Indefinite Integration 1 Question 47
48. If $I=\sum _{k=1}^{98} \int _k^{k+1} \frac{k+1}{x(x+1)} d x$, then
(a) $I>\log _e 99$
(b) $I<\log _e 99$
(c) $I<\frac{49}{50}$
(d) $I>\frac{49}{50}$
(2017 Adv.)
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Solution:
- $I=\sum _{k=1}^{98} \int _k^{k+1} \frac{(k+1)}{x(x+1)} d x$
Clearly, $\quad I>\sum _{k=1}^{98} \int _k^{k+1} \frac{(k+1)}{(x+1)^{2}} d x$
$\Rightarrow \quad I>\sum _{k=1}^{98}(k+1) \int _k^{k+1} \frac{1}{(x+1)^{2}} d x$
$\Rightarrow \quad I>\sum _{k=1}^{98}(-(k+1)) \frac{1}{k+2}-\frac{1}{k+1} \Rightarrow I>\sum _{k=1}^{98} \frac{1}{k+2}$
$\Rightarrow \quad I>\frac{1}{3}+\ldots+\frac{1}{100}>\frac{98}{100} \Rightarrow I>\frac{49}{50}$
Also, $\quad I<\sum _{k=1}^{98} \int _k^{k+1} \frac{k+1}{x(k+1)} d x=\sum _{k=1}^{98}\left[\log _e(k+1)-\log _e k\right]$ $I<\log _e 99$
