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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
$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$  is equal to
  • ✓
    $\frac{\pi}{8}+\frac{1}{4} \log _{ e } 2$
  • B
    $\frac{\pi}{4}+\frac{1}{8} \log _{ e } 2$
  • C
    $\frac{\pi}{4}-\frac{1}{8} \log _{ e } 2$
  • D
    $\frac{\pi}{8}+\log _{ e } \sqrt{2}$
✓

Answer

Correct option: A.
$\frac{\pi}{8}+\frac{1}{4} \log _{ e } 2$
a
$\lim _{n \rightarrow \infty}\left(\sum \limits_{r=1}^{n} \frac{n^{2}}{\left(n^{2}+r^{2}\right)(n+r)}\right)$

$=\lim _{n \rightarrow \infty}\left(\sum \limits_{r=1}^{n} \frac{1}{n\left(1+\left(\frac{r}{n}\right)^{2}\right)\left(1+\left(\frac{r}{n}\right)\right)}\right)$

$=\int_{0}^{1} \frac{d x}{\left(1+x^{2}\right)(1+x)}=\frac{1}{2} \int_{0}^{1} \frac{1-x}{1+x^{2}} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{1+x} d x$

$=\frac{1}{2} \int\left(\frac{1}{1+x^{2}}-\frac{x}{1+x^{2}}\right) d x+\frac{1}{2}(\ln (1+x))_{0}^{1}$

$=\frac{1}{2}\left[\tan ^{-1} x -\frac{1}{2} \ln \left(1+ x ^{2}\right)\right]_{0}^{1}+\frac{1}{2} \ell \operatorname{nn} 2$

$=\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{2} \ell \operatorname{nn} 2\right]+\frac{1}{2} \ln 2$

$=\frac{\pi}{8}+\frac{1}{4} \ln 2$

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