The trapezoidal rule and thus also the composite variant are symmetric. Thus the error order is even, which means that it is $2$. Which implies that the extrapolated value is
$$
N_2(h)=\frac13(4N_1(\tfrac h2)-N_1(h))
$$
Next, if $n·h=b-a$, then also $2n·\frac h2=b-a$, so that the explicit formulas are
\begin{align}
N_1(h)&=h·\left(\tfrac12 f(a)+\sum_{k=1}^{n-1}f(a+kh)+\tfrac12 f(b)\right)
\\
N_1(\tfrac h2)&=\tfrac h2·\left(\tfrac12 f(a)+\sum_{k=1}^{2n-1}f(a+k\tfrac h2)+\tfrac12 f(b)\right)
\\
N_2(h)&=\tfrac h6·\left(f(a)+4f(a+\tfrac h2)+\sum_{k=1}^{n-1}\left[2f(a+kh)+4f(a+(k+\tfrac12)h)\right]+f(b)\right)
\end{align}
which is the composite Simpson rule (also symmetric) of error order $4$.
To experiment one could use this python code
from math import exp
def f(x):
return exp(-x)
def trapezint(f,a,b,N):
h=(b-a)/N
sum=(f(a)+f(b))/2;
for k in range(1,N):
sum = sum + f(a+k*h)
return sum*h
last = 0;
for k in range(10):
N = 1<<k;
intval = trapezint(f, 0.0, 1.0, N);
print "N=%3d, I(N) = %.12f, N*N*(I(N/2)-I(N))= %f, (4*I(N)-I(N/2))/3 = %.12f" \
% (N, intval, N*N*(last-intval), (4*intval-last)/3)
last=intval
with output
N= 1, I(N) = 0.683939720586, N*N*(I(N/2)-I(N))= -0.683940, (4*I(N)-I(N/2))/3 = 0.911919627448
N= 2, I(N) = 0.645235190149, N*N*(I(N/2)-I(N))= 0.154818, (4*I(N)-I(N/2))/3 = 0.632333680004
N= 4, I(N) = 0.635409429028, N*N*(I(N/2)-I(N))= 0.157212, (4*I(N)-I(N/2))/3 = 0.632134175321
N= 8, I(N) = 0.632943418210, N*N*(I(N/2)-I(N))= 0.157825, (4*I(N)-I(N/2))/3 = 0.632121414605
N= 16, I(N) = 0.632326313844, N*N*(I(N/2)-I(N))= 0.157979, (4*I(N)-I(N/2))/3 = 0.632120612389
N= 32, I(N) = 0.632172000094, N*N*(I(N/2)-I(N))= 0.158017, (4*I(N)-I(N/2))/3 = 0.632120562177
N= 64, I(N) = 0.632133419302, N*N*(I(N/2)-I(N))= 0.158027, (4*I(N)-I(N/2))/3 = 0.632120559038
N=128, I(N) = 0.632123773957, N*N*(I(N/2)-I(N))= 0.158029, (4*I(N)-I(N/2))/3 = 0.632120558842
N=256, I(N) = 0.632121362611, N*N*(I(N/2)-I(N))= 0.158030, (4*I(N)-I(N/2))/3 = 0.632120558829
N=512, I(N) = 0.632120759774, N*N*(I(N/2)-I(N))= 0.158030, (4*I(N)-I(N/2))/3 = 0.632120558829
Best Answer
You should be careful with this expression:
$$ {\rm err} = -\frac{b-a}{12}h^2f''(\mu) \tag{1} $$
The meaning is: there is a point $\mu \in (a,b)$ such that the error is given by this expression. To show this is true I calculate $S(h)$ for various values of $h$ and the absolute error $\epsilon$. I then find the value of $\mu$ guaranteed by Eq. (1), that is, the value of $\mu$ such that ${\rm err} = \epsilon$