$\int_a^b f = \int_a^b g$, then there exists a point $c \in [a,b]$ such that $f(c) =g(c)$. 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?
 A: 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
$$\int_a^b h(x)dx=0\implies h(x_0)=0\text{ for some }x_0\in[a,b].$$
A: 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
$$h(c)=0$$
A: 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 $f(c)=\frac{S}{b-a}.$
A: You can't conclude that $\forall x\in [a,b], h(x) = 0 $. Take $h(x) = x$ on $[-1,1]$ as a counterexemple.
But if $\int_a^b h(x) = 0$, then $H$ a primitive of $h$ verify $H(a) = H(b)$. Then you use Rolle theorem to show that there exist $c\in ]a,b[$ such that $H'(c) = 0$
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 to $h$ to conclude that
$$
0 = \int_a^b h = h(c) (b-a)
$$
which implies $h(c)=0$, that is, $f(c)=g(c)$.
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 $\int_a^b f \ne \int_a^b g$.
