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:

*

*$f$ 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 $f(c)=0$.


*$f$ is $0$ for all $x$ in $[a,b]$ hence it is trivial.
Is this argument correct?
 A: 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 $$f\left(c\right)=0.$$
A: 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 $F^{\prime}(c)=f(c)=0.$ 
A: 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...
