Changes
/* Composite Simpson's Rule */ Added a continuation of example 2
:The formulas above do not work unless <math>N</math> is even.
=== Example 2 Continued ===
As our estimation was quite off, we should use the Composite Rule to get a better approximation.
<br />For this example, we will use 8 subintervals.
:<math>N = 8</math>
:<math>h = {8 \over 8} = 1</math>
:<math>x_0 = 4, x_1 = 5, x_2 = 6, x_3 = 7, x_4 = 8, x_5 = 9, x_6 = 10, x_7 = 11, x_8 = 12</math>
:<math>\int_4^{12} {3x+11 \over x^2 - x - 6}dx \approx \left({1\over3}\right ) \bigg[ f(4)+4f(5)+2f(6)+4f(7)+2f(8)+4f(9)+2f(10)+4f(11)+f(12) \bigg]</math>
:<math>= \left ( { 1 \over 3 } \right ) \bigg[ {23\over6} + 4\left ( {13\over7} \right) + 2\left ( {29\over24} \right) + 4\left ( {8\over9} \right) + 2\left ( {7\over10} \right) + 4\left ( {19\over33} \right) + 2\left ( {41\over84} \right) + 4 \left ( {11\over26} \right) + {47\over126} \bigg]</math>
:<math>=\left ( { 1 \over 3 } \right ) \left( {68579\over2860} \right) = {68579\over8580}</math>
:<math>\approx 7.99289...</math>
<br />This new answer is much closer than the original; it is off by about <math>.05</math>. If the number of subintervals is increased, it can be shown that the estimated value approaches the real value:
{| class="wikitable"
|-
!<math>N</math>
!Estimate
|-
|<math>2</math>
|<math>9.34179</math>
|-
|<math>4</math>
|<math>8.26138</math>
|-
|<math>8</math>
|<math>7.99289</math>
|-
|<math>10</math>
|<math>7.96783</math>
|-
|<math>100</math>
|<math>7.94161</math>
|-
|<math>1000</math>
|<math>7.94160</math>
|}
After we reach 106 subintervals, the approximation is correct for five digits.
==Other Versions of Simpson's Rule==
