If 16 counters are 4/9, then what is the ONE?
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For f(x) = 12/(1+x²), and subinterval width 4, you are to evaluate f(1), f(5), and f(9) and combine them according to the rule
... Integral ≈ (4/3)(f(1) + 4·f(5) + f(9)) = (4/3)(6.0000 + 4·0.4615 + 0.1463) ≈ 10.66
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Simpson's rule has you combine values of f(x) with coefficients 1, 4, 2, 4, ..., 2, 4, 1, where those values are evenly spaced at the edges of an even number of subintervals. Since we have only 3 values to combine, there are no terms that have a coefficient of 2. The entire sum is multiplied by 1/3 the subinterval width.
