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 =∫_0 ^1 x^n e^x^2 dx ⇒U_n =∫_0 ^1 x^(n−1) (xe^x^2 )dx by parts f=x^(n−1) and g^′ =xe^(x^2 ) ⇒ U_n =[(1/2)x^(n−1) e^x^2 ]_0 ^1 −(1/2)∫_0 ^1 (n−1)x^(n−2) e^x^2 dx =(1/2)(e)−((n−1)/2) ∫_0 ^1 x^(n−2) e^x^2 dx =(e/2)−((n−1)/2) U_(n−2) ⇒ U_(n+2) =(e/2)−((n+1)/2) U_n the relation is proved.](Q90482.png)
$${U}_{{n}} \:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{{x}^{\mathrm{2}} } {dx}\:\:\Rightarrow{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} \:\left({xe}^{{x}^{\mathrm{2}} } \right){dx} \\ $$$${by}\:{parts}\:{f}={x}^{{n}−\mathrm{1}} \:{and}\:{g}^{'} ={xe}^{{x}^{\mathrm{2}} \:} \:\Rightarrow \\ $$$${U}_{{n}} =\left[\frac{\mathrm{1}}{\mathrm{2}}{x}^{{n}−\mathrm{1}} \:{e}^{{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} \:{e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({e}\right)−\frac{{n}−\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{2}} \:{e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$=\frac{{e}}{\mathrm{2}}−\frac{{n}−\mathrm{1}}{\mathrm{2}}\:{U}_{{n}−\mathrm{2}} \:\Rightarrow \\ $$$${U}_{{n}+\mathrm{2}} =\frac{{e}}{\mathrm{2}}−\frac{{n}+\mathrm{1}}{\mathrm{2}}\:{U}_{{n}} \:\:\:{the}\:{relation}\:{is}\:{proved}. \\ $$
Commented by Hassen_Timol last updated on 24/Apr/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much},\:\mathrm{it}'\mathrm{s}\:\mathrm{very}\:\mathrm{nice}... \\ $$$$\mathrm{Be}\:\mathrm{blessed} \\ $$
Commented by Hassen_Timol last updated on 24/Apr/20
$$\mathrm{But}\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{pass}\:\mathrm{from}\:\mathrm{2nd}\:\mathrm{to}\:\mathrm{3rd}\:\mathrm{line}\:\mathrm{please}\:? \\ $$
Commented by mathmax by abdo last updated on 24/Apr/20
$${integration}\:{by}\:{parts} \\ $$
Commented by Hassen_Timol last updated on 24/Apr/20
$$\mathrm{okay}\:\mathrm{thank}\:\mathrm{you} \\ $$