integrates up to $\displaystyle 1/5$)

RonL

(note Maxima unlike Mathematica is freeware - and does not just produce plain ASCII output as above either)