Question

$\stackrel{e^2}{\underset{1}{\int }}\left[lo{g}_{e}x\right]dx,x>0$ and $[\cdot ]$ is greatest integer function, is equal to

• A. $e^2-e$
• B. $e^2-1$
• C. $e-1$
• D. $e^2+1$

Answer 1

Answer: (A)

$e^2-e$

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

${\int }_1^{e^2}\left[{\log }_{e}x\right].\Rightarrow {\int }_1^e\left[{\log }_{e}x\right]d_{2}t{\int }_e^{e^2}\left[{\log }_{e}x\right]dx$

Now, ${\log }_{e}x$ 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 $\left(1toe\right){\log }_{e}x$ will be between $(0,1)$

$\therefore \left[{\log }_{e}x\right]=0$

and for $\left(eto{e}^2\right),{\log }_{e}x$ will be between $(1,2)$

$\therefore \left[{\log }_{e}x\right]=1$

$\therefore {\int }_1^{e^2}\left[{\log }_{e}x\right]=0+{\int }_e^{e^2}.1dx=x{|}_e^{e^2}$

$\therefore {\int }_1^{e^2}\left[{\log }_{e}x\right]=e^2-e$