Questions: Newton-Cotes Quadrature Formulas

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

Simpson's rule uses a quadratic (degree-2) interpolating polynomial, yet it integrates cubic polynomials exactly. Why?

AA quadratic polynomial can implicitly represent a cubic through the midpoint evaluation, which doubles its effective degree
BThe error term in Simpson's rule involves a fourth derivative, which vanishes for cubic polynomials—a consequence of the symmetric three-point arrangement
CSimpson's rule uses adaptive step sizing that automatically improves for polynomial integrands
DThis is a coincidence specific to the interval [−1, 1] and does not hold for general intervals
Question 2 Multiple Choice

A student wants very high accuracy for a smooth integrand over [0, 1] and decides to use a single 10-point Newton-Cotes formula instead of applying Simpson's rule repeatedly over many small subintervals. What problem are they likely to encounter?

AThe 10-point formula requires solving a 10×10 linear system, making it computationally prohibitive
BHigher-order Newton-Cotes formulas with equally-spaced points suffer from Runge's phenomenon: large oscillations near the endpoints cause the weights to become large and alternating in sign, potentially making accuracy worse
CThe 10-point formula has lower convergence order than Simpson's rule, so it always produces less accurate results
DNewton-Cotes formulas are only defined for intervals of the form [0, 1] and cannot be applied to general domains
Question 3 True / False

Halving the step size h in the trapezoidal rule reduces the integration error by a factor of 4, because the trapezoidal rule has O(h²) global accuracy.

TTrue
FFalse
Question 4 True / False

Higher-order Newton-Cotes formulas—those using more equally-spaced evaluation points—typically produce more accurate results than lower-order formulas like Simpson's rule for the same interval.

TTrue
FFalse
Question 5 Short Answer

Why are composite quadrature methods preferred over single high-order Newton-Cotes formulas in practice, even when high accuracy is required?

Think about your answer, then reveal below.