Integral to derive Simpson's Rule error expression I have this question from an old Numerical Analysis exam:
Let $h>0$ and $f$ be a sufficiently differentiable function. Prove that
\begin{align*}
I:=\frac{1}{6}\int_0^h[f'''(-t)-f'''(t)]t(t-h)^2dt=-\frac{h^5}{90}f^{(4)}(\xi),
\end{align*}
for some $\xi\in \left]-h,h\right[$.
My attempt: Since the function $t(h-t)^2$ doesn't change sign in $[0,h]$, we can apply the Mean Value Theorem to guarantee the existence of some number $\eta\in \left]0,h\right[$ such that
$$I=\frac{f'''(-\eta)-f'''(\eta)}{6}\int_0^ht(t-h)^2dt.$$
We can easily calculate the integral above and obtain that
$$I=\frac{h^4}{72}[f'''(-\eta)-f'''(\eta)].$$
Now, applying the standard version of the Mean Value Theorem we get
$$I=-\frac{h^4}{72}f^{(4)}(\xi)(\eta - (-\eta))=-\frac{h^4}{36}\eta f^{(4)}(\xi).$$
However, I don't know how to continue.
Any help would be appreciated!
PD: I think the integral in this exercise appears when trying to deduce the error in Simpson's Rule.
 A: Let $\,h>0\,$ and let $\,f:[-h,h]\to\Bbb R\,$ a function such that $\,f’’’(x)\,$ is continuous on $\,[-h,h]\,$ and differentiable on $\,]-h,h[\,,\,$ that is, $\,\exists\,f^{(4)}(x)$ for any $\,x\!\in\,]\!-\!h,h[\,.$
Let $\,\phi:[0,h]\to\Bbb R\,$ be the function defined as :
$\phi(t)=\begin{cases}\dfrac{f'''(-t)-f'''(t)}t\quad&\text{for any }\,t\in\,]0,h]\\-2f^{(4)}(0)&\text{for }\,t=0\end{cases}$
It results that the function $\,\phi\,$ is continuous on $\,[0,h]\,$ and
$f’’’(-t)-f’’’(t)=t\,\phi(t)\quad$ for all $\,t\in[0,h]\,.$
Moreover ,
$\displaystyle I:=\frac{1}{6}\int_0^h\left[f'''(-t)-f'''(t)\right]t(t-h)^2dt=$
$=\displaystyle\frac{1}{6}\int_0^h\phi(t)\,t^2(t-h)^2dt\,.$
Since the function $\;\psi(t)=t^2(t−h)^2\,$ is nonnegative on $\,[0,h]\,,\,$ we can apply the integral version of the Mean Value Theorem, hence, $\;\exists c\in[0,h]\,$ such that
$\displaystyle I=\frac{1}{6}\phi(c)\!\int_0^h t^2(t-h)^2dt=\dfrac{h^5}{180}\phi(c)\,.$
There two possible cases : $\;c=0\;$ or $\;c\in\,]0,h]\;.$
If $\;c=0\,,\,$ then $\;I=\dfrac{h^5}{180}\phi(0)=-\dfrac{h^5}{90}f^{(4)}(0)\;.$
If $\;c\in\,]0,h]\,,\,$ then $\;\phi(c)=\dfrac{f’’’(-c)-f’’’(c)}c\;,$ moreover,
by applying the Mean Value Theorem to the function $f’’’$ on the interval $\,[-c,c]\subseteq[-h,h]\,,\,$ we get that there exists $\;\xi\in\,]\!-\!c,c[\,\subseteq\,]\!-\!h,h[\,$ such that
$f’’’(c)-f’’’(-c)=2c\,f^{(4)}(\xi)\;.$
Consequently,
$I=\dfrac{h^5}{180}\phi(c)=\dfrac{h^5}{180}\dfrac{f’’’(-c)-f’’’(c)}c=-\dfrac{h^5}{90}f^{(4)}(\xi)\;.$
In any case, we have proved that there exists $\;\xi\in\,]\!-\!h,h[\;$ such that
$I=-\dfrac{h^5}{90}f^{(4)}(\xi)\;.$
A: I just had to apply first the Mean Value Theorem under the integral and then apply the integral version of the Mean Value Theorem:
We know that for each $t\in \left[0,h\right]$ there exists some $\xi_t\in \left]-t,t\right[$ such that
$$f'''(t)-f'''(-t)=2tf^{(4)}(\xi_t).$$
Therefore,
$$I=-\frac{1}{3}\int_0^hf^{(4)}(\xi_t)\,t^2(t-h)^2dt.$$
Since $t^2(t-h)^2$ doesn't change sign in $[0,h]$ there exists some $\xi\in \left]-h,h\right[$ such that
$$I=-\frac{1}{3}f^{(4)}(\xi)\int_0^ht^2(t-h)^2dt.$$
Calculate the last integral to obtain the desired result.
