Given: $\int_{a}^{b}f=\int_{a}^{b}g$, Can we always find some point $\theta$ in $[a,b]$ such that $f(\theta)=g(\theta)$ I noted something today, I don't know whether the result holds in general setting or not.

Suppose $f$ and $g$ are continuous function on $[a,b]$ such that
  $\int_{a}^{b}f=\int_{a}^{b}g$. 
Can we always find some point $\theta$ in $[a,b]$ such that
  $f(\theta)=g(\theta)$

Example: Take $f=x^2$ and $[a,b]=[0,1]$ then $\int f=1/3$. Take $g=1/3$ then $\int_{[0,1]} g=1/3$
Now we can easily find some points which satisfy $x^2=1/3$. 
There are lot of examples that I checked and every time I found that resullt holds good.
Can we write a proof?
Thanks for reading and helping out.

Let $F(x):= \int_a^x f(t) dt, G(x):= \int_a^x g(t) dt$ and $H(x)=F(x)-G(x).$ Then we have:
$H(a)=H(b)=0$ and $H'(x)=f(x)-g(x).$  By Rolle, there is $\theta \in [a,b]$ such that $H'( \theta)=0.$
