Mathematics

# $\displaystyle \overset{e^2}{\underset{1}{\int}} [log_e \,x]dx, x > 0$ and $[\cdot]$ is greatest integer function, is equal to

$e^2 - e$

##### SOLUTION
Breaking limits in 1 to e and e to $e^2$

$\displaystyle \int_1^{e^2} [\log_ex] . \Rightarrow \int_1^e [\log_ex ] d_2t \int_e^{e^2} [\log_ex] dx$

Now, $\log_ex$ for $x = 1 \Rightarrow \log = 0$

$\log_e^x$ for $x = e \Rightarrow \log_e^e = 1 \,\, \log_e^x$ for $x = e^2 \Rightarrow \log_e^{e^2} = 2 \log_e^{e = 2}$

$\therefore$ for $(1 to e) \log_ex$ will be between $(0, 1)$

$\therefore [\log_ex] = 0$

and for $(e \, to \, e^2), \, \log_ex$ will be between $(1, 2)$

$\therefore [\log_e x] = 1$

$\therefore \displaystyle \int_1^{e^2} [\log_ex] = 0 + \int_e^{e^2} . 1 dx = x|^{e^2}_e$

$\therefore \displaystyle \int_1^{e^2} [\log_ex] = e^2 - e$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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