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Question Number 88902 by M±th+et£s last updated on 13/Apr/20

$${\int }_{\frac{1}{e}}^{e}ln\mid x\mid dx$$

Commented by abdomathmax last updated on 13/Apr/20

$${\int }_{\frac{1}{e}}^{e}ln\mid x\mid dx={[xlnx-x]}_{\frac{1}{e}}^e=(e-e)-(-\frac{1}{e}-\frac{1}{e})$$$$=\frac{2}{e}$$

Commented by abdomathmax last updated on 13/Apr/20

$$anotherwaylet\lambda >0wehave\mid x\mid =x\Rightarrow$$$${\int }_{\frac{1}{\lambda }}^{\lambda }ln\left(x\right)dx={[xlnx-x]}_{\frac{1}{\lambda }}^{\lambda }=\lambda ln\lambda -\lambda -(\frac{1}{\lambda }ln\left(\frac{1}{\lambda }\right)-\frac{1}{\lambda })$$$$=\lambda ln\left(\lambda \right)-\lambda +\frac{ln\left(\lambda \right)}{\lambda }+\frac{1}{\lambda }$$$$\lambda =e\Rightarrow {\int }_{\frac{1}{e}}^{e}ln\left(x\right)dx=e-e+\frac{1}{e}+\frac{1}{e}=\frac{2}{e}$$

Commented by M±th+et£s last updated on 13/Apr/20

$$thankyousir$$

Commented by turbo msup by abdo last updated on 13/Apr/20

$$youarewelcome$$

Answered by TANMAY PANACEA. last updated on 13/Apr/20

$$t=\mid x\mid$$$$dt=dxsincex>0$$$${\int }_{\frac{1}{e}}^{e}lntdt$$$$=\mid tlnt-t{\mid }_{\frac{1}{e}}^e$$$$=elne-e-\frac{1}{e}ln\left(\frac{1}{e}\right)+\frac{1}{e}$$$$=-\frac{1}{e}\times -1+\frac{1}{e}=\frac{2}{e}$$$$$$$$$$

Commented by M±th+et£s last updated on 13/Apr/20

$$thankyousir$$

Commented by TANMAY PANACEA. last updated on 13/Apr/20

$$mostwelcomesir$$

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