$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $$     equals :

Solution :
$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{{\frac{r}{n}}}{{\sqrt {1 + {{\left( {\frac{r}{n}} \right)}^2}} }}} \cr & = \int_\alpha ^\beta {\frac{x}{{\sqrt {1 + {x^2}} }}dx} \cr & = \int_0^2 {\frac{{x\,dx}}{{\sqrt {1 + {x^2}} }}} \cr & = \frac{1}{2}\int_0^2 {\frac{{d\left( {1 + {x^2}} \right)}}{{\sqrt {1 + {x^2}} }}} \cr & = \left[ {\frac{1}{2}.2\sqrt {1 + {x^2}} } \right]_0^2 \cr & = \sqrt 5 - 1 \cr} $$