If $f\left(x\right)=\frac{1}{2^n}$when $\frac{1}{2^{n+1}}<x\le \frac{1}{2^n},n=0,1,2\dots$

then $l{t}_{n\to \mathrm{\infty }}{\int }_{1/2^n}^1 f\left(x\right)\mathrm{d}x$

$\begin{array}{l}{\int }_{1/2^n}^1 f\left(x\right)dx\\ ={\int }_{1/2^n}^{1/2^{n-1}} f\left(x\right)dx+{\int }_{1/2^{n-1}}^{1/2^{n-2}} f\left(x\right)dx+\cdots +{\int }_{1/2}^1 f\left(x\right)dx\end{array}$

$=\frac{1}{2^{n-1}}\left[\frac{1}{2^{n-1}}-\frac{1}{2^n}\right]+\frac{1}{2^{n-2}}\left[\frac{1}{2^{n-2}}-\frac{1}{2^{n-1}}\right]+\cdots +1\left[1-\frac{1}{2}\right]$

$=\frac{1}{2^{2n-1}}+\frac{1}{2^{2n-3}}+\cdots +\frac{1}{2}=\frac{1}{2}\left[\frac{1-{\left(1/2^2\right)}^n}{1-1/2^2}\right]=\frac{2}{3}\left[1-{\left(\frac{1}{4}\right)}^n\right]$

$\underset{n\to \mathrm{\infty }}{lim} {\int }_{1/2^n}^1 f\left(x\right)dx=\frac{2}{3}$