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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
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
    $\pi /2$
✓

Answer

Correct option: A.
$\frac{1}{2}\log 2$
a
(a) Let $I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\,\frac{k}{{{n^2} + {k^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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