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Assume $f$ is continuous on $[a,b]$, if $\int_a^b f(x)\,dx=0$, prove that $f(c)=0$ for at least one $c$ in $[a,b]$.

The problem didn't state anything about the function $f$, is it safe to assume either:

  1. $f$ is an odd function and implies that there is some $x_1$ $x_2$ in $[a,b]$ such that $f(x_1)<0$, $f(x_2)>0$ and apply Bolzano's Theorem to conclude that there is at least a $c$ in $[a,b]$ such that $f(c)=0$.

  2. $f$ is $0$ for all $x$ in $[a,b]$ hence it is trivial.

Is this argument correct?

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  • $\begingroup$ You cannot assume $f$ is an odd function, it need not be. $\endgroup$ Apr 5, 2015 at 6:35
  • $\begingroup$ That is why I'm doubtful with my argument. So what will I assume about f? $\endgroup$
    – shinobi20
    Apr 5, 2015 at 6:37

3 Answers 3

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By first mean value theorem for integration we have that exists $c\in\left[a,b\right]$ such that (assuming $a<b$ ) $$0=\int_{a}^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right)$$ then $$f\left(c\right)=0.$$

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  • $\begingroup$ Oh! I think this is what Apostol wants me to see. Thanks for the short but intuitive proof! $\endgroup$
    – shinobi20
    Apr 5, 2015 at 7:32
  • $\begingroup$ You're welcome. $\endgroup$ Apr 5, 2015 at 7:41
  • $\begingroup$ good answer @MarcoCantarini $\endgroup$
    – user210387
    Apr 5, 2015 at 7:53
  • $\begingroup$ Cool approach! I love seeing multiple ways of getting at the same result. $\endgroup$
    – Kaj Hansen
    Apr 5, 2015 at 8:13
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Consider the function $$ F(t)=\int_{a}^{t}f(x)dx $$ then $F(a)=0,$ $F(b)=0$ and by Fundamental theorem of calculus $F(t)$ is continuous on $[a,b],$ differentiable on $(a,b)$ and $F^{\prime}(t)=f(t).$

Apply Rolle's theorem to $F(t),$ we obtain there exists at least one $c\in(a,b)$ such that $F^{\prime}(c)=f(c)=0.$

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You are on roughly the right track thinking about areas under the curve (though "odd" is not the appropriate term). However, proving the contrapositive instead would yield a cleaner argument without needing cases. That is, suppose $f(x) \neq 0$ for all $x \in [a, b]$.

Then $f(x) > 0$ or $f(x) < 0$ for all $x \in [a, b]$ by the intermediate value theorem. From there, you can apply the very definition of $\displaystyle \int_a^b f(x) dx$ to finish up. In particular, for any partition $\mathcal{P}$ of $[a, b]$, we will have: $$\displaystyle \int_a^b f(x) dx \geq L(f, \mathcal{P}) = \sum_{i} (x_{i+1} - x_i)\inf \Big( \{f(x) \ | \ x \in [x_i, x_{i+1}] \} \Big)$$ where $x_i$'s $\in \mathcal{P}$.

So...

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  • $\begingroup$ Suppose f>0 for all x in [a,b]. By the Intermediate Value Theorem, since f is continuous in [a,b], we choose two arbitrary points in x1<x2 in [a,b] such that f(x1) is not equal to f(x2). Then f takes on every value between f(x1) and f(x2). Since f>0, the integral of f>0 for all x in [a,b]. The proof for the case f<0 is the same. Is this correct? $\endgroup$
    – shinobi20
    Apr 5, 2015 at 7:02
  • $\begingroup$ Essentially yes. In terms of area under the curve, if $f(x) > 0$ everywhere, then certainly the area under the curve is positive and so the integral will be positive. That is more or less what I think you're saying, and more or less what I'm conveying in my post above (with added rigor). $\endgroup$
    – Kaj Hansen
    Apr 5, 2015 at 7:20

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