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I saw some statements of Simpson's Rules like this:

Let the f be a function with a continuous fourth derivative on [a, b] with $$ |f^{(4)}(x)| \le K $$ Then $$ |\int_{a}^{b} f(x) dx - S_{2N}| \le \frac{K(b-a)^{5}}{180(2N)^{4}}$$

I am confused about why "fourth derivative continuity" should be applied here. The reason I saw it applies "fourth derivative continuity" is that in the proof, it uses $$ \frac{f^{(4)}(w_{0})+f^{(4)}(w_{1})+...+f^{(4)}(w_{N-1})}{N}=f^{(4)}(w) $$ for some $$ w \in (a, b)$$ because $$ f^{(4)}(x) $$ is continuous And eventually sum up each subinterval.

Hence $$ \int_{a}^{b} f(x)dx = S_{2N} - \frac{h^{5}}{90}Nf^{(4)}(w)$$

However, I can use the fact that fourth derivative is bounded instead of being continuous for each subinterval: $$ |f^{(4)}(w_{i})| \le K_{i} $$

Then for each subinterval $$[x_{i}, x_{i+1}]$$ It is suffice to say $$ \int_{x_{2i}}^{x_{2i+2}}f(x)dx = \frac{h}{3}(f_{2i}+4f_{2i+1}+f_{2i+2}) - \frac{h^{5}}{90}f^{(4)}(w_{i})$$ Then
$$ |\int_{x_{2i}}^{x_{2i+2}}f(x)dx - \frac{h}{3}(f_{2i}+4f_{2i+1}+f_{2i+2})| = |\frac{h^{5}}{90}f^{(4)}(w_{i})|$$

$$ |\int_{x_{2i}}^{x_{2i+2}}f(x)dx - \frac{h}{3}(f_{2i}+4f_{2i+1}+f_{2i+2})| \le |\frac{h^{5}}{90}K_{i}| = \frac{h^{5}}{90}K_{i}$$ Eventually sum up each subinterval and get the same result But this methods does not need to require fourth derivative continuity. This method works as long as fourth derivative exists and is bounded.

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    Boundedness isn't enough for that equality (with $N$ in the denominator on the LHS) to hold. The equality basically says there's some point $x=w$ at which $f^{(4)}(x)$ is equal to its average of $N$ other values in the interval in question. Without continuity, you aren't guaranteed that such a $w$ will exist. – 2017-02-28
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    sorry, I didn't claim my question clear enough. I have updated i just now. – 2017-02-28

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