Show that the integral of an odd function over symmetric interval is always $0$ for a cubic function in the Simpson's rule?

Question

I know that the definite integral of a cubic function $f(x)$ over a symmetric interval is $0$. I just need some clarity on why that is.

Do you mean why is Simpson's exact for cubics or why cubics (as an odd function) integrates to zero over a symmetric interval? I think the former question is answered already on the site and has to do with the error term of Simpson's rule being expressed in terms of a 4th derivative. — Karl, Apr 12 '20 at 17:45
I guess I am asking both, since I believe defining why Simpson's rule is exact for cubics is also incorporating the latter. — ttsega, Apr 12 '20 at 18:23
Differentiate a cubic 4 times and you get zero. Linked to a proof. — Karl, Apr 12 '20 at 19:43

score 1 · Answer 1 · answered Apr 12 '20 at 17:18

1

Since $f$ is odd, one has that $f(-x) = -f(x)$. Consequently,

\begin{align*} \int_{-a}^{a}f(x)\mathrm{d}x & = \int_{-a}^{0}f(x)\mathrm{d}x + \int_{0}^{a}f(x)\mathrm{d}x\\\\ & = \int_{0}^{a}f(-x)\mathrm{d}x + \int_{0}^{a}f(x)\mathrm{d}x\\\\ & = -\int_{0}^{a}f(x)\mathrm{d}x + \int_{0}^{a}f(x)\mathrm{d}x = 0 \end{align*}

Hopefully this helps.

answered Apr 12 '20 at 17:18

user0102

21,867

Thank you, the integration set up helps a lot in clarification. – ttsega Apr 12 '20 at 18:28

score 0 · Answer 2 · answered Apr 12 '20 at 17:24

0

Does this plot of a cubic (odd) function over a symmetric interval help?

answered Apr 12 '20 at 17:24

David G. Stork

30,409

Yes thank you for the visual representation! – ttsega Apr 12 '20 at 18:25

Show that the integral of an odd function over symmetric interval is always $0$ for a cubic function in the Simpson's rule?

2 Answers2