# Trapezoid rule error analysis

How can I prove that the max error of the trapezoid rule for the integral $\int_{a}^{b}{f(x)\, \mathrm{d}x}$ is: $$\Delta=-\frac{1}{12n^2}f''(c)(b-a)^3 \text{for } c \in (a,b) \ ?$$

I know that to obtain that result first you have to prove that $$\exists \;c \in (a,b); \int_{a}^{b}f(x)\,dx = \frac{b - a}{2}\{f(a) + f(b)\} - \frac{1}{12}f''(c)(b-a)^3$$ But I'm stuck here, I tried using the mean value theorem but got nowhere. Anyone got any ideas?

If it helps: $\forall x_0 \in (a,b) \;\exists\;\xi_0 \in (a,b);\; f(x_0) - p(x_0) = f''(\xi_0)\frac{(x_0 - a)(x_0 - b)}{2}$, where p(x) is the linear function that interpolates f(x) in the points a and b ($p(x) = f(a) + \frac{f(b) - f(a)}{b-a}(x-a)$)

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Guess that, if true, when ∫f(x)dx relates to f″(c), it basically says that (just for purpose, assuming that f is C^3), for any such function there exists c such that : f'''(c) =-12(b-a)^(-3)*(f(b)-f(a)) – Newben Feb 23 '13 at 23:45
Btw, I think that finding that c such that ∫f(x)dx=−1/12*f″(c)(b−a)^3, fails for the identity function – Newben Feb 23 '13 at 23:54
It doesn't fail beacause when you're dealing with linear functions, $\int{f-p} = 0$, since p = f, $\forall x \in \mathbb R$ – Pedro Amorim Feb 24 '13 at 0:16

## 2 Answers

Let $p = (a + b)/2$ and $2h = b - a$ so that $a = p - h, b = p + h$. We further define the functions $g(t)$ and $r(t)$ by $$g(t) = \int_{p - t}^{p + t}f(x)\,dx - t\{f(p - t) + f(p + t)\},\,\, r(t) = g(t) - \left(\frac{t}{h}\right)^{3}g(h)$$ Then we can see that $$g'(t) = -t\{f'(p + t) - f'(p - t)\},\,\, r'(t) = g'(t) - \frac{3t^{2}}{h^{3}}g(h)$$ By Mean Value theorem we can see that $$g'(t) = -2t^{2}f''(t')$$ for some $t' \in (p - t, p + t)$. Thus we have $$r'(t) = -t^{2}\left(2f''(t') + \frac{3}{h^{3}}g(h)\right)$$ Clearly we can see that $r(0) = r(h) = 0$ so that (by Rolle's Theorem) there is some point $t_{0} \in (0, h)$ such that $r'(t_{0}) = 0$. This means that $$-t_{0}^{2}\left(2f''(t') + \frac{3}{h^{3}}g(h)\right) = 0$$ and therefore we have $$g(h) = -\frac{2h^{3}}{3}f''(t')$$ where $t' \in (p - t_{0}, p + t_{0}) \subset (p - h, p + h) = (a, b)$. We finally arrive at (by putting values of $h = (b - a)/2,\, p - h = a,\, p + h = b$ and definition of $g(t)$) $$\int_{a}^{b}f(x)\,dx = \frac{b - a}{2}\{f(a) + f(b)\} - \frac{(b - a)^{3}}{12}f''(t')$$ where $t' \in (a, b)$

Note: This is based on an exercise problem in G. H. Hardy's "A Course of Pure Mathematics". Compared to all the usual proofs given on Numerical Analysis books (primarily based on various interpolation formulas and Taylor series) I find this proof by Hardy to be the simplest one.

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Hint: Use the Lagrange Remainder of the Taylor Expansion: $$f(x)\sim f(x_1) +f'(x_1)(x-x_1)+f''(x^*)\frac{(x-x_1)^2}{2!}$$ For some $x^*$ in $[x_1,x]$, and integrate the error term over $[x_1,x_2]$.

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I still can't solve it. You've got another hint? – Pedro Amorim Feb 24 '13 at 19:58
Why the downvote? – nbubis Aug 21 '13 at 17:03