Continuous function with zero integral Let $f$ be a continuous function on $[a,b]$ ($a<b$), such that $\int_{a}^{b}{f(t)dt}=0$
Show that $\exists c\in[a,b], f(c)=0$.
 A: Let $m=\min\{f(x)|x\in[a,b]\}, m=\max\{f(x)|x\in[a,b]\}$(We can get the minimum and maximum because $f$ is continuous on a closed interval) . If $m,M$ have the same sign it can be shown that the integral cant be zero (for example if both are positive, the the integral will be positive). If $m,M$ have different signs apply the intermediate value theorem
A: By the Fundamental Theorem of Calculus, the function
$$
F(x):=\int_a^xf(t)dt
$$
is continuous on $[a,b]$ with derivative $F'=f$ on $(a,b)$.
Now $F(a)=F(b)=0$.
By Rolle, there exists $c\in (a,b)$ such that $F'(c)=f(c)=0$.
A: Proceed by contradiction: assume that $f(x)\not=0$ for all $a\le x\le b$. Then either $f(x)>0$ for all $a\le x\le b$, or $f(x)<0$ for all $a\le x\le b$. Because, if there existed $x_1,x_2\in[a,b]$ with $f(x_1)<0<f(x_2)$, then, by the intermediate value theorem, we would have a $c$ such that $f(c)=0$.
In the first case let $d=\inf\{f(x);a\le x\le b\}=\min\{f(x);a\le x\le b\}$ ($d$ is a minimum because $f$ is continuous and $[a,b]$ is compact. Furthermore $d>0$), therefore $\int_a^bf(t)dt\ge d(b-a)>0$, which contradicts our assumption.
The second case is similar.
