Discussion
Let \(y = f(x)\) take the values \(y_{0}, y_{1}, \ldots, y_{n}\) for \(a = x_{0}\), \(x_{1} = x_{0} + h\), ..., \(x_{n} = x_{0} + nh = b\). Then the value of :
\[ \int_{a}^{b} f(x)\, dx \]
by Simpson's \(\frac{1}{3}\)-rule is :
Answer: A
Simpson's \(\frac{1}{3}\) rule is \(\frac{h}{3}[y_{0} + 4(y_{1} + y_{3} + \ldots + y_{n-1})\) \(+ 2(y_{2} + y_{4} + \ldots + y_{n-2}) + y_{n}]\).
The rule fits a parabola through each pair of strips, giving \(\frac{h}{3}(y_{0} + 4y_{1} + y_{2})\) on the first pair. Adding over all pairs (with n even), the end ordinates appear once, the odd-numbered ordinates get weight 4, and the interior even-numbered ordinates, shared by two pairs, get weight 2.
The \(\frac{h}{8}\) forms are not Simpson's rules (the \(\frac{3}{8}\) rule has factor \(\frac{3h}{8}\) and different weights). Option C puts the weight 4 on the even ordinates, including the ends, which reverses the pattern.
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