# If $\int_a^b f(x)\,dx=0$, prove that $f(c)=0$ for at least one $c$ in $[a,b]$

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?

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

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 $$f\left(c\right)=0.$$

• Oh! I think this is what Apostol wants me to see. Thanks for the short but intuitive proof! Apr 5, 2015 at 7:32
• You're welcome. Apr 5, 2015 at 7:41
• good answer @MarcoCantarini
– user210387
Apr 5, 2015 at 7:53
• Cool approach! I love seeing multiple ways of getting at the same result. Apr 5, 2015 at 8:13

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 $F^{\prime}(c)=f(c)=0.$

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...

• 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? Apr 5, 2015 at 7:02
• 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). Apr 5, 2015 at 7:20