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We already know that if $f(x)\le g(x)$ for $x\in[a,b]$, then $\int_a^b f(x)dx\le \int_a^b g(x)dx.$

Can we have example of $f$ and $g$ such that $f(x)<g(x)$ for all $x\in[a,b]$, but $\int_a^b f(x) dx=\int_a^b g(x) dx$?

ZeroToZero
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    Yes , If youTake a=b – MAHI Nov 03 '21 at 09:06
  • Smart answer, hehehehehe. What if a<b? – ZeroToZero Nov 03 '21 at 09:11
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    See https://math.stackexchange.com/questions/351157/is-the-riemann-integral-of-a-strictly-positive-function-positive – Clement C. Nov 03 '21 at 09:21
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    ZeroToZero I think if a<b it will not possible atleast in $\Bbb R^1$. But if we consider the Extended Real Number space it might be possible. But I'm not sure whether we can consider the Extended Real Number space as such . – MAHI Nov 03 '21 at 13:39
  • Thank you @Clement C, so: "If a function is strictly positive, then the integral is always strictly positive", but it still haven't answered, the possibility whether the integral could be 0, I think? – ZeroToZero Nov 05 '21 at 03:58
  • Thank you for the remark @Debjit Mullick. Yes, I am just considering the real $\mathbb{R}$, not the extended real one. – ZeroToZero Nov 05 '21 at 03:59
  • @ZeroToZero This says that if $h(x) > 0$ for all $x\in[a,b]$, then $\int_a^b h(x) dx >0$. This answers your question, by using $h=g-f>0$ (saying we cannot have an example like the one you want). – Clement C. Nov 05 '21 at 04:58

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