Originally Posted by

**romsek** I assume that h is supposed to be getting pretty small. If that's the case we can expand f as a taylor series

$f(x+t)\approx f(x)+f'(x)t$ so

$\displaystyle{\int_{-h}^h}t f(x+t)~dt \approx

\displaystyle{\int_{-h}^h}t \left(f(x)+f'(x)t\right)~dt =

\displaystyle{\int_{-h}^h}t f(x)+f'(x)t^2~dt$

$\displaystyle{\int_{-h}^h}t f(x)+f'(x)t^2~dt= 0 + \dfrac 2 3 h^3 f'(x) = \dfrac 2 3 h^3 f'(x) $

$f'(x) \approx \alpha \dfrac 2 3 h^3 f'(x)$

so $\alpha = \dfrac 3 {2 h^3}$

probably not a bad idea to get another set of eyeballs to check this.