Question: Let $f : [a,b] \to \mathbb R$ and $g : [a,b] \to \mathbb R$ be continuous on $[a,b]$ and $\int\_a^b f = \int\_a^b g$, then there exists a point $c \in [a,b]$ such that $f(c) =g(c)$.

If we let $h(x) = f(x) - g(x)$ then we have

$$\int\_a^b h = \int\_a^b \bigl(f-g\bigr) = \int\_a^b f - \int\_a^b g = 0$$

From these we can conclude that $h(x) = 0 \ \ \forall x \in [a,b]$. So why does the problem say we get one point? Please help me to find mistake and how to do it?

Answer: 
It is not true that
$$\int\_a^bh(x)dx=0\implies h(x)=0,\forall x\in[a,b].$$
For example
$$\int\_0^{2\pi}\sin x\,dx=0.$$
However, if $\int\_a^bh(x)dx=0$ and $h$ is continuous, then $h(x)=0$ at one point. We can prove it by contrapositive. Suppose $h(x)\neq 0$ for all $x\in[a,b]$. By continuity and the Inetermediate Value Theorem, either $h(x)>0$ for all $x$ or $h(x)<0$ for all $x$. If $h>0$ then $\int\_a^bh(x)dx>0$ and if $h<0$ then $\int\_a^bh(x)dx<0$. So in either case $\int\_a^bh(x)dx\neq 0$.

So we proved that
$$h(x)\neq 0,\forall x\in[a,b]\implies \int\_a^bh(x)dx\neq 0$$
and this the contrapositive equivalent of
$$\int\_a^b h(x)dx=0\implies h(x\_0)=0\text{ for some }x\_0\in[a,b].$$
Consider proving the contrapositive: if $f(x) \ne g(x)$ for all $x$ in $[a, b]$, then $\int\_a^b f \ne \int\_a^b g$. Because $f - g$ is continuous, it has *constant sign* on $[a, b]$: since it's never zero, IVT prevents it from being positive somewhere and negative elsewhere. So, swapping the roles of $f$ and $g$ if necessary, we may assume that $f - g > 0$. Positive continuous functions have positive integrals, so $\int\_a^b f - g > 0$ and therefore $\int\_a^b f \ne \int\_a^b g$.