Question Number 40136 by maxmathsup by imad last updated on 16/Jul/18
$${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:{e}^{−{x}} {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{\mathrm{1}} \:\:\:{and}\:{A}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:\:{A}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){A}_{{n}} \:−\frac{\mathrm{1}}{{e}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{A}_{\mathrm{3}} \:\:\:,\:{A}_{\mathrm{4}} ,\:{and}\:{A}_{\mathrm{5}} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{x}^{\mathrm{3}} \:+\mathrm{2}{x}^{\mathrm{2}\:} \:−{x}\right){e}^{−{x}} \:{dx} \\ $$
Answered by math khazana by abdo last updated on 21/Jul/18
![1) A_1 = ∫_0 ^1 x e^(−x) dx =[−xe^(−x) ]_0 ^1 +∫_0 ^1 e^(−x) dx =−e^(−1) +[−e^(−x) ]_0 ^1 =−e^(−1) +1−e^(−1) =1−2e^(−1) A_1 =1−(2/e) A_2 = ∫_0 ^1 x^2 e^(−x) dx =[−x^2 e^(−x) ]_0 ^1 +∫_0 ^1 2x e^(−x) dx =−e^(−1) +2A_1 =−e^(−1) +2(1−(2/e) )=−e^(−1) +2−(4/e) A_2 =2−(5/e) 2) by parts u^′ =x^n and v=e^(−x) ⇒ A_n =[(1/(n+1))x^(n+1) e^(−x) ]_0 ^1 +∫_0 ^1 (x^(n+1) /(n+1)) e^(−x) dx = (1/((n+1)e)) +(1/(n+1)) A_(n+1) ⇒ (n+1)A_n = (1/e) +A_(n+1) ⇒ A_(n+1) =(n+1)A_n −(1/e) 3) A_3 =3A_2 −(1/e) =3(2−(5/e))−(1/e) A_3 = 6−((16)/e) A_4 =4A_3 −(1/e) =4{6−((16)/e)}−(1/e) A_4 =24 −((65)/e) A_5 = 5A_4 −(1/e) =5{24−((65)/e)}−(1/e) =120 −((5.65+1)/e) =120−((326)/e)](https://www.tinkutara.com/question/Q40377.png)
$$\left.\mathrm{1}\right)\:{A}_{\mathrm{1}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{e}^{−{x}} {dx}\:=\left[−{xe}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{−{x}} {dx} \\ $$$$=−{e}^{−\mathrm{1}} \:\:+\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} =−{e}^{−\mathrm{1}} \:+\mathrm{1}−{e}^{−\mathrm{1}} =\mathrm{1}−\mathrm{2}{e}^{−\mathrm{1}} \\ $$$${A}_{\mathrm{1}} =\mathrm{1}−\frac{\mathrm{2}}{{e}} \\ $$$${A}_{\mathrm{2}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}} \:{e}^{−{x}} {dx}\:=\left[−{x}^{\mathrm{2}} {e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{2}{x}\:{e}^{−{x}} {dx} \\ $$$$=−{e}^{−\mathrm{1}} \:+\mathrm{2}{A}_{\mathrm{1}} =−{e}^{−\mathrm{1}} \:+\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{2}}{{e}}\:\right)=−{e}^{−\mathrm{1}} +\mathrm{2}−\frac{\mathrm{4}}{{e}} \\ $$$${A}_{\mathrm{2}} =\mathrm{2}−\frac{\mathrm{5}}{{e}} \\ $$$$\left.\mathrm{2}\right)\:{by}\:{parts}\:{u}^{'} ={x}^{{n}} \:\:\:{and}\:{v}={e}^{−{x}} \:\Rightarrow \\ $$$${A}_{{n}} =\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:{e}^{−{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:{e}^{−{x}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){e}}\:\:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{A}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\left({n}+\mathrm{1}\right){A}_{{n}} =\:\frac{\mathrm{1}}{{e}}\:+{A}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$${A}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){A}_{{n}} \:−\frac{\mathrm{1}}{{e}} \\ $$$$\left.\mathrm{3}\right)\:{A}_{\mathrm{3}} =\mathrm{3}{A}_{\mathrm{2}} \:−\frac{\mathrm{1}}{{e}}\:=\mathrm{3}\left(\mathrm{2}−\frac{\mathrm{5}}{{e}}\right)−\frac{\mathrm{1}}{{e}} \\ $$$${A}_{\mathrm{3}} =\:\mathrm{6}−\frac{\mathrm{16}}{{e}} \\ $$$${A}_{\mathrm{4}} =\mathrm{4}{A}_{\mathrm{3}} \:−\frac{\mathrm{1}}{{e}}\:=\mathrm{4}\left\{\mathrm{6}−\frac{\mathrm{16}}{{e}}\right\}−\frac{\mathrm{1}}{{e}} \\ $$$${A}_{\mathrm{4}} \:=\mathrm{24}\:−\frac{\mathrm{65}}{{e}} \\ $$$${A}_{\mathrm{5}} =\:\mathrm{5}{A}_{\mathrm{4}} \:−\frac{\mathrm{1}}{{e}}\:=\mathrm{5}\left\{\mathrm{24}−\frac{\mathrm{65}}{{e}}\right\}−\frac{\mathrm{1}}{{e}} \\ $$$$=\mathrm{120}\:−\frac{\mathrm{5}.\mathrm{65}+\mathrm{1}}{{e}}\:=\mathrm{120}−\frac{\mathrm{326}}{{e}} \\ $$