14. Let $f:[0,1] \rightarrow[0,1]$ be a continuous function such that $f(f(x))=1$ for all $x \in[0,1]$. Determine the set of possible values of $\int_{0}^{1} f(x) d x$.
Answer: $\left(\frac{3}{4}, 1\right]$ Since the maximum value of $f$ is $1, \int_{0}^{1} f(x) d x \leq 1$.
By our condition $f(f(x))=1, f$ is 1 at any point within the range of $f$. Clearly, 1 is in the range of $f$, so $f(1)=1$. Now $f(x)$ is continuous on a closed interval so it attains a minimum value $c$. Since $c$ is in the range of $f, f(c)=1$.
If $c=1, f(x)=1$ for all $x$ and $\int_{0}^{1} f(x) d x=1$.
Now assume $cc^{2}$, and the equality is strict because $f$ is continuous and thus cannot be $c$ for all $xc^{2}+(1-c)=\left(c-\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4} .
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Therefore $\frac{3}{4}<\int_{0}^{1} f(x) d x \leq 1$, and it is easy to show that every value in this interval can be reached.