Calculate integral with cantor measure

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: $$ \int_{0}^{1} x^2\, d\mu_F = \lim_{n\to +\infty}\int_{0}^{1} x^2\, d\mu_{P_n} = \lim_{n\to +\infty}\sum_{k=0}^{n}\frac{1}{3^{2k+1}}=\color{red}{\frac{3}{8}} .\tag{2}$$