# Probability distribution function proof

[SOLVED] Probability distribution function proof

## Homework Statement

Prove the function p(x) = $$\frac{3}{4b^{3}}$$(b$$^{2}$$-x$$^{2}$$) for -b $$\leq$$ x $$\leq$$ +b is a valid probability distribution function.

## Homework Equations

I'm not sure if it's as simple as this, but I've been using $$\int$$ p(x) dx (between b and -b) = 1 if the function is valid

## The Attempt at a Solution

$$\int$$ p(x) dx (between b and -b) = $$\frac{3x}{4b}$$ - $$\frac{x^{3}}{12b^{3}}$$ between b and -b = $$\frac{3}{4}$$ - $$\frac{1}{12}$$ + $$\frac{3}{4}$$ - $$\frac{1}{12}$$ = $$\frac{4}{3}$$

I treated b as a constant as I was integrating with respect to x, but clearly $$\frac{4}{3}$$ is not equal to 1.... please help!

HallsofIvy
You seem to have forgotten the "3" in "3/4" when you multiplied $$\frac{x^3}{3}$$ by $$\frac{3}{4b^3}$$. You should have
$$\frac{3}{4}- \frac{1}{4}+ \frac{3}{4}- \frac{1}{4}= 1$$