# Order of Accuracy of Trapezoidal-Rule for $\int_{-1}^{1}x^3dx$

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 $$s = 4$$ ?

• 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. Aug 22 '20 at 13:03

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.
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 $$\frac12(f(x)+f(a+b-x)).$$