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

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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.

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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.

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

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