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Question Number 90465 by Hassen_Timol last updated on 23/Apr/20

Commented by Hassen_Timol last updated on 23/Apr/20

$$\mathrm{Proof}\mathrm{for}\mathrm{the}\mathrm{last}\mathrm{line}$$$$\mathrm{Please}...\mathrm{its}\mathrm{quite}\mathrm{urgent},\mathrm{please}...$$

Commented by abdomathmax last updated on 23/Apr/20

$$U_n={\int }_0^1{x}^n{e}^{x^2}dx\Rightarrow U_n={\int }_0^1{x}^{n-1}\left({xe}^{x^2}\right)dx$$$$bypartsf=x^{n-1}and{g}^{{}^{\prime }}={xe}^{x^2}\Rightarrow$$$$U_n={\left[\frac{1}{2}{x}^{n-1}{e}^{x^2}\right]}_0^1-\frac{1}{2}{\int }_0^1(n-1)x^{n-2}{e}^{x^2}dx$$$$=\frac{1}{2}\left(e\right)-\frac{n-1}{2}{\int }_0^1{x}^{n-2}{e}^{x^2}dx$$$$=\frac{e}{2}-\frac{n-1}{2}{U}_{n-2}\Rightarrow$$$$U_{n+2}=\frac{e}{2}-\frac{n+1}{2}{U}_{n}therelationisproved.$$

Commented by Hassen_Timol last updated on 24/Apr/20

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{very}\mathrm{nice}...$$$$\mathrm{Be}\mathrm{blessed}$$

Commented by Hassen_Timol last updated on 24/Apr/20

$$\mathrm{But}\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{pass}\mathrm{from}2\mathrm{nd}\mathrm{to}3\mathrm{rd}\mathrm{line}\mathrm{please}?$$

Commented by mathmax by abdo last updated on 24/Apr/20

$$integrationbyparts$$

Commented by Hassen_Timol last updated on 24/Apr/20

$$\mathrm{okay}\mathrm{thank}\mathrm{you}$$

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