Question Number 146157 by henderson last updated on 11/Jul/21
Answered by gsk2684 last updated on 11/Jul/21
![∫_(1/e) ^λ (1/2)(1−ln x)((d(1−ln x))/(−1)) −(1/2)[(((1−ln x)^2 )/2)]_(1/e) ^λ −(1/4)[(1−ln λ)^2 −(1−ln (1/e))^2 ] −(1/4)[(1−ln λ)^2 −4]](https://www.tinkutara.com/question/Q146159.png)
$$\underset{\frac{\mathrm{1}}{{e}}} {\overset{\lambda} {\int}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{ln}\:{x}\right)\frac{{d}\left(\mathrm{1}−\mathrm{ln}\:{x}\right)}{−\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\mathrm{1}−\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right]_{\frac{\mathrm{1}}{{e}}} ^{\lambda} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\mathrm{1}−\mathrm{ln}\:\lambda\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{ln}\:\frac{\mathrm{1}}{{e}}\right)^{\mathrm{2}} \right] \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\mathrm{1}−\mathrm{ln}\:\lambda\right)^{\mathrm{2}} −\mathrm{4}\right] \\ $$
Answered by hknkrc46 last updated on 11/Jul/21
$$\clubsuit\:\int_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\:\boldsymbol{\lambda}} \:\frac{\mathrm{1}\:−\:\mathrm{ln}\:\boldsymbol{{x}}}{\mathrm{2}\boldsymbol{{x}}}\:\boldsymbol{{dx}}\:=\:\int_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\:\boldsymbol{\lambda}} \:\left(\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{x}}}\:−\:\frac{\mathrm{ln}\:\boldsymbol{{x}}}{\mathrm{2}\boldsymbol{{x}}}\right)\boldsymbol{{dx}} \\ $$$$=\:\int_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\:\boldsymbol{\lambda}} \:\frac{\boldsymbol{{dx}}}{\mathrm{2}\boldsymbol{{x}}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\:\boldsymbol{\lambda}} \:\frac{\mathrm{ln}\:\boldsymbol{{x}}}{\boldsymbol{{x}}}\:\boldsymbol{{dx}}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\:\boldsymbol{\lambda}} \:\frac{\boldsymbol{{dx}}}{\boldsymbol{{x}}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\:\boldsymbol{\lambda}} \:\mathrm{ln}\:\boldsymbol{{x}}\:\centerdot\:\frac{\boldsymbol{{d}}\left(\mathrm{ln}\:\boldsymbol{{x}}\right)}{\boldsymbol{{dx}}}\:\boldsymbol{{dx}} \\ $$$$=\:\left(\frac{\mathrm{ln}\:\boldsymbol{{x}}}{\mathrm{2}}\right)_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\boldsymbol{\lambda}} −\left(\frac{\mathrm{ln}\:^{\mathrm{2}} \boldsymbol{{x}}}{\mathrm{4}}\right)_{\frac{\mathrm{1}}{\boldsymbol{{e}}}} ^{\boldsymbol{\lambda}} \:=\:\frac{−\mathrm{ln}\:^{\mathrm{2}} \boldsymbol{\lambda}\:+\:\mathrm{2ln}\:\boldsymbol{\lambda}\:+\:\mathrm{3}}{\mathrm{4}} \\ $$
Answered by mathmax by abdo last updated on 11/Jul/21
![A_λ =∫_(1/e) ^λ ((1−logx)/(2x))dx changement logx=t give x=e^t ⇒ A_λ =∫_(−1) ^(logλ) ((1−t)/(2e^t ))e^t dt =∫_(−1) ^(logλ) (1−t)dt=(1/2)[t−(t^2 /2)]_(−1) ^(logλ) =(1/2)(logλ−((log^2 λ)/2) +1+(1/2))=(1/2)logλ−(1/4)log^2 λ+(3/4)](https://www.tinkutara.com/question/Q146163.png)
$$\mathrm{A}_{\lambda} =\int_{\frac{\mathrm{1}}{\mathrm{e}}} ^{\lambda} \:\frac{\mathrm{1}−\mathrm{logx}}{\mathrm{2x}}\mathrm{dx}\:\mathrm{changement}\:\mathrm{logx}=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{e}^{\mathrm{t}} \:\Rightarrow \\ $$$$\mathrm{A}_{\lambda} =\int_{−\mathrm{1}} ^{\mathrm{log}\lambda} \:\frac{\mathrm{1}−\mathrm{t}}{\mathrm{2e}^{\mathrm{t}} }\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:=\int_{−\mathrm{1}} ^{\mathrm{log}\lambda} \left(\mathrm{1}−\mathrm{t}\right)\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{t}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{1}} ^{\mathrm{log}\lambda} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\lambda−\frac{\mathrm{log}^{\mathrm{2}} \lambda}{\mathrm{2}}\:+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\lambda−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}^{\mathrm{2}} \lambda+\frac{\mathrm{3}}{\mathrm{4}} \\ $$