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: 
The integral of a function being zero does not imply that the function is zero. But you can apply the mean value theorem to $h$ to show that it must be zero at a point.

The integral mean value theorem for continuous function s$\mathbb R\to \mathbb R$ states that if $$\int\_a^b f(x)\,dx=S,$$ then there exists a point $c\in [a,b]$ such that $f(c)=\frac{S}{b-a}.$
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 $\int\_a^b f \ne \int\_a^b g$.