Calculate integral with cantor measure 
Calculate the integral $$\int_{[0,1]}x^2d\mu_F$$ where F is the cantor function. Use the following hints about the cantor function:

*

*$F(1-x)=1-F(x)$

*$F(\frac x 3)=\frac{F(x)}{2}\quad\forall x\in[0,1]$

*$F(0)=0$

I thought that $$\int_{[0,1]}x^2d\mu_F=\int_{[1,0]}(1-x)^2d\mu_{F}=\int_{[0,1]}x^2d\mu_{1-F(x)}$$ but here I'm stuck and I don't know how to continue calculating this integral. Furthermore, how do we use the second and third properties when given the cantor function above?

Let $C_1=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]$, $C_2=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9},\frac{3}{9}\right]\cup\left[\frac{6}{9},\frac{7}{9}\right]\cup\left[\frac{8}{9},\frac{9}{9}\right]$ and so on the usual sets used to define the Cantor set. Then $\mu_F$ is the limit as $n\to +\infty$ of the probability measure $\mu_{P_n}$ on $C_n$. Let $I=[a,a+3b]$ be any closed interval of the real line and $J$ the same interval without its middle third, $J=[a,a+b]\cup[a+2b,a+3b]$. Then:
$$ \int_I x^2 d\mu = \frac{1}{3}\left((a+3b)^3-a^3\right)=3b(a^2+3ab+3b^2), $$
$$\frac{3}{2}\int_J x^2 d\mu = 3b(a^2+3ab+3b^2)+b^3, $$
so:
$$ \frac{3}{2}\int_J x^2 d\mu = \int_I x^2 d\mu + \frac{\mu(I)^3}{27},\tag{1}$$
giving immediately:
$$ \int_{0}^{1} x^2\, d\mu_F = \lim_{n\to +\infty}\int_{0}^{1} x^2\, d\mu_{P_n} = \lim_{n\to +\infty}\sum_{k=0}^{n}\frac{1}{3^{2k+1}}=\color{red}{\frac{3}{8}} .\tag{2}$$
