MCQ
$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{{n^2} + {k^2}}}} $is equals to
- ✓$\frac{1}{2}\log 2$
- B$log\ 2$
- C$\pi /4$
- D$\pi /2$
$ = \mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{k = 1}^n {} \frac{1}{n}\frac{{\left( {\frac{k}{n}} \right)}}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}$
$I = \int\limits_0^1 {\frac{x}{{1 + {x^2}}}dx} $
$ = \frac{1}{2}[\log (1 + {x^2})]_{\,0}^{\,1}$$ = \frac{1}{2}\left[ {\log 2} \right]$.
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The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$ is:
$\text{y}=\text{e}^{\text{x}-\text{y}}-\text{x}^2\text{e}^{-\text{y}}+\text{c}$
$\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{c}$
$\text{e}^{\text{x}}+\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$
$\text{e}^{\text{x}}-\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$