Riemann Sum of $\int_1^b x^{-1/2}\,dx$ with $x^*_i=\frac{(\sqrt{x_{i-1}}+\sqrt{x_i})^2}{4}$

Question

I am trying to evaluate the integral $$\int_1^b x^{-1/2}\,dx$$ with $b\in\mathbb{Z}_{>1}$ using Riemann sums. I consider the uniform partition $$\mathcal{P}_n=\left\{1,1+\frac{b}{n},1+\frac{2b}{n},\dots,1+\frac{b(n-1)}{n},b\right\}$$ on $[1,b]$ and the Riemann sum $$ R_n=\sum_{i=1}^n(x_i-x_{i-1})f(x^*_i) $$ where $$x^*_i=\frac{(\sqrt{x_{i-1}}+\sqrt{x_i})^2}{4}.$$ I am having difficulty evaluating $f(x^*_i)$. We have \begin{align} f(x^*_i)=\frac{2}{\sqrt{x_{i-1}}+\sqrt{x_i}}, \end{align} but how do we represent $x_{i-1}$ and $x_i$ to simplify this expression? Is there something wrong with my partition?