Suppose $h$ is a $2a$-periodic function ($a > 0)$ that is integrable on $[-a,a]$, is it true that $\int_{[-a,a]} h(x) = \int_{[-a,a]} h(-x) ?$
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2026-04-04 02:16:52.1775269012
$\int_{[-a,a]} h(x) = \int_{[-a,a]} h(-x) ?$
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Sub in the last integral: $$\int_{[-a,a]} h(-x)=-\int_{[a,-a]} h(x)=\int_{[-a,a]} h(x)$$