Help computing $\lim_{n\to \infty}\int_{0}^{1}f_{n}(x)\,dx$ We have the function $f(x)=2x(1-x)$ and we define $f^{\circ 2}=f\circ f$, and $f^{\circ n}=f\circ f^{\circ(n-1)}$ for $n>2$. We need to compute $$\lim_{n\to\infty}\int_{0}^{1}f_{n}(x)dx$$. 
To do that, I found that $f^{\circ n}(x)\leq \frac{1}{2}$ for all $x\in[0,1]$ and for all $n$. Furthermore, I found that $$f(x)\leq f^{\circ 2}(x)\leq f^{\circ 3}(x)\leq \cdots \leq f^{\circ n}(x)\leq \frac{1}{2}.$$
With this in mind I obtained that $$\lim_{n\to \infty }\int_{0}^{1}f^{\circ n}(x)\,dx\leq \frac{1}{2}$$. In addition, I guess that the sequence of functions $\{f^{\circ n}(x)\}_{n\in \mathbf{N}}$ converges pointwise to the function $g(x)=1/2$ for $x\neq 0,1$ and $0$ for $x=0,1,$ but I couldn't prove it.
Please any suggestion for this problem would be really appreciated. 

You can verify that for $x\in (0,1)$,
$$\left|f(x)-f\left(\frac 1 2\right)\right| = \left|f(x)-\frac 1 2\right|  \leq \|f^\prime\|_{\infty}\left|x-\frac 12\right|=2\left|x-\frac 12\right|$$
Thus
$$\left|f_n(x)-\frac 1 2\right|\leq 2^n\left|x-\frac 12\right|^n$$
So you can now see that for all $x\in (0,1)$, you have pointwise convergence.
The integral converges towards to $\int_0^1\frac 12dx=\frac 12$ by the dominated convergence theorem.
