Question Number 49902 by rahul 19 last updated on 12/Dec/18
$${If}\:{F}\left({t}\right)=\:\int_{\mathrm{0}} ^{\:{t}} {e}^{{t}−{y}} .{ydy}. \\ $$$${Prove}\:{that}\:{F}\left({t}\right)=\:{e}^{{t}} −\left(\mathrm{1}+{t}\right). \\ $$
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).](https://www.tinkutara.com/question/Q49928.png)
$${F}\left({t}\right)\:={e}^{{t}} \:\int_{\mathrm{0}} ^{{t}} \:{y}\:{e}^{−{y}} {dy}\:\:{by}\:{parts}\:\int_{\mathrm{0}} ^{{t}} \:{y}\:{e}^{−{y}} {dy} \\ $$$$=\left[−{y}\:{e}^{−{y}} \right]_{\mathrm{0}} ^{{t}} \:+\int_{\mathrm{0}} ^{{t}} \:\:{e}^{−{y}} {dy}\:\:=−{t}\:{e}^{−{t}} \:+\left[−{e}^{−{y}} \right]_{\mathrm{0}} ^{{t}} \\ $$$$=−{t}\:{e}^{−{t}} \:+\mathrm{1}−{e}^{−{t}} \:\Rightarrow \\ $$$${F}\left({t}\right)={e}^{{t}} \left(−{t}\:{e}^{−{t}} \:+\mathrm{1}\:−{e}^{−{t}} \right)=−{t}\:+{e}^{{t}} \:−\mathrm{1} \\ $$$$={e}^{{t}} \:−\left(\mathrm{1}+{t}\right). \\ $$
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](https://www.tinkutara.com/question/Q49912.png)
$${F}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} {e}^{{t}} {e}^{−{y}} {ydy}={e}^{{t}} \int_{\mathrm{0}} ^{{t}} {ye}^{−{y}} {dy} \\ $$$${by}\:{parts} \\ $$$${u}={y}\Rightarrow{u}'=\mathrm{1} \\ $$$${v}'={e}^{−{y}} \Rightarrow{v}=−{e}^{−{y}} \\ $$$${F}\left({t}\right)={e}^{{t}} \left(−\left[{ye}^{−{y}} \right]_{\mathrm{0}} ^{{t}} +\int_{\mathrm{0}} ^{{t}} {e}^{−{y}} {dy}\right) \\ $$$$={e}^{{t}} \left(−{te}^{−{t}} −\left[{e}^{−{y}} \right]_{\mathrm{0}} ^{{t}} \right) \\ $$$$={e}^{{t}} \left(−{te}^{−{t}} −{e}^{−{t}} +\mathrm{1}\right) \\ $$$$=−{t}−\mathrm{1}+{e}^{{t}} \\ $$$${F}\left({t}\right)=−\left(\mathrm{1}+{t}\right)+{e}^{{t}} \\ $$
Commented by rahul 19 last updated on 12/Dec/18
$${thank}\:{you}\:{sir}! \\ $$