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  • The integral  <img alt="\mathrm{\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} d x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B

The integral  \mathrm{\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} d x}, where \mathrm{\left [ \cdot \right ]} denotes the greater integer function, is equal to

Option: 1

\mathrm{1+6\log_{e}\left ( \frac{6}{7} \right )}


Option: 2

\mathrm{1-6\log_{e}\left ( \frac{6}{7} \right )}


Option: 3

\mathrm{\log_{e}\left ( \frac{7}{6} \right )}


Option: 4

\mathrm{1-7\log_{e}\left ( \frac{6}{7} \right )}


Answers (1)

best_answer

\mathrm{I =\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} d x} \\

\mathrm{= \int_{1 / 2}^{1 } \frac{1}{7} d x+\int_{1 / 3}^{1 / 2} \frac{1}{7^{2}} d x+\int_{1 / 4}^{1 / 3} \frac{1}{7^{3}} d x+\int_{1 / 5}^{7^{4}} \frac{1}{7^{4}} d x+\cdots} \\

\mathrm{=\left(1-\frac{1}{2}\right)\frac{1}{7}+\left(\frac{1}{2}-\frac{1}{3}\right) \frac{1}{7^{2}}+\left(\frac{1}{3}-\frac{1}{4}\right) \frac{1}{7^{3}}+\left(\frac{1}{4}-\frac{1}{5}\right) \frac{1}{7^{4}}+\cdots}

= \mathrm{\left(\frac{1}{7}+\frac{1}{2} \cdot \frac{1}{7^{2}}+\frac{1}{3} \cdot \frac{1}{7^{3}}+\frac{1}{4} \cdot \frac{1}{7^{4}}+\cdots\right)}\\

\mathrm{-\left(\frac{1}{2} \cdot \frac{1}{7}+\frac{1}{3} \cdot \frac{1}{7^{2}}+\frac{1}{4} \cdot \frac{1}{7^{3}}+\frac{1}{5} \cdot \frac{1}{7^{4}}+\cdots\right)}\\

\mathrm{=\log _{e}\left(1-\frac{1}{7}\right)-7\left(\frac{-1}{7} - \log _{e}\left(1-\frac{1}{7}\right)\right)}\\

\mathrm{=1+6 \log _{e} \frac{6}{7}}

Hence the correct answer is option 1.

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mansi

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