Range of continuous function $f:[0,1] \rightarrow [0,1]$ satisfying $f(f(x))=1$. 
Continuous function $f:[0,1] \rightarrow [0,1]$ satisfies $${f(f(x))=1},{\forall x \in [0,1]}.$$
  Find range of $$\int_{0}^{1} f(x) dx.$$

I found that $f(x) \in [0,1]$ so $$f(f(f(x)))=f(1)$$ and $$f(1)=1$$.
And also maximum is $1$, when $f(x)=1$ . But I couldn't imagine another case and minimum case.....

My opinion: 

Thm.Continuous function $f$ that $f:[0,1] \rightarrow [0,1]$ satisfied $${f(f(x))=1},{\forall x \in [0,1]}$$
   Than $$f(x)=1$$.

Proof) $f:[0,1] \rightarrow [0,1]$ is continuous, So $\delta_h (\epsilon)>0$ is exists such that $$\mid {x-h} \mid < \delta_h (\epsilon) \Rightarrow \mid {f(x)-f(h)} \mid <\epsilon$$
$f(f(h))\in [0,1]$ and there are  $\delta_{f(h)}(\epsilon)>0$ 
 such that $$\mid x-f(h)\mid <\delta_{f(h)}(\epsilon) \Rightarrow\mid f(x)-f(f(h))\mid <\epsilon$$
So $$\mid f(x)-f(f(h))\mid = \mid f(x) -1 \mid <\epsilon$$

I found out it that makes my opinion wrong, thanks for help
 A: Let $m$ be the minimum of $f$. Then $f([0,1])=[m,1]$, so $f=1$ on $[m,1]$; and $f: [0,m] \rightarrow [m,1]$ is continuous surjective.
Thus, the integral of $f$ is more than $(1-m)*1+m*m=1-m+m^2$ and less than $(1-m)*1+m*1=1$.
Any value between can be reached by taking $f(x)=m+c_tx^t$ between $0$ and $m$ where $c_t=(1-m)/m^t$, for $t > 0$, and the integral of $f$ is $1-m+m^2+((1-m)m)/(1+t)=1-\frac{t}{1+t}m(1-m)$. 
So the possible integrals of $f$ range from $3/4$ to $1$, $1$ included, $3/4$ excluded.
A: Well, as $f$ is a continuous function, it follows that Im$(f[0,1]) = [a,b]$ for some $a,b$, and as $f(f(x)) = 1$ it follows that there exists an $x$ s.t. $f(x)$ is 1, so $b$ must be 1. So Im$(f) = [a,1]$ and furthermore, for each $x \in [a,1]$, it follws that $f(x) = 1$ and for each $x \le a$, that $f(x) \ge a$.
So for some $a \in [0,1]$, it follows that $f$ must satisfy
$$\int_0^1 f(x) dx  \geq (a \times a) \ + \ (1-a) \times 1 \ = \ 1-a+a^2 \doteq h(a) $$
However, $h(a) \geq \frac{3}{4}$ and $h(1/2)=3/4$, so $\int_0^1 f(x)dx$ must be at least $\frac{3}{4}$.
And this bound is "almost" tight: for each such $a$ and $\epsilon >0$ there is indeed an $g$ that satisfies $\in_0^1 g(x) dx \le 1-a+a^2$ and $g(g(x))=1$; set $g(x) = a$ for all $x<a-\epsilon$; $g(x) = a+\frac{(x-a+\epsilon)}{\epsilon}$ for all $x \in [a-\epsilon, a]$ [i.e., $f(x)$ ramps up rapidly from $a$ to 1 in the interval $[a-\epsilon, a]$ ], and $g(x)=1$ for all $x \in [a,1]$.
So $\int f dx \in (\frac{3}{4}, 1]$.
