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We consider the Trapezoidal Rule

$$Q[f] = \frac{b-a}{2}(f(a) + f(b))$$

We notice that $$I[x^3] = \int_{-1}^{1}x^3dx = 0 = Q[x^3] = \frac{1-(-1)}{2}(1^3 + (-1)^3) = 0$$

Does that imply that the Trapezoidal Rule has order of accuracy of at least $s = 4$ ?

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  • $\begingroup$ No. The Trapezoidal Rule has order of accuracy of $s=1$. That is, it s exact for all $a, b$ only for $x^0=1, x^1 $. That the result is exact for $$\int_{-1}^{1}x^3dx$$ is only a coincidence, nothing more. This could be considered as a trap question in exams. $\endgroup$
    – Ryukyu
    Aug 22 '20 at 13:03
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No. The Trapezoidal Rule has order of accuracy of $s=1$. That is, it s exact for all $a, b$ only for $x^0=1, x^1 $. That the result is exact for $$\int_{-1}^{1}x^3dx$$ is only a coincidence, nothing more. This could be considered as a trap question in exams.

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The trapezoidal rule, in this simple and also composite form with evenly spaced subdivisions, is correct for any function that has odd symmetry at the center of the integration interval $[a,b]$. Indeed it will compute the same value for $f(x)$ as for its even symmetric part $$\frac12(f(x)+f(a+b-x)).$$

This does not change the fact that there will be an error for quadratic functions. If a quadratic function is open above and positive, it is convex and the trapezes that are used to approximate it have consistently a larger area than that inside them under the parabola.

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