Question

$\underset{e}{\overset{e^2}{\int }}\log x.dx=$

• A. $e$
• B. $\frac{1}{e}$
• C. $e^2$
• D. $e^{-2}$

Answer 1

Answer: (C)

$e^2$

We have to find ${\int }_e^{e^2}logxdx$

Using integrating by parts

${\int }_e^{e^2}logx.1dx=logx{\int }_e^{e^2}1.dx-{\int }_e^{e^2}\frac{dlogx}{dx}\int 1.dx$

$\Rightarrow \left(xlogx\right)\frac{e_2}{e}-\left(x\right)\frac{e^2}{e}$

$\Rightarrow e^{2}log{e}^2-eloge-(e^2-e)$

$\Rightarrow 2e^2-e-e^2+e=e^2$