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For $a$\int_{a}^{b}f(x)dx=0$. Prove that there is at least one number in $(a,b)$ such that $f(c)=0$.

I attempted this

Let $c$ be any number in $(a,b)$

$\int_{a}^{b}f(x)dx=0$ implies $\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=0$ implies $\int_{a}^{c}f(x)dx=-\int_{c}^{b}f(x)dx$

Since $a

(3) If the function is zero, we are done as f(c)=0.

(1),(2) That is, either $f(a)>0 $ and $f(b)<0$ or $f(a)<0 $ and $f(b)>0$, and since $f(x)$ is continuous, by the Intermediate Value Theorem, there exists a $c$ such that

$$f(c)=0$$

I am very unsure if I would get my marks for showing this. Is it a correct proof? I am also looking for more elegant proofs. Thanks in advance!

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    Let $f(x)=x^2-1/3$. Then $\int_{-1}^1f(x)\,dx=0$ but $f(-1)=f(1)=2/3>0$.2012-10-29
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    I like the spirit of your proof. It is not completely right (as shown by the former comment), but it is in the good direction. Think about what you can say about the function knowing it integral in some interval, in terms of positiveness or negativeness.2012-10-29
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    wow! I guess that counterexample killed it! damnnn...2012-10-29

3 Answers 3

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Put

$$ F(x) = \int_{a}^{x}f(x)dx $$

and use Rolle's theorem.

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Use the mean value theorem on the function $g(x) = \int_a^{x}f(t)dt$. This problem is a special case of the mean value theorem for integrals.

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If $f$ is never $0$ in the $[a,b]$ interval, then, by continuity, either $f>0$ or $f<0$. Wlog suppose the former. Now, by $[a,b]$ being compact, $f$ has a minimum $m$ in $[a,b]$, which is also $>0$, hence $\int_a^b f \ge m(b-a) >0$.