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Question Number 143702 by henderson last updated on 17/Jun/21 | ||
$${n}\:\in\:\mathrm{IN}. \\ $$$${I}_{{n}} \:=\:\int_{\mathrm{1}} ^{\:\mathrm{e}} {x}^{{n}+\mathrm{1}} {lnx}\:{dx}. \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\left(\boldsymbol{{I}}_{\boldsymbol{{n}}} \right)\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{increasing}}. \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{part}}−\boldsymbol{\mathrm{by}}−\boldsymbol{\mathrm{part}}\:\boldsymbol{\mathrm{integration}},\:\boldsymbol{\mathrm{calculate}}\:\boldsymbol{{I}}_{\boldsymbol{{n}}} . \\ $$ | ||
Answered by mindispower last updated on 17/Jun/21 | ||
$${ln}\left({x}\right)\geqslant\mathrm{0},\forall{x}\geqslant\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{{e}} {x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\geqslant\int_{\mathrm{1}} ^{{e}} \mathrm{0}.{dx}=\mathrm{0} \\ $$$$\mathrm{2}\:{I}_{{n}} =\left[\frac{{x}^{{n}+\mathrm{2}} }{{n}+\mathrm{2}}{ln}\left({x}\right)\right]_{\mathrm{1}} ^{{e}} −\frac{\mathrm{1}}{{n}+\mathrm{2}}\int_{\mathrm{1}} ^{{e}} \frac{{x}^{{n}+\mathrm{2}} }{{x}}{dx} \\ $$$$\frac{{e}^{{n}+\mathrm{2}} }{{n}+\mathrm{2}}−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }\left[{x}^{{n}+\mathrm{2}} \right]_{\mathrm{1}} ^{{e}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right){e}^{{n}+\mathrm{2}} +\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$ \\ $$ | ||
Answered by mathmax by abdo last updated on 17/Jun/21 | ||
$$\left.\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{e}\:\Rightarrow\mathrm{x}^{\mathrm{n}+\mathrm{1}} \mathrm{logx}\geqslant\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{I}_{\mathrm{n}+\mathrm{1}} −\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{1}} ^{\mathrm{e}} \:\left(\mathrm{x}^{\mathrm{n}+\mathrm{2}} −\mathrm{x}^{\mathrm{n}+\mathrm{1}} \right)\mathrm{logx}\:\mathrm{dx}\:=\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{x}^{\mathrm{n}+\mathrm{1}} \left(\mathrm{x}−\mathrm{1}\right)\mathrm{logx}\:\mathrm{dx}\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} \mathrm{is}\:\mathrm{increazing} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{by}\:\mathrm{parts}\:\mathrm{I}_{\mathrm{n}} =\left[\frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\mathrm{logx}\right]_{\mathrm{1}} ^{\mathrm{e}} −\int_{\mathrm{1}} ^{\mathrm{e}} \:\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{2}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\mathrm{x}^{\mathrm{n}+\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{e}} \:=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }\left(\mathrm{e}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{n}+\mathrm{2}−\left(\mathrm{e}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }\:=\frac{\mathrm{n}+\mathrm{3}−\mathrm{e}^{\mathrm{n}+\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$ | ||