$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $$ is equal to :

<p>$$\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $$</p> <p>$$ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}} $$</p> <p>$= \int_a^b {f(x)dx}$</p> <p>$a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0$</p> <p>$b = \mathop {\lim }\limits_{n \to \alpha } \left( {{n \over n}} \right) = 1$</p> <p>and ${r \over n} \to x$</p> <p>$= \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx}$</p> <p>$= \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx}$</p> <p>$= \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx}$</p> <p>$= \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx}$</p> <p>$= \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx}$</p> <p>$= \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1$</p> <p>$$ = - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]$$</p> <p>$$ = - {3 \over 2}\log \left( {{1 \over 3}} \right) + 2\log \left( {{1 \over 2}} \right)$$</p> <p>$= + {3 \over 2}\log 3 - 2\log 2$</p> <p>$= \log \sqrt {{3^3}} - \log 4$</p> <p>$= \log {{\sqrt {{3^2} \times 3} } \over 4}$</p> <p>$= \log {{3\sqrt 3 } \over 4}$</p>