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Question Number 146157 by henderson last updated on 11/Jul/21

Answered by gsk2684 last updated on 11/Jul/21

$$\underset{\frac{1}{e}}{\stackrel{\lambda }{\int }}\frac{1}{2}(1-\ln x)\frac{d(1-\ln x)}{-1}$$$$-\frac{1}{2}{\left[\frac{{(1-\ln x)}^2}{2}\right]}_{\frac{1}{e}}^{\lambda }$$$$-\frac{1}{4}[{(1-\ln \lambda )}^2-{(1-\ln \frac{1}{e})}^2]$$$$-\frac{1}{4}[{(1-\ln \lambda )}^2-4]$$

Answered by hknkrc46 last updated on 11/Jul/21

$$\mathrm{♣}{\int }_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}\frac{1-\ln \mathit{x}}{2\mathit{x}}\mathit{d}\mathit{x}={\int }_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}(\frac{1}{2\mathit{x}}-\frac{\ln \mathit{x}}{2\mathit{x}})\mathit{d}\mathit{x}$$$$={\int }_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}\frac{\mathit{d}\mathit{x}}{2\mathit{x}}-\frac{1}{2}{\int }_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}\frac{\ln \mathit{x}}{\mathit{x}}\mathit{d}\mathit{x}$$$$=\frac{1}{2}{\int }_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}\frac{\mathit{d}\mathit{x}}{\mathit{x}}-\frac{1}{2}{\int }_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}\ln \mathit{x}\cdot \frac{\mathit{d}\left(\ln \mathit{x}\right)}{\mathit{d}\mathit{x}}\mathit{d}\mathit{x}$$$$={\left(\frac{\ln \mathit{x}}{2}\right)}_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}-{\left(\frac{\ln {}^2\mathit{x}}{4}\right)}_{\frac{1}{\mathit{e}}}^{\mathit{\lambda }}=\frac{-\ln {}^2\mathit{\lambda }+2\ln \mathit{\lambda }+3}{4}$$

Answered by mathmax by abdo last updated on 11/Jul/21

$${\mathrm{A}}_{\lambda }={\int }_{\frac{1}{\mathrm{e}}}^{\lambda }\frac{1-\mathrm{logx}}{2\mathrm{x}}\mathrm{dx}\mathrm{changement}\mathrm{logx}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{e}}^{\mathrm{t}}\Rightarrow$$$${\mathrm{A}}_{\lambda }={\int }_{-1}^{\log \lambda }\frac{1-\mathrm{t}}{{2\mathrm{e}}^{\mathrm{t}}}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}={\int }_{-1}^{\log \lambda }(1-\mathrm{t})\mathrm{dt}=\frac{1}{2}{[\mathrm{t}-\frac{{\mathrm{t}}^2}{2}]}_{-1}^{\log \lambda }$$$$=\frac{1}{2}(\log \lambda -\frac{{\log }^2\lambda }{2}+1+\frac{1}{2})=\frac{1}{2}\log \lambda -\frac{1}{4}{\log }^2\lambda +\frac{3}{4}$$

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