Integral to derive Simpson's Rule error expression

Question

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.

Answer 1

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

$I=-\dfrac{h^5}{90}f^{(4)}(\xi)\;.$

Answer 2

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.