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Line 94: − |- − !width="200"|Explanation − |- − |Note how much less "error" area that there is. This graph represents the first parabola we use for Example 2 Continued. − |- − |The parabola evaluates the integral from 4 to 6. − |- − |By reducing the area that each individual parabola approximates, our total approximation becomes more accurate. Line 114:
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→Example 2 Continued: reformatting and changing text order
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|[[Image:Simpsonsrule2cd7.png|250px|Click on the image to get a better view.]]
|[[Image:Simpsonsrule2cd7.png|250px|Click on the image to get a better view.]]
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:<math>=\left ( { 1 \over 3 } \right ) \left( {68579\over2860} \right) = {68579\over8580}</math>
:<math>=\left ( { 1 \over 3 } \right ) \left( {68579\over2860} \right) = {68579\over8580}</math>
:<math>\approx 7.99289...</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:
<br />This new answer is much closer than the original; it is off by about <math>.05</math>. Note how much less "error" area that there is on the picture. The graph represents the first parabola we use for Example 2 Continued, evaluating from 4 to 6. By reducing the area that each individual parabola approximates, our total approximation became more accurate.
If the number of subintervals is increased, it can be shown that the estimated value approaches the real value:
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