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I have to show that

Let $f$ be continuous on $[a,b]$. If $\displaystyle \int \limits _a^b|f|=0$, then $f(x)=0$ for all $x\in [a,b]$.

I know the theorem that says that if $f\in \mathcal{R}[a,b]$, then $|f|\in \mathcal{R}[a,b]$ and $\displaystyle \left |\int \limits _a^bf\right |\leq \int \limits _a^b|f|$. Here $\mathcal{R}[a,b]$ denotes the collection of all Riemann integrable functions on $[a,b]$.
Using this theorem, I can figure out that $\displaystyle \int \limits _a^bf=0$.
However I don't know what the next step should be. I don't know how to prove that if $\displaystyle \int \limits _a^bf=0$, then $f=0$.
Any suggestions, please?

commie trivial
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superdumb
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  • It is not true that $\int_a^b f = 0$ implies $f = 0$ if $f$ may take on negative values. Your method removes this restriction on the integrand, so it kind of takes you in the wrong direction. – Stephen Donovan May 31 '22 at 17:25
  • Assume that $f\in C^0([a,b])$ takes a non-zero value for some $c\in[a,b]$. Since $|f|$ is continuous, $|f|$ is positive in a neighbourhood of $c$. So $\int|f|$ is positive. – Jack D'Aurizio May 31 '22 at 17:25

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