find all $ a \in \mathbb{R}$? 
find  all $a \in \mathbb{R}$ such that   if $f$  is  any contnious   function on $[1,3]$  with  $\int_{1}^{ 3}f(x) dx= 1$ , then  there  exist  some  $x_0 \in (1,3)$  with $f(x_0) = a$

My attempt :
I take $f(x) = \frac{x}{4}$  then  $\int_{1}^{ 3}f(x) dx= 1$
It  is given that  $x_0 \in (1,3)$  and we  have   $f(x) = \frac{x}{4}$
so  $'a'$  will be   $\frac{1.1 }{4}, \frac{1.2}{ 4},............,\frac{2.9}{4}$
Is its  true ?
 A: Since $f$ is any continuous function, with the condition that $\displaystyle\int_1^3 f(x) dx=1$, then $f(x)=\frac{1}{2}$ is an example of such a function.
So there must be some $x_0 \in (1,3)$ such that $f(x_0)=a$ but since this $f$ is a constant function then $\forall x\in (1,3)$, $f(x)=\frac{1}{2}$ thus $a=\frac{1}{2}$
It remains to prove that for any other function with these condition we can find an $x_0\in (1,3)$ such that $f(x_0)=\frac{1}{2}$, we can see this clearly by knowing the integral gives us the area under the curve, if the function gets below the horizantal line $y=\frac{1}{2}$ then for sure in some other point the curve will be above this line in order for the area to stay equal to $1$; which means that definetly at some point $x_0 \in (0,3)$ we will have $f(x_0)=\frac{1}{2}$
A: Find a non-trivial continuous function $f: [1,2] \rightarrow \mathbb{R}$ such that $\int_1^2 f(x) d\text{x} = 0$ and $f(1) = f(2) = 0$ (a spike function would do, for example), then extend $f$ to $[1,3]$ such that $\int_1^3 f(x)d\text{x} = 1$, for example by having $f(x) = 2(x-2)$ for $x \in [2,3]$. This proves that the set consists of the whole $\mathbb{R}$.
