Which is the highest polynomial order that allows the trapezoidal rule to obtain an exact value for integration?

Question

BlackTom AI · Accepted Answer

The trapezoidal rule is based on linear interpolation, so it is exact for all polynomials of degree ≤ 1 (constant and linear functions).

Relevant concepts:
- Degree of precision (algebraic accuracy) of a quadrature rule is the highest degree n such that all polynomials of degree ≤ n are integrated exactly.
- The trapezoidal rule on a single interval [x0, x1] approximates ∫_{x0}^{x1} f(x) dx by (x1 − x0)·(f(x0) + f(x1))/2. Because this uses the straight line joining (x0,f(x0)) and (x1,f(x1)), it exactly integrates linear functions.

Step-by-step verification for a general linear polynomial f(x) = m x + c on [x0,x1]:
1. Exact integral:
   ∫_{x0}^{x1} (m x + c) dx = (m/2)(x1^2 − x0^2) + c(x1 − x0).
2. Trapezoidal approximation:
   (x1 − x0)·(f(x0) + f(x1))/2
   = (x1 − x0)/2 · (m x0 + c + m x1 + c)
   = (x1 − x0)/2 · (m(x0 + x1) + 2c)
   = (m/2)(x1 − x0)(x0 + x1) + c(x1 − x0).
3. Note (x1^2 − x0^2) = (x1 − x0)(x1 + x0), so the trapezoidal expression equals the exact integral. Thus degree of precision ≥ 1.

Counterexample showing it is not exact for degree 2:
- Take f(x) = x^2 on [0,1].
  Exact integral = ∫_0^1 x^2 dx = 1/3 ≈ 0.3333.
  Trapezoid gives (1 − 0)·(f(0) + f(1))/2 = (0 + 1)/2 = 0.5 ≠ 1/3.
So the trapezoidal rule is not exact for general quadratics (degree 2).

Why other choices are incorrect:
- a. 4, c. 3, d. 5: These claim higher degrees of exactness than the trapezoidal rule has; they are false because the rule is derived from linear interpolation and fails on degree ≥ 2.
- e. 2: Quadratic (degree 2) is not always integrated exactly (counterexample shown), so degree 2 is incorrect.

Conclusion: The highest polynomial order for which the trapezoidal rule is exact is 1. This matches the provided answer b. 1.

Which is the highest polynomial order that allows the trapezoidal rule to obtain an exact value for integration?单项选择题

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