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Question Number 100956 by bobhans last updated on 29/Jun/20

Answered by john santu last updated on 29/Jun/20

$$\mathrm{set}\ln \left(\mathrm{x}\right)=\mathrm{z}\to \{\begin{array}{l}\mathrm{z}=1\\ \mathrm{z}=\mathrm{e}\end{array}$$$${\mathrm{I}=\underset{1}{\stackrel{\mathrm{e}}{\int }}\ln \left(\mathrm{z}\right).\mathrm{z}\mathrm{dz}=\frac{1}{2}{\mathrm{z}}^2.\ln \left(\mathrm{z}\right)]}_1^{\mathrm{e}}-\frac{1}{2}{\stackrel{\mathrm{e}}{\int }}_1\mathrm{z}\mathrm{dz}$$$$=\frac{{\mathrm{e}}^2}{2}-{\left[\frac{1}{4}{\mathrm{z}}^2\right]}_1^{\mathrm{e}}=\frac{{\mathrm{e}}^2}{2}-[\frac{{\mathrm{e}}^2}{4}-\frac{1}{4}]$$$$=\frac{{\mathrm{e}}^2+1}{4}.$$

Commented by bobhans last updated on 29/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 29/Jun/20

$$\mathrm{I}={\int }_{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{e}}}\frac{\ln \left(\mathrm{x}\right)\ln \left(\mathrm{lnx}\right)}{\mathrm{x}}\mathrm{dx}\mathrm{changement}\mathrm{lnx}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{e}}^{\mathrm{t}}\Rightarrow$$$$\mathrm{I}={\int }_1^{\mathrm{e}}\frac{\mathrm{tln}\left(\mathrm{t}\right)}{{\mathrm{e}}^{\mathrm{t}}}{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}={\int }_1^{\mathrm{e}}\mathrm{tln}\left(\mathrm{t}\right)\mathrm{dt}{=}_{\mathrm{by}\mathrm{parts}}{\left[\frac{{\mathrm{t}}^2}{2}\ln \left(\mathrm{t}\right)\right]}_1^{\mathrm{e}}+{\int }_1^{\mathrm{e}}\frac{{\mathrm{t}}^2}{2}\frac{\mathrm{dt}}{\mathrm{t}}$$$$=\frac{{\mathrm{e}}^2}{2}+\frac{1}{2}{\left[\frac{{\mathrm{t}}^2}{2}\right]}_1^{\mathrm{e}}=\frac{{\mathrm{e}}^2}{2}+\frac{1}{4}({\mathrm{e}}^2-1)=\frac{3}{4}{\mathrm{e}}^2-\frac{1}{4}$$

Commented by john santu last updated on 29/Jun/20

$$\mathrm{why}{\left[\frac{{\mathrm{t}}^2}{2}\ln \left(\mathrm{t}\right)\right]}_1^{\mathrm{e}}+{\int }_1^{\mathrm{e}}\frac{{\mathrm{t}}^2}{2}\frac{\mathrm{dt}}{\mathrm{t}}?$$$$$$

Commented by john santu last updated on 29/Jun/20

$$\mathrm{by}\mathrm{parts}\int \mathrm{u}\mathrm{dv}=\mathrm{u}.\mathrm{v}-\int \mathrm{v}\mathrm{du}?$$

Commented by mathmax by abdo last updated on 29/Jun/20

$$\mathrm{integration}\mathrm{by}\mathrm{parts}$$

Commented by mathmax by abdo last updated on 29/Jun/20

$$\mathrm{sorry}\mathrm{error}\mathrm{of}\mathrm{sign}\mathrm{I}={\left[\frac{{\mathrm{t}}^2}{2}\mathrm{lnt}\right]}_1^{\mathrm{e}}-{\int }_1^{\mathrm{e}}\frac{{\mathrm{t}}^2}{2\mathrm{t}}\mathrm{dt}$$$$=....=\frac{{\mathrm{e}}^2}{2}-\frac{1}{4}({\mathrm{e}}^2-1)=\frac{1}{4}{\mathrm{e}}^2+\frac{1}{4}=\frac{{\mathrm{e}}^2+1}{4}$$

Commented by bobhans last updated on 29/Jun/20

$$\mathrm{yes}\mathrm{sir}$$

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