SATHEE: Indefinite Integration 1 Question 47

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) $I < \log_{e} 99$

(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, $I > \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{(k + 1)}{(x + 1)^{2}} d x$

$\Rightarrow I > \sum_{k = 1}^{98} (k + 1) \int_{k}^{k + 1} \frac{1}{(x + 1)^{2}} d x$

$\Rightarrow I > \sum_{k = 1}^{98} (- (k + 1)) \frac{1}{k + 2} - \frac{1}{k + 1} \Rightarrow I > \sum_{k = 1}^{98} \frac{1}{k + 2}$

$\Rightarrow I > \frac{1}{3} + \dots + \frac{1}{100} > \frac{98}{100} \Rightarrow I > \frac{49}{50}$

Also, $I < \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{k + 1}{x (k + 1)} d x = \sum_{k = 1}^{98} [\log_{e} (k + 1) - \log_{e} k]$ $I < \log_{e} 99$

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Indefinite Integration 1 Question 47

48. If I=∑k=198∫kk+1k+1x(x+1)dx, then

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48. If $I = \sum_{k = 1}^{98} \int_{k}^{k + 1} \frac{k + 1}{x (x + 1)} d x$ , then