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 = ∫


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Question Number 40136 by maxmathsup by imad last updated on 16/Jul/18

$l e t A_{n} = \int_{0}^{1} x^{n} e^{- x} d x$ $1) c a l c u l a t e A_{1} a n d A_{2}$ $2) p r o v e t h a t A_{n + 1} = (n + 1) A_{n} - \frac{1}{e}$ $3) c a l c u l a t e A_{3}, A_{4}, a n d A_{5}$ $4) c a l c u l a t e I = \int_{0}^{1} (- x^{3} + 2 x^{2} - x) e^{- x} d x$
	Answered by math khazana by abdo last updated on 21/Jul/18
	$1) A_1 = ∫_0 ^1 x e^(−x) dx =[−xe^(−x) ]_0 ^1 +∫_0 ^1 e^(−x) dx =−e^(−1) +[−e^(−x) ]_0 ^1 =−e^(−1) +1−e^(−1) =1−2e^(−1) A_1 =1−(2/e) A_2 = ∫_0 ^1 x^2 e^(−x) dx =[−x^2 e^(−x) ]_0 ^1 +∫_0 ^1 2x e^(−x) dx =−e^(−1) +2A_1 =−e^(−1) +2(1−(2/e) )=−e^(−1) +2−(4/e) A_2 =2−(5/e) 2) by parts u^′ =x^n and v=e^(−x) ⇒ A_n =[(1/(n+1))x^(n+1) e^(−x) ]_0 ^1 +∫_0 ^1 (x^(n+1) /(n+1)) e^(−x) dx = (1/((n+1)e)) +(1/(n+1)) A_(n+1) ⇒ (n+1)A_n = (1/e) +A_(n+1) ⇒ A_(n+1) =(n+1)A_n −(1/e) 3) A_3 =3A_2 −(1/e) =3(2−(5/e))−(1/e) A_3 = 6−((16)/e) A_4 =4A_3 −(1/e) =4{6−((16)/e)}−(1/e) A_4 =24 −((65)/e) A_5 = 5A_4 −(1/e) =5{24−((65)/e)}−(1/e) =120 −((5.65+1)/e) =120−((326)/e)$
	$1) A_{1} = \int_{0}^{1} x e^{- x} d x = {[- {x e}^{- x}]}_{0}^{1} + \int_{0}^{1} e^{- x} d x$ $= - e^{- 1} + {[- e^{- x}]}_{0}^{1} = - e^{- 1} + 1 - e^{- 1} = 1 - 2 e^{- 1}$ $A_{1} = 1 - \frac{2}{e}$ $A_{2} = \int_{0}^{1} x^{2} e^{- x} d x = {[- x^{2} e^{- x}]}_{0}^{1} + \int_{0}^{1} 2 x e^{- x} d x$ $= - e^{- 1} + 2 A_{1} = - e^{- 1} + 2 (1 - \frac{2}{e}) = - e^{- 1} + 2 - \frac{4}{e}$ $A_{2} = 2 - \frac{5}{e}$ $2) b y p a r t s u^{^{'}} = x^{n} a n d v = e^{- x} \Rightarrow$ $A_{n} = {[\frac{1}{n + 1} x^{n + 1} e^{- x}]}_{0}^{1} + \int_{0}^{1} \frac{x^{n + 1}}{n + 1} e^{- x} d x$ $= \frac{1}{(n + 1) e} + \frac{1}{n + 1} A_{n + 1} \Rightarrow$ $(n + 1) A_{n} = \frac{1}{e} + A_{n + 1} \Rightarrow$ $A_{n + 1} = (n + 1) A_{n} - \frac{1}{e}$ $3) A_{3} = 3 A_{2} - \frac{1}{e} = 3 (2 - \frac{5}{e}) - \frac{1}{e}$ $A_{3} = 6 - \frac{16}{e}$ $A_{4} = 4 A_{3} - \frac{1}{e} = 4 {6 - \frac{16}{e}} - \frac{1}{e}$ $A_{4} = 24 - \frac{65}{e}$ $A_{5} = 5 A_{4} - \frac{1}{e} = 5 {24 - \frac{65}{e}} - \frac{1}{e}$ $= 120 - \frac{5.65 + 1}{e} = 120 - \frac{326}{e}$
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