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 MVT states that if
$$\int\_a^bh(t)dt=k$$ then there exists some $c\in[a,b]$ such that
$$k(b-a)=h(c)$$

Since in this case $k=0$ we get
$$h(c)=0$$
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 $f(c)=\frac{S}{b-a}.$