Examples:

1. Estimate the integral of the function ${\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}\sin \left(x\right)dx}$ using Simpson’s Method with N=4 and thus h=$\pi$ /4. This integral is of course equal to 1.

   f(0)	0
   f($\pi$ /4)	0.3535533
   f( $\pi$ /2)	0.5
   f(3 $\pi$ /4)	0.3535533
   f( $\pi$ )	0

   $\begin{array}{l}{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}\sin \left(x\right)dx}\approx \frac{\pi /4}{3}\left[0+4\left(0.3535533\right)+2\left(0.5\right)+4\left(0.3535533\right)+0\right]\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}\sin \left(x\right)dx}\approx 1.002279\end{array}$

   Thus the error is $\frac{1.002279-1}{1}=.002279=.2279\%$

2. Use the same intergral but now use N=8 subintervals and h=$\pi$ /8.

   f(0)	0
   f( $\pi$ /8)	0.1913417
   f( $\pi$ /4)	0.3535533
   f(3 $\pi$ /8)	0.4619397
   f( $\pi$ /2)	0.5
   f(5 $\pi$ /8)	0.46193975
   f(3 $\pi$ /4)	0.3535533
   f(7 $\pi$ /8)	0.19134174
   f( $\pi$ )	0

   $\begin{array}{l}{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}\sin \left(x\right)dx}\approx \frac{\pi /8}{3}\left[\begin{array}{l}0+4\left(0.1913417\right)+2\left(0.3535533\right)+4\left(0.4619397\right)+2\left(0.5\right)\\ +4\left(0.4619397\right)+2\left(0.3535533\right)+4\left(0.1913417\right)+0\end{array}\right]\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{1}{2}\sin \left(x\right)dx}\approx 1.000134\end{array}$

3. We expect that when we increase the number of intervals by a factor of 2 (8/4=2), that our error will be reduced by a factor of 2⁴=16. That means that our answer will be 16 times more accurate, or our error will be 1/16 of the error in the first approximation.

   Let's see if that's the case for this example.

   Recall, that the error when N=4 was $\frac{1.002279-1}{1}=.002279=.2279\%$

   We can now compute the error when N=8. It is ~0.01% $\frac{1.000134-1}{1}=.000134=.0134\%$

   which is approximately $\frac{0.002279}{0.000134}=17.0074$ times better than the approximation with N=4.

   This result is comparable to the $2^4=16$ improvement we would expect when we increase the number of intervals from 4 to 8.