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Old exam question: Let $a,b\in\mathbb{R} : a<b$ and let $f$ be a continuous function, $[a,b]\rightarrow\mathbb{R}$.

Show that if $$\int_a^bf(x)dx=0$$ then there is $x_0\in [a,b]$ such that $f(x_0)=0$

Hint: $f$ can only have a sign, if $f$ has no roots.

My approach:$f$ is continuous so by the fundamental theorem there is $F$ such that $$\int_a^bf(x)dx = F(b)-F(a)=0$$ and $F'=f$

Then by the intermediate value theorem for derivatives we have $$0=\frac{F(b)-F(a)}{b-a}=F'(\zeta)=f(\zeta)$$ for some $\zeta\in(a,b)$

Now for my question: why the hint? I'm afraid I may have misunderstood the question as I don't even know how to use the hint.

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  • $\begingroup$ @MatthewDaly Thanks, edited $\endgroup$ – Ruben Kruepper Aug 14 at 10:26
  • $\begingroup$ By the hint it suffices to consider $f>0$ or $f<0$ everywhere, which is a lot easier. $\endgroup$ – trisct Aug 14 at 10:36
  • $\begingroup$ Your solution is OK. $\endgroup$ – Dr Zafar Ahmed DSc Aug 14 at 13:21
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Assume $a<b$. Proof suggested by the hint: If $f$ has no zeros then it has the same sign throughout the interval. Let us say $f(x) >0$ for all $x$. Since $f$ is continuous it attains its minimum value $m$ so $m >0$. This gives $\int_a^{b}f(x)dx >m(b-a)$ so the integral cannot be $0$.

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    $\begingroup$ Advantage with this proof: no derivatives used. $\endgroup$ – Kavi Rama Murthy Aug 14 at 10:28
  • $\begingroup$ Thank you very much, sir, very elegant! For peace of mind: was mine wrong or just lengthy? $\endgroup$ – Ruben Kruepper Aug 14 at 10:30
  • $\begingroup$ Nothing wrong at all with your proof. $\endgroup$ – Kavi Rama Murthy Aug 14 at 10:31

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