Question Number 66330 by mathmax by abdo last updated on 12/Aug/19
$${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{−{x}} \:{dx}\:\:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}_{\mathrm{0}} \:,\:{I}_{\mathrm{1}} \:{and}\:{I}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){find}\:{arelation}\:{between}\:{I}_{{n}} \:{and}\:{I}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{I}_{{n}} \:{interms}\:{of}\:{n}. \\ $$
Commented by mathmax by abdo last updated on 18/Aug/19
![1)I_n =∫_0 ^1 x^n e^(−x) dx ⇒I_0 =∫_0 ^1 e^(−x) dx =[−e^(−x) ]_0 ^1 =1−e^(−1) I_1 =∫_0 ^1 x e^(−x) dx =_(by parts) [−xe^(−x) ]_0 ^1 +∫_0 ^1 e^(−x) dx=−e^(−1) + 1−e^(−1) =1−2e^(−1) I_2 =∫_0 ^1 x^2 e^(−x) dx =_(by psrts) [−x^2 e^(−x) ]_0 ^1 +∫_0 ^1 2x e^(−x) dx =−e^(−1) +2(1−2e^(−1) ) =2−3e^(−1) 2) by parts u =x^n and v^′ =e^(−x) ⇒I_n =[−x^n e^(−x) ]_0 ^1 +∫_0 ^1 nx^(n−1) e^(−x) dx =−e^(−1) +n∫_0 ^1 x^(n−1) e^(−x) dx =n I_(n−1) −(1/e) ⇒I_n =nI_(n−1) −(1/e) 3)let V_n =(I_n /(n!)) we have V_(n+1) −V_n =(I_(n+1) /((n+1)!))−(I_n /(n!)) =(((n+1)I_n −(1/e))/((n+1)!)) −(I_n /(n!)) =(I_n /(n!))−(1/(e(n+1)!))−(I_n /(n!)) =−(1/(e(n+1)!)) ⇒ Σ_(k=0) ^(n−1) (V_(k+1) −V_k ) =−(1/e)Σ_(k=0) ^(n−1) (1/((k+1)!)) =−(1/e)Σ_(k=1) ^n (1/(k!)) ⇒ V_n −V_0 =−(1/e)Σ_(k=1) ^n (1/(k!)) but V_0 =I_0 =1−(1/e) ⇒ V_n =−(1/e)(Σ_(k=1) ^n (1/(k!))+1)+1 =1−(1/e)Σ_(k=0) ^n (1/(k!)) ⇒ I_n =n!{1−(1/e)Σ_(k=0) ^n (1/(k!))} (n≥1)](Q66686.png)
$$\left.\mathrm{1}\right){I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{−{x}} \:{dx}\:\Rightarrow{I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {dx}\:=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1}−{e}^{−\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{e}^{−{x}} {dx}\:=_{{by}\:{parts}} \left[−{xe}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}} {dx}=−{e}^{−\mathrm{1}} +\:\mathrm{1}−{e}^{−\mathrm{1}} \\ $$$$=\mathrm{1}−\mathrm{2}{e}^{−\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}} \:{e}^{−{x}} {dx}\:=_{{by}\:{psrts}} \:\:\left[−{x}^{\mathrm{2}} {e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{x}\:{e}^{−{x}} {dx} \\ $$$$=−{e}^{−\mathrm{1}} \:+\mathrm{2}\left(\mathrm{1}−\mathrm{2}{e}^{−\mathrm{1}} \right)\:=\mathrm{2}−\mathrm{3}{e}^{−\mathrm{1}} \: \\ $$$$\left.\mathrm{2}\right)\:{by}\:{parts}\:{u}\:={x}^{{n}} \:{and}\:{v}^{'} \:={e}^{−{x}} \:\Rightarrow{I}_{{n}} =\left[−{x}^{{n}} \:{e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}−\mathrm{1}} \:{e}^{−{x}} {dx} \\ $$$$=−{e}^{−\mathrm{1}} \:+{n}\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} \:{e}^{−{x}} {dx}\:={n}\:{I}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{{e}}\:\Rightarrow{I}_{{n}} ={nI}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{{e}} \\ $$$$\left.\mathrm{3}\right){let}\:{V}_{{n}} =\frac{{I}_{{n}} }{{n}!}\:\:{we}\:{have}\:{V}_{{n}+\mathrm{1}} −{V}_{{n}} =\frac{{I}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}−\frac{{I}_{{n}} }{{n}!} \\ $$$$=\frac{\left({n}+\mathrm{1}\right){I}_{{n}} −\frac{\mathrm{1}}{{e}}}{\left({n}+\mathrm{1}\right)!}\:−\frac{{I}_{{n}} }{{n}!}\:=\frac{{I}_{{n}} }{{n}!}−\frac{\mathrm{1}}{{e}\left({n}+\mathrm{1}\right)!}−\frac{{I}_{{n}} }{{n}!}\:=−\frac{\mathrm{1}}{{e}\left({n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({V}_{{k}+\mathrm{1}} −{V}_{{k}} \right)\:=−\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\:=−\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\Rightarrow \\ $$$${V}_{{n}} −{V}_{\mathrm{0}} =−\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\:{but}\:{V}_{\mathrm{0}} ={I}_{\mathrm{0}} =\mathrm{1}−\frac{\mathrm{1}}{{e}}\:\Rightarrow \\ $$$${V}_{{n}} =−\frac{\mathrm{1}}{{e}}\left(\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}+\mathrm{1}\right)+\mathrm{1}\:=\mathrm{1}−\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\Rightarrow \\ $$$${I}_{{n}} ={n}!\left\{\mathrm{1}−\frac{\mathrm{1}}{{e}}\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\right\}\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$