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.

An equivalent formulation (via $h=f-g$) is 

Suppose $h$ is a continuous function on $[a,b]$ such that
  $\int_{a}^{b}h(x) \, dx=0$. 
Can we always find some point $\theta$ in $[a,b]$ such that
  $h(\theta)=0$?

and that is true. If  $h$ has no zeros then the intermediate value theorem implies that $h$ is strictly positive or strictly negative in the interval, and consequently 
$\int_{a}^{b}h(x) \, dx \ne 0$.
(Strictly speaking, we must assume that $a \ne b$, because otherwise  $\int_a^b h(x) \, dx = 0$ for arbitrary functions $h$.)
