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Question Number 143702 by henderson last updated on 17/Jun/21

$$n\in \mathrm{IN}.$$$$I_n={\int }_1^{\mathrm{e}}{x}^{n+1}lnxdx.$$$$1.\mathbf{prove}\mathbf{that}\left({\mathit{I}}_{\mathit{n}}\right)\mathbf{is}\mathbf{positive}\mathbf{and}\mathbf{increasing}.$$$$2.\mathbf{using}\mathbf{a}\mathbf{part}-\mathbf{by}-\mathbf{part}\mathbf{integration},\mathbf{calculate}{\mathit{I}}_{\mathit{n}}.$$

Answered by mindispower last updated on 17/Jun/21

$$ln\left(x\right)⩾0,\mathrm{\forall }x⩾1$$$$\Rightarrow {\int }_1^e{x}^{n+1}ln\left(x\right)⩾{\int }_1^{e}0.dx=0$$$$2I_n={\left[\frac{x^{n+2}}{n+2}ln\left(x\right)\right]}_1^e-\frac{1}{n+2}{\int }_1^e\frac{x^{n+2}}{x}dx$$$$\frac{e^{n+2}}{n+2}-\frac{1}{{(n+2)}^2}{\left[x^{n+2}\right]}_1^e$$$$=\frac{(n+1)e^{n+2}+1}{{(n+2)}^2}$$$$$$

Answered by mathmax by abdo last updated on 17/Jun/21

$$1)\mathrm{we}\mathrm{have}1⩽\mathrm{x}⩽\mathrm{e}\Rightarrow {\mathrm{x}}^{\mathrm{n}+1}\mathrm{logx}⩾0\mathrm{and}$$$${\mathrm{I}}_{\mathrm{n}+1}-{\mathrm{I}}_{\mathrm{n}}={\int }_1^{\mathrm{e}}({\mathrm{x}}^{\mathrm{n}+2}-{\mathrm{x}}^{\mathrm{n}+1})\mathrm{logx}\mathrm{dx}={\int }_1^{\mathrm{e}}{\mathrm{x}}^{\mathrm{n}+1}(\mathrm{x}-1)\mathrm{logx}\mathrm{dx}⩾0\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}\mathrm{is}\mathrm{increazing}$$$$2)\mathrm{by}\mathrm{parts}{\mathrm{I}}_{\mathrm{n}}={\left[\frac{{\mathrm{x}}^{\mathrm{n}+2}}{\mathrm{n}+2}\mathrm{logx}\right]}_1^{\mathrm{e}}-{\int }_1^{\mathrm{e}}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+2}\mathrm{dx}$$$$=\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+2}{\left[\frac{1}{\mathrm{n}+2}{\mathrm{x}}^{\mathrm{n}+2}\right]}_1^{\mathrm{e}}=\frac{1}{\mathrm{n}+2}-\frac{1}{{(\mathrm{n}+2)}^2}({\mathrm{e}}^{\mathrm{n}+2}-1)$$$$=\frac{\mathrm{n}+2-({\mathrm{e}}^{\mathrm{n}+2}-1)}{{(\mathrm{n}+2)}^2}=\frac{\mathrm{n}+3-{\mathrm{e}}^{\mathrm{n}+2}}{{(\mathrm{n}+2)}^2}$$

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