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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).

$${If}\:{F}\left({t}\right)=\:\int_{\mathrm{0}} ^{\:{t}} {e}^{{t}−{y}} .{ydy}. \\ $$$${Prove}\:{that}\:{F}\left({t}\right)=\:{e}^{{t}} −\left(\mathrm{1}+{t}\right). \\ $$

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).

$${F}\left({t}\right)\:={e}^{{t}} \:\int_{\mathrm{0}} ^{{t}} \:{y}\:{e}^{−{y}} {dy}\:\:{by}\:{parts}\:\int_{\mathrm{0}} ^{{t}} \:{y}\:{e}^{−{y}} {dy} \\ $$$$=\left[−{y}\:{e}^{−{y}} \right]_{\mathrm{0}} ^{{t}} \:+\int_{\mathrm{0}} ^{{t}} \:\:{e}^{−{y}} {dy}\:\:=−{t}\:{e}^{−{t}} \:+\left[−{e}^{−{y}} \right]_{\mathrm{0}} ^{{t}} \\ $$$$=−{t}\:{e}^{−{t}} \:+\mathrm{1}−{e}^{−{t}} \:\Rightarrow \\ $$$${F}\left({t}\right)={e}^{{t}} \left(−{t}\:{e}^{−{t}} \:+\mathrm{1}\:−{e}^{−{t}} \right)=−{t}\:+{e}^{{t}} \:−\mathrm{1} \\ $$$$={e}^{{t}} \:−\left(\mathrm{1}+{t}\right). \\ $$

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

$${F}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} {e}^{{t}} {e}^{−{y}} {ydy}={e}^{{t}} \int_{\mathrm{0}} ^{{t}} {ye}^{−{y}} {dy} \\ $$$${by}\:{parts} \\ $$$${u}={y}\Rightarrow{u}'=\mathrm{1} \\ $$$${v}'={e}^{−{y}} \Rightarrow{v}=−{e}^{−{y}} \\ $$$${F}\left({t}\right)={e}^{{t}} \left(−\left[{ye}^{−{y}} \right]_{\mathrm{0}} ^{{t}} +\int_{\mathrm{0}} ^{{t}} {e}^{−{y}} {dy}\right) \\ $$$$={e}^{{t}} \left(−{te}^{−{t}} −\left[{e}^{−{y}} \right]_{\mathrm{0}} ^{{t}} \right) \\ $$$$={e}^{{t}} \left(−{te}^{−{t}} −{e}^{−{t}} +\mathrm{1}\right) \\ $$$$=−{t}−\mathrm{1}+{e}^{{t}} \\ $$$${F}\left({t}\right)=−\left(\mathrm{1}+{t}\right)+{e}^{{t}} \\ $$

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

thank you sir!

$${thank}\:{you}\:{sir}! \\ $$

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