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Is right that $\int_0^x f(t)\,dt>0$ if $x>0$ and $f(t)$ is positive and also if $x<0$ and $f$ is negative?

My doubt is in the second: but I have thought that since if $x<0$ then $\int_0^x f(t)dt=-\int_x^0 f(t)dt$ and if $f(t)$ si negative then $\int_x^0 f(t)dt<0$ and so $-\int_x^0 f(t)dt>0$. Is it right?

Sisi
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  • Use the definition of definite integral. – Mostafa Ayaz Apr 27 '21 at 16:03
  • @MostafaAyaz by definition I have $F(x)-F(0)$ and then? (I have also edited the question with my attempt) – Sisi Apr 27 '21 at 16:04
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    Yes, this is right. – Theo Bendit Apr 27 '21 at 16:05
  • The integral sign only depends on the behavior of $f$ in the subyacent domain of integration – Pablo Valentin Cortes Castillo Apr 27 '21 at 16:07
  • @PabloValentinCortesCastillo ok so it is as I have written? – Sisi Apr 27 '21 at 16:08
  • The second one indeed follows quite quickly from the first one, since if $f$ is negative, then $-f$ is positive which lets you exchange the limits of the integral. The first one is true (given that we assume $f$ is Riemann integrable) but it's quite hard to prove properly if you've only just started real analysis! There is some more discussion here. The accepted answer relies on a result about continuity of integrable functions. The answer after that is more explicit. – Izaak van Dongen Apr 27 '21 at 16:13

4 Answers4

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Notice that, if $m \leq f$ on $[0,x]$, then $$\int_0^xf(t) \ dt \geq mx.$$ If $f$ is positive ($m>0$) and $x>0$, then $mx>0$. If $f$ is negative ($m<0$) and $x<0$, then $mx>0$.

Edit: As pointed out in a comment, one can have $f$ intregable with $f>0$ on $[0,x]$, but $f \not\geq m$ for any $m>0$. An example of this might be $$f(t) = \begin{cases} t & 0<t\leq x\\ 1 & t=0 \end{cases}.$$ In that case, as long as we have $f\geq m > 0$ on some small subinterval $I \subset [0,x]$ of length $\varepsilon>0$, then $$\int_0^x f(t) \ dt \geq \int_I f(t) \ dt \geq m\varepsilon > 0.$$

Gary Moon
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  • Are you assuming that $f$ is continuous here to get your lower bound? I think an integrable function may be positive on a compact interval while not being bounded below by any positive constant. – Izaak van Dongen Apr 27 '21 at 16:19
  • @IzaakvanDongen My explanation was a bit loose, but I don't think that you need continuity to get the lower bound. If the function is just integrable, you probably need to do a bit more work but I think you could get the same type of estimate. – Gary Moon Apr 27 '21 at 16:30
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The definition of integral states that $$ \int_0^xf(t)dt=\lim_{n\to\infty}\frac{x}{n}\sum_{k=1}^{n} f(\frac{kx}{n}), $$ for $x>0$ and $\int_0^xf(t)dt=-\int_x^0f(t)dt$. If $f(t)>0$ for $t>0$ we obtain $\int_0^xf(t)dt\ge 0$.

Mostafa Ayaz
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$$\int_0^xf(x)\,dx=-\int_x^0f(x)\,dx$$ answers the question.

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okay so if we have $f(t)>0\forall t,\,x>0$ then we can say: $$\int_0^xf(t)dt>0$$ however if we are to take: $$\int_0^{-x}f(t)dt=-\int_{-x}^0f(t)dt<0$$ and this is due to the fundamental theorem of calculus (switching limits) and due to the fact $f(t)>0$ and so the integral is too. Hope this is what you were looking for :)

Henry Lee
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