Answer
$\int_{1}^{2}x^{-2}\,dx=\frac{1}{2}$
Work Step by Step
We have $x_{i}^* = \sqrt{x_{i-1}x_{i}}$ and $x_{i}=1+\frac{i}{n}$. We can find the definite integral using the definition:
$\int_{1}^{2}x^{-2}\,dx=\lim\limits_{n \to \infty}\frac{1}{n}\sum_{i=1}^n x_{i}^{*-2} = \lim\limits_{n \to \infty}\frac{1}{n}\sum_{i=1}^n \frac{1}{(1+\frac{i-1}{n})(1+\frac{i}{n})}$
$=\lim\limits_{n \to \infty}\frac{1}{n}\sum_{i=1}^n \frac{n^2}{n(1+\frac{i-1}{n})n(1+\frac{i}{n})}$
$=\lim\limits_{n \to \infty}\frac{1}{n}\sum_{i=1}^n \frac{n^2}{(n+i-1)(n+i)}$
$=\lim\limits_{n \to \infty}n\sum_{i=1}^n \frac{1}{(n+i-1)(n+i)}$
$=\lim\limits_{n \to \infty}n\sum_{i=1}^n(\frac{1}{n+i-1}-\frac{1}{n+i})=\lim\limits_{n \to \infty}n(\sum_{i=1}^n(\frac{1}{n+i-1})-\sum_{i=1}^n(\frac{1}{n+i}))$
$=\lim\limits_{n \to \infty}n[(\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n-1})-(\frac{1}{n+1}+...+\frac{1}{2n-1}+\frac{1}{2n})]$
$=\lim\limits_{n \to \infty}n(\frac{1}{n}-\frac{1}{2n})=\lim\limits_{n \to \infty}(1-\frac{1}{2})=\frac{1}{2}$
