Composite trapezoidal rule's error term representation has $k_1h^2 + k_2 h^4 +... $ to apply Romberg integration I am trying to apply the Romberg method but when trying to evaluate the Error, I can't eliminate the appearance of the term $\frac{1}{12}f^{(3)}(\psi)h^3$, When I wish to get $E = K_1h^2 +K_2h^4 +K_3 h^6...$
What I tried to do was: 
For each interval $[x_i, x_{i+1}]$ the error element is of the form: $\frac{1}{12}h^{3}f^{(2)}(s_i)$ when $s_i\in[x_{i},x_{i+1}]$ and $h=[x_{i+1}-x_i]$ (the equi-distanced partition).
Using Taylor polynomial at $x_i$ $f^{(2)}(s_i) = f^{(2)}(x_i) + f^{(3)}(x_i)(s_i-x_i) + \frac{1}{2}f^{(4)}(\psi_i)(s_i-x_i)^2$.
So $|E_i| \le |\frac{1}{12}h^3[  f^{(2)}(x_i) + f^{(3)}(x_i)(h) + \frac{1}{2}f^{(4)}(\psi_i)(h)^2]|$
Thus the error sum is bounded by the absolute value of: $\frac{1}{12}h^2 \cdot h\cdot \sum [  f^{(2)}(x_i) + f^{(3)}(x_i)(h) + \frac{1}{2}f^{(4)}(\psi_i)(h)^2]$ Now applying intermediate value theorem, we get (for example for $\sum f^{(2)}(x_i)$: $\min\{f(x_i)\} \le h\cdot\sum_{i=0} ^{n-1} f^{(2)}(x_i) =\frac{1}{n}\cdot\sum_{i=0} ^{n-1} f^{(2)}(x_i) \le \max\{f(x_i)\} $ .
Thus the error can be represented:
$\frac{1}{12}h^2 \cdot [  f^{(2)}(\varphi) + f^{(3)}(\psi)(h) + \frac{1}{2}f^{(4)}(\chi)(h)^2]$ for $\varphi , \psi,\chi \in [x_0,x_n]$ derived from the intermediate value theorem (and the assumption $f^{(4)}$ is continuous.
So the error is bounded by $\frac{1}{12} \cdot [  f^{(2)}(\varphi)h^2 + f^{(3)}(\psi)h^3 + \frac{1}{2}f^{(4)}(\chi)h^4]$.
In Burden's book it tells that there is a detailed explanation in Ralston and Rabinowitz book, at pages 136-140. But I don't have access to this book for the next few days. 
 A: It's about even/odd symmetry. Translate that interval $[x_i,x_{i+1}]$ to $\left[-\frac{h}{2},\frac{h}{2}\right]$ and split $f$ into even and odd parts; $f(x)=g(x)+q(x)$ on that interval with $g$ even and $q$ odd.
Then $\int_{-h/2}^{h/2}f(x)\,dx=\int_{-h/2}^{h/2}g(x)\,dx$ since the integral of an odd function is zero, and the trapezoid rule estimate is $\frac{h}{2}\left(g(\frac h2)+q(\frac h2)+g(-\frac h2)+q(-\frac h2)\right) = hg(\frac h2)$.
Now, since we've eliminated the odd part, expand $g$ by its power series $$g(x)=g(0)+\frac{g''(0)}{2}x^2+\frac{g^{(4)}(0)}{4!}x^4+\frac{g^{(6)}(0)}{6!}x^6+\cdots$$
Because we're working with an even function, there aren't any odd derivatives involved. Integrate that power series, and
$$\int_{-h/2}^{h/2}g(x)\,dx = hg(0)+\frac{h^3}{24}g''(0)+\frac{h^5}{5!\cdot 2^4}g^{(4)}(0)+\frac{h^7}{7!\cdot 2^6}g^{(6)}(0)+\cdots$$
compared to a trapezoid rule estimate of
$$hg\left(\frac h2\right) = hg(0)+\frac{h^3}{8}g''(0)+\frac{h^5}{4!\cdot 2^4}g^{(4)}(0)+\frac{h^7}{6!\cdot 2^6}g^{(6)}(0)+\cdots$$
All of those even derivatives of $g$ at zero are equal to the even derivatives of $f$ there, of course.
On an arbitrary interval, we would get the same effect, with a lot more clutter, by expanding the Taylor series around the center of the interval. That's the key to it. It's just clearer to move that symmetry point to the origin for illustration purposes.
