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Question Number 49902 by rahul 19 last updated on 12/Dec/18

If F(t)= ∫_0 ^( t) e^(t−y) .ydy. Prove that F(t)= e^t −(1+t).

IfF(t)=∫0tet−y.ydy.ProvethatF(t)=et−(1+t).

Commented by Abdo msup. last updated on 12/Dec/18

F(t) =e^t ∫_0 ^t y e^(−y) dy by parts ∫_0 ^t y e^(−y) dy =[−y e^(−y) ]_0 ^t +∫_0 ^t e^(−y) dy =−t e^(−t) +[−e^(−y) ]_0 ^t =−t e^(−t) +1−e^(−t) ⇒ F(t)=e^t (−t e^(−t) +1 −e^(−t) )=−t +e^t −1 =e^t −(1+t).

F(t)=et∫0tye−ydybyparts∫0tye−ydy=[−ye−y]0t+∫0te−ydy=−te−t+[−e−y]0t=−te−t+1−e−t⇒F(t)=et(−te−t+1−e−t)=−t+et−1=et−(1+t).

Answered by Smail last updated on 12/Dec/18

F(t)=∫_0 ^t e^t e^(−y) ydy=e^t ∫_0 ^t ye^(−y) dy by parts u=y⇒u′=1 v′=e^(−y) ⇒v=−e^(−y) F(t)=e^t (−[ye^(−y) ]_0 ^t +∫_0 ^t e^(−y) dy) =e^t (−te^(−t) −[e^(−y) ]_0 ^t ) =e^t (−te^(−t) −e^(−t) +1) =−t−1+e^t F(t)=−(1+t)+e^t

F(t)=∫0tete−yydy=et∫0tye−ydybypartsu=y⇒u′=1v′=e−y⇒v=−e−yF(t)=et(−[ye−y]0t+∫0te−ydy)=et(−te−t−[e−y]0t)=et(−te−t−e−t+1)=−t−1+etF(t)=−(1+t)+et

Commented by rahul 19 last updated on 12/Dec/18

thank you sir!

thankyousir!

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