Let g be an odd real function defined on a neighbourhood of 0, and 5 times differentiable on this neighbourhood. Show that
g(x)=3x(g′(x)+2g′(0))−180x5g(5)(ξ)(ξ=θx,0<θ<1) (same method as in exerc. 9).
Deduce that if f is a real function defined on [a,b] and 5 times differentiable on this interval, then
f(b)−f(a)=6b−a[f′(a)+f′(b)+4f′(2a+b)]−2880(b−a)5f(5)(ξ) with
a<ξ<b ("Simpson's formula").