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The trapezoidal rule with \( n \) strips of width \( h \) estimates the integral as

  • A.\( \dfrac{h}{2}\left[y_{0} + y_{n}\right] \)
  • B.\( \dfrac{h}{2}\left[y_{0} + y_{n} + 2(y_{1} + \cdots + y_{n-1})\right] \)
  • C.\( \dfrac{h}{3}\left[y_{0} + y_{n} + 4(y_{1} + \cdots + y_{n-1})\right] \)
  • D.\( h\left[y_{0} + y_{1} + \cdots + y_{n}\right] \)

Answer: B

Each strip contributes the area of a trapezium, \( \tfrac{h}{2}(y_{i} + y_{i+1}) \). Summing, every interior ordinate is counted twice and the two end ordinates once:

\[ I \;\approx\; \frac{h}{2}\left[y_{0} + y_{n} + 2\sum_{i=1}^{n-1} y_{i}\right]. \]

The divisor three and the weight four belong to Simpson's rule

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