Proving error bound on Simpson's Rule, Numerical Integration

The approximation from "Simpson's Rule" for $\int_a^b f(x)\, dx$ is, \begin{equation} S_{[a,b]}f = \bigg[\frac{2}{3}f\Big(\frac{a+b}{2}\Big) + \frac{1}{3}\Big(\frac{f(a) + f(b)}{2}\Big)\bigg](b-a). \end{equation} If $f$ has continuous derivatives up to order three, prove that \begin{equation} \bigg|\int_a^b f(x)\, dx - S_{[a,b]}f\bigg| \leq C(b-a)^4 \max_{[a,b]} |f^{(3)}(x)|, \end{equation} where $C$ does not depend on $f.$

I've seen many other error estimations for Simpson's Rule, but this one has given me some trouble. I've attempted using some type of majorizing interpolation, but my result was complicated and I think it is over thinking it. Any hints or solutions would be appreciated.

Let $F$ be an anti-derivative of $f$, $F'=f$. W.l.o.g. $a+b=0$, set $x=(b-a)/2$ then we are interested in the error expression $$ g(x)=F(x)-F(-x)-\frac{x}3(f(x)+4f(0)+f(-x)). $$ This has derivatives \begin{alignat}{2} g'(x)&=\frac23(f(x)-2f(0)+f(-x))&&-\frac x3 (f'(x)-f'(-x)) \\ g''(x)&=\frac13(f'(x)-f'(-x))&&-\frac x3(f''(x)+f''(-x)) \\ g'''(x)&=&&-\frac x3(f'''(x)-f'''(-x)). \end{alignat} Consequently, by the extended mean value theorem $$ \frac{g(x)}{x^4}=…=\frac{g'''(x_3)}{24x_3}=-\frac{f'''(x_3)-f'''(-x_3)}{72} $$ which gives an error bound of $$ |g(x)|\le\max_{s\in[a,b]}|f'''(s)|·\frac{x^4}{36}=\max_{s\in[a,b]}|f'''(s)|·\frac{(b-a)^4}{576}. $$

Of course, the more common error term is obtained via $$ \frac{g(x)}{x^5}=\frac{g'''(x_3)}{60x_3^2}=-\frac1{90}·\frac{f'''(x_3)-f'''(-x_3)}{2x_3}=-\frac1{90}·f^{(4)}(x_4). $$

$x_3$ above is the middle point in the third iterative application of the mean value theorem. As $g(0)=g'(0)=g''(0)=g'''(0)=0$ you get for $m\ge 3$ $$\frac{g(x)}{x^m}=\frac{g'(x_1)}{mx_1^{m-1}}=\frac{g''(x_2)}{m(m-1)x_2^{m-2}}=\frac {g'''(x_3)}{m(m-1)(m-2)x_3^{m-3}}$$ with $0<x_3<x_2<x_1<x$.