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22 hours ago comment added ryang Indeed, every antiderivative of an even function is a vertical translation of a common odd function.
23 hours ago comment added Nate Eldredge Indeed, the argument above correctly shows that $G(x) = -G(u) + C$. Since $u = -x$, we have $G(x) = -G(-x) + C$, i.e. the antiderivative is an odd function plus a constant, which is true.
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yesterday vote accept Samuel Ho
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yesterday history answered ryang CC BY-SA 4.0