Combining error terms in Simpson's rule My numerical analysis textbook (Burden and Faires) derives Simpson's rule as
$$\begin{align}
\int_{x_0}^{x_2}f(x)\,dx&=2hf(x_1)+\frac{h^3}{3}f''(x_1)+\frac{h^5}{60}f^{(4)}(\xi_1)
\\&=\frac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]-\frac{h^5}{12}\left[\frac{1}{3}f^{(4)}(\xi_2)-\frac{1}{5}f^{(4)}(\xi_1)\right]
\end{align}$$
for some $\xi_1,\xi_2\in(x_0,x_2)$. Here $x_0=x_1-h$ and $x_2=x_1+h$.
Note that $\xi_1$ comes from the error term when integrating the Taylor series of $f(x)$ while $\xi_2$ comes from the error term in rewriting $f''(x_1)$.

How can we show that we can replace $\xi_1$ and $\xi_2$ in the above formula with some $\xi\in(x_0,x_2)$?

My textbook leaves this part as an exercise, but I am unconvinced the exercise demonstrates what we want. The following is the exercise verbatim:
"Derive Simpson's rule with error term by using
$$\int_{x_0}^{x_2}f(x)\,dx=a_0f(x_0)+a_1f(x_1)+a_2f(x_2)+kf^{(4)}(\xi)$$
Find $a_0$, $a_1$, and $a_2$ from the fact that Simpson's rule is exact for $x^n$ when $n=1$, $2$, and $3$. Then find $k$ by applying the integration formula with $f(x)=x^4$."
Performing this exercise, we certainly get the correct coefficients, but I don't see how this tells us we can combine $\xi_1$ and $\xi_2$ - it seems that we just assumed it.
 A: Let $F$ be an anti-derivative of $f$, $F'=f$. W.l.o.g. $x_1=0$, set $x=h$ 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
$$
\frac{g(x)}{x^m}
=\frac{g'(x_1)}{mx_1^{m-1}}
=\frac{g''(x_2)}{m(m-1)x_2^{m-2}}
=\frac {g'''(x_3)}{m(m-1)(m-2)x_3^{m-3}}
$$ 
with $0<x_3<x_2<x_1<x$.
Using $m=5$ this gives
$$
\frac{g(x)}{x^5}=\frac{g'''(x_3)}{60x_3^2}=-\frac1{90}·\frac{f'''(x_3)-f'''(-x_3)}{2x_3}=-\frac1{90}·f^{(4)}(x_4).
$$
with $|x_4|<x_3<x$.
This results in the error formula
$$
g(x)=-\frac1{90}·f^{(4)}(x_4)·x^5.
$$
or after translating the initial simplifications back,
\begin{align}
\int_{x_0}^{x_2}f(x)\,dx
&=\frac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]-\frac{h^5}{90}f^{(4)}(\xi)
\end{align}
with $\xi\in(x_0,x_2)$
