$\,\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\frac{1}{n}} \right)}^a} + {{\left( {\frac{2}{n}} \right)}^a} + ..... + {{\left( {\frac{n}{n}} \right)}^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {{n^a} + \frac{{n\left( {n + 1} \right)}}{2}} \right]}}.$
$ = \,\frac{{\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^a}} }}{{{{\left( {1 + \frac{1}{n}} \right)}^{a - 1}}\left[ {a + \frac{1}{2}\left( {1 + \frac{1}{n}} \right)} \right]}}\, = \frac{1}{{60}}$
$ = \frac{{\int\limits_0^1 {{x^n}dx} }}{{\left( {a + \frac{1}{2}} \right)}} = \frac{1}{{60}} = \frac{{\frac{1}{{a + 1}}}}{{a + \frac{1}{2}}} = \frac{1}{{60}}$
$ \Rightarrow \frac{{\frac{1}{{a + 1}}}}{{\left( {a + \frac{1}{2}} \right)}} = \frac{1}{{60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$
$ \Rightarrow 2{a^2} + 3a - 119 = 0$
$ \Rightarrow 2{a^2} + 17a - 14a - 119 = 0$
$ \Rightarrow \left( {a - 7} \right)\left( {2a + 17} \right) = 0$
$a = 7, - \frac{{17}}{2}$
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and $g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in R .$ Then the area (in sq. units) of the region bounded by the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2 \mathrm{x}=1$ and $2 \mathrm{x}=\sqrt{3},$ is