Definite Integration Question 47
Question 47
- 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, $\quad I<\sum_{k=1}^{98} \int_{k}^{k+1} \frac{k+1}{x(k+1)} d x=\sum_{k=1}^{98}\left[\log _{e}(k+1)-\log _{e} k\right]$ $I<\log _{e} 99$
