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If $f(x)\geq 0$, is it true that $\int f(x)dx \geq 0$?

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closed as off-topic by RRL, Lee David Chung Lin, Leucippus, verret, Cesareo Apr 17 at 7:14

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    $\begingroup$ It’s kind of meaningless to ask whether a function is nonnegative when it’s only defined up to an arbitrary constant. You need to clarify what you mean by “the integral”. $\endgroup$ – Erick Wong Apr 15 at 17:14
  • $\begingroup$ If f is a non negative function for all values of x in a domain D then is it necessary that the integral of f over D is nonnegative $\endgroup$ – Farah Apr 15 at 21:08
  • $\begingroup$ As I said, “the integral” can mean many things. With no explicit reference to a domain of integration, the most common interpretation would be the indefinite integral. $\endgroup$ – Erick Wong Apr 15 at 21:28
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To clarify, $\int f(x) dx$ denotes the primitive of $f$. That is, any function $F(x)$ such that $F'(x)=f(x)$. There are infinitely many such functions for any given $f$, since you may add any constant to one primitive to get another, and the constant you add can be negative and have arbitrarily large magnitude, so there exist infinitely many primitives of a given $f$ that are negative at some point. If you are asking about $\int_a^bf(x)dx$, then if $b\geq a$ and $f(x)\geq0$ for all $x\in [a,b]$ then $\int_a^bf(x) dx\geq0$ and $\int_b^a f(x)dx\leq 0$.

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  • $\begingroup$ Indeed, for the typical case, every primitive will be negative somewhere, like $x+C$. $\endgroup$ – Erick Wong Apr 16 at 17:18

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