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Question Number 73037 by aliesam last updated on 05/Nov/19

Commented by mathmax by abdo last updated on 06/Nov/19

1) I_0 =∫_0 ^1 e^(1−x) dx =[−e^(1−x) ]_0 ^1 =e−1 I_1 =∫_0 ^1 x e^(1−x) dx =_(by parts) [−xe^(1−x) ]_0 ^1 −∫_0 ^1 (−e^(1−x) )dx =−1+e−1 =e−2 2) I_(n+1) =∫_0 ^1 x^(n+1) e^(1−x) dx =e ∫_0 ^1 x^(n+1) e^(−x) dx by parts u=x^(n+1) and v^, =e^(−x) ⇒∫_0 ^1 x^(n+1) e^(−x ) dx =[−x^(n+1) e^(−x) ]_0 ^1 −∫_0 ^1 (n+1)x^n (−e^(−x) )dx =−e^(−1) +(n+1) ∫_0 ^1 x^n e^(−x) dx ⇒I_(n+1) =e(−e^(−1) +(n+1)∫_0 ^1 x^n e^(−x) dx) =−1+(n+1)∫_0 ^1 x^n e^(1−x) dx =(n+1)I_n −1 ⇒ I_(n+1) =(n+1)I_n −1 remrk we have I_n =nI_(n−1) −1 for n≥1 let V_n =(I_n /(n!)) ⇒V_(n+1) =(I_(n+1) /((n+1)!)) =(((n+1)I_n )/((n+1)!)) −1 =(I_n /(n!))−1 =V_n −1 ⇒V_(n+1) −V_n =−1 ⇒Σ_(k=0) ^(n−1) (V_(k+1) −V_k )=−n ⇒ V_n −V_0 =−n ⇒V_n =V_0 −n =e−1−n ⇒(I_n /(n!)) =e−1−n ⇒ I_n =n!{e−1−n}

1)I0=∫01e1−xdx=[−e1−x]01=e−1I1=∫01xe1−xdx=byparts[−xe1−x]01−∫01(−e1−x)dx=−1+e−1=e−22)In+1=∫01xn+1e1−xdx=e∫01xn+1e−xdxbypartsu=xn+1andv,=e−x⇒∫01xn+1e−xdx=[−xn+1e−x]01−∫01(n+1)xn(−e−x)dx=−e−1+(n+1)∫01xne−xdx⇒In+1=e(−e−1+(n+1)∫01xne−xdx)=−1+(n+1)∫01xne1−xdx=(n+1)In−1⇒In+1=(n+1)In−1remrkwehaveIn=nIn−1−1forn⩾1letVn=Inn!⇒Vn+1=In+1(n+1)!=(n+1)In(n+1)!−1=Inn!−1=Vn−1⇒Vn+1−Vn=−1⇒∑k=0n−1(Vk+1−Vk)=−n⇒Vn−V0=−n⇒Vn=V0−n=e−1−n⇒Inn!=e−1−n⇒In=n!{e−1−n}

Answered by mind is power last updated on 05/Nov/19

I_0 =∫_0 ^1 e^(1−x) dx=[−e^(1−x) ]_0 ^1 =e−1 I_1 =∫_0 ^1 xe^(1−x) dx=[−xe^(1−x) ]_0 ^1 +∫e^(1−x) dx =−1+e−1=e−2 I_(n+1) =∫_0 ^1 x^(n+1) e^(1−x) =[−x^(n+1) e^(1−x) ]_0 ^1 +(n+1)∫_0 ^1 x^n e^(1−x) dx integration by part =−1+(n+1)∫_0 ^1 x^n e^(1−x) dx=−1+(n+1)I_n I_3 =3I_2 −1 I_2 =2I_1 −1=2(e−2)−1=2e−5 I_3 =3(2e−5)−1=6e−16

I0=∫01e1−xdx=[−e1−x]01=e−1I1=∫01xe1−xdx=[−xe1−x]01+∫e1−xdx=−1+e−1=e−2In+1=∫01xn+1e1−x=[−xn+1e1−x]01+(n+1)∫01xne1−xdxintegrationbypart=−1+(n+1)∫01xne1−xdx=−1+(n+1)InI3=3I2−1I2=2I1−1=2(e−2)−1=2e−5I3=3(2e−5)−1=6e−16

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