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Define $f^{(k)}\left|[0,1]\rightarrow [0,1]\right., k=0,1,\ldots $ iteratively as follows,
$$
f^{(0)}(x)=2x(1-x)
$$
and for $k\geq 1$
$$
f^{(k)}(x)=f\circ f^{(k-1)}(x).
$$
The problem is to evaluate
$$
I_k:=\int_0^1f^{(k)}(x)\,\mathrm{d}x.
$$
To evaluate the integral introduce a new independent variable $\eta:=x-\frac{1}{2}$.  One has

$$
f^{(0)}(x)=f^{(0)}\left(\eta+\frac{1}{2}\right)=\frac{1}{2}-2\eta^2
$$
Define $g^{(k)}(\eta)\left|\left[-\frac{1}{2},\frac{1}{2}\right]\rightarrow \left[-\frac{1}{2},\frac{1}{2}\right]\right., k=0,1\ldots$ iteratively as follows
$$
g^{(0)}(\eta)=\frac{1}{2}-2\eta^2

$$
and for $k\geq 1$
$$
g^{(k)}(\eta)=g\circ g^{(k-1)}(\eta)
$$
With this the problem becomes, evaluate 
$$
I_k:=\int_{-\frac{1}{2}}^{\frac{1}{2}}g^{(k)}(\eta)\,\mathrm{d}\eta
$$
The utility in the above transformation comes from the fact that it is easier to decern the pattern for $g^{(k)}$ than $f^{(k)}$. This can be seen from computing the first few of these functions.
$$
g^{(1)}(\eta)=\frac{1}{2}-2\left[\frac{1}{2}-2\eta^2\right]^2=2^2\left[\eta^2-\eta^4\right]
$$

$$
g^{(2)}(\eta)=\frac{1}{2}-2\left[2^2\left(\eta^2-\eta^4\right)\right]^2=\frac{1}{2}-2^5\left[\eta^2-\eta^4\right]^2
$$

$$
g^{(3)}(\eta)=\frac{1}{2}-2\left[\frac{1}{2}-2^5\left[\eta^2-\eta^4\right]^2\right]^2=2^8\left[\eta^2-\eta^4\right]^2-2^{11}\left[\eta^2-\eta^4\right]^4
$$
I will leave the rest for you to do.

Sat, 15 Feb 2025 21:39 GMT