$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {\frac{1}{{\left( {1 + \frac{{{r^2}}}{{{n^2}}}} \right)}}} $ Using $D.I.$ as limit of sum, we get
$ = \int\limits_0^2 {\frac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}2} $
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$2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
$0$
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
$(a-c) x^2+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix
$\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right]$
is singular. Then the value of
$\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$