Calculus Proof Problem Let $f$ be a continuous function whose domain includes $[0,1]$, such that $0 \le f(x) \le 1$ for all $x \in [0,1]$, and such that $f(f(x)) = 1$ for all $x \in [0,1]$. Prove that $\int_0^1 f(x)\,dx &gt; \frac34$.

I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help me?

Thanks!
While not a general proof of the proposition, it can be seen as true of a certain class of functions satisfying the conditions.

Let $f(x)$ be a continuous non-decreasing function defined on the interval $[0,1]$ with $a\in[0,1]$ as shown in the diagram and satisfying


$f(0)=a$
$f(a)=1$ for $x\ge a$


Then the integral $\int_0^1 f(x)\,dx$ must be greater than $a^2+(1-a)$, the sum of the two shaded rectangles.

But the minimum area of the two rectangles is $\frac{3}{4}$ and occurs when $a=\frac{1}{2}$. 

It remains to be determined whether there is any other class of function satisfying the conditions.

Addendum: Actually we can drop the non-decreasing part and merely require that $f(x)\ge a$ on $[0,a)$.
Edit: This proof does not work if f (x) does not have range [0,1] for all x in [0,1]. The simple way to prove this:

We have function f(x) which is continuous along the interval [0,1]. 

f (x) is defined such that                                                              f(x) has domain [0,1] , and range [0,1].                  (a)

f(f (x)) has domain [0,1] and range [1].                 (b)

--Note: The domain of f(f(x)) comes from the range of f(x) along the interval [0,1].--

Now we want to prove that the integral of f(x) over [0,1] is > 3/4

In order to prove this we can break the problem into two parts. First we can show that f(f(x)) and f(x) have the same range for domain [0,1]. Second we can show that because the function f's range is [1] for all x in [0,1] that the integral over [0,1] must be greater than 1.

First Part:

Since f is a function it cannot have two different outputs for one input, so (a) and (b) cannot have contradictory ranges for the same domain. From (a) we know that the range of f over domain [0,1] is 0&lt;=f(x)&lt;=1, from (b) we also know the range over domain [0,1] is f(x)=1. So (a) and (b)'s range do not contradict each other  in so much as f(f(x))=1 is within the range of f(x). But since f is a function it must be the case that f(x)=1 for all x in the interval [0,1]. Otherwise the function f would have multiple outputs for the same inputs and would not be a function. Therefore it must be the case that for all input for f in the interval [0,1], f has an output of 1, and thus for the function f the domain [0,1] has range [1].

The Second Part:
Since the function f on domain [0,1] has range [1], then any integral within [0,1] of inputs must be greater than or equal to 1. Therefore since the integral is greater >=1 and 1 is > 3/4, the integral is > 3/4.

Proof complete.