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let-A-n-0-1-x-n-e-x-dx-1-calculate-A-1-and-A-2-2-prove-that-A-n-1-n-1-A-n-1-e-3-calculate-A-3-A-4-and-A-5-4-calculate-I-0-1-x-3-2x-2-x-e-x-dx-

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Question Number 40136 by maxmathsup by imad last updated on 16/Jul/18
let A_n = ∫_0 ^1 x^n e^(−x) dx 1) calculate A_1 and A_2 2) prove that A_(n+1) =(n+1)A_n −(1/e) 3) calculate A_3 , A_4 , and A_5 4) calculate I = ∫_0 ^1 (−x^3 +2x^(2 ) −x)e^(−x) dx
letAn=∫01xne−xdx1)calculateA1andA22)provethatAn+1=(n+1)An−1e3)calculateA3,A4,andA54)calculateI=∫01(−x3+2x2−x)e−xdx
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)
1)A1=∫01xe−xdx=[−xe−x]01+∫01e−xdx=−e−1+[−e−x]01=−e−1+1−e−1=1−2e−1A1=1−2eA2=∫01x2e−xdx=[−x2e−x]01+∫012xe−xdx=−e−1+2A1=−e−1+2(1−2e)=−e−1+2−4eA2=2−5e2)bypartsu′=xnandv=e−x⇒An=[1n+1xn+1e−x]01+∫01xn+1n+1e−xdx=1(n+1)e+1n+1An+1⇒(n+1)An=1e+An+1⇒An+1=(n+1)An−1e3)A3=3A2−1e=3(2−5e)−1eA3=6−16eA4=4A3−1e=4{6−16e}−1eA4=24−65eA5=5A4−1e=5{24−65e}−1e=120−5.65+1e=120−326e

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