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 $f(c)=\frac{S}{b-a}.$
Others have explained your mistake. Here is a proof of the fact that starts with your approach:

Apply the [mean value theorem for integrals](https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration) to $h$ to conclude that
$$
0 = \int\_a^b h = h(c) (b-a)
$$
which implies $h(c)=0$, that is, $f(c)=g(c)$.