Originally Posted by

**awkward** Hi geton,

One of the requirements for a pdf is that $\displaystyle f(x) \geq 0$ for all values of x. Now

$\displaystyle x + x^2 < 0 \text{ for } -1 < x < 0$

and

$\displaystyle x + x^2 > 0 \text{ for } 0 < x < 1$,

so if $\displaystyle k > 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle -1 < x < 0$, and if $\displaystyle k < 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle 0 < x < 1$, so no value of $\displaystyle k \neq 0$ will make f suitable for a pdf.

The only remaining possibility is $\displaystyle k = 0$ but then $\displaystyle \int f(x) dx = 0$, so that won't work either.