Consider $ \mathrm{I}(\alpha)=\int_{\alpha}^{\alpha^{2}} \frac{d x}{x} $ (where $ \alpha0 $ ), then the value of $ \sum_{r=2}^{5} \mathrm{I}(r)+\sum_{k=2}^{5} \mathrm{I}\left(\frac{1}{k}\right) $ is

Question

Consider $ \mathrm{I}(\alpha)=\int_{\alpha}^{\alpha^{2}} \frac{d x}{x} $ (where $ \alpha>0 $ ), then the value of $ \sum_{r=2}^{5} \mathrm{I}(r)+\sum_{k=2}^{5} \mathrm{I}\left(\frac{1}{k}\right) $ is, $ 0 $, $ 1 $, $ \ln 2 $, $ \ln 4 $

MARKS App · Accepted Answer

Ι α = ln ⁡ x α α 2 = ln ⁡ α 2 - ln ⁡ α = ln ⁡ α 2 α = ln ⁡ α Thus, ∑ r = 2 5 Ι r = ln ⁡ 2 + ln ⁡ 3 + ln ⁡ 4 + ln ⁡ 5 and ∑ r = 2 5 Ι 1 k = ln ⁡ 1 2 + ln ⁡ 1 3 + ln ⁡ 1 4 + ln ⁡ 1 5 = - ln ⁡ 2 + ln ⁡ 3 + ln ⁡ 4 + ln ⁡ 5 Hence, their sum equals to zero

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