Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 6042 by FilupSmith last updated on 10/Jun/16

Can you solve the indefinite integral:  ∫e^(−u) u^n du

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{indefinite}\:\mathrm{integral}: \\ $$$$\int{e}^{−{u}} {u}^{{n}} {du} \\ $$

Commented by Yozzii last updated on 11/Jun/16

Define I(n)=∫e^(−u) u^n du (n≥0).  ∴ Integrating by parts gives  I(n)=−e^(−u) u^n −∫−ne^(−u) u^(n−1) du  I(n)=−e^(−u) u^n +n∫e^(−u) u^(n−1) du  I(n)=−e^(−u) u^n +nI(n−1)  I(n)=−e^(−u) u^n −ne^(−u) u^(n−1) +n(n−1)I(n−2)  I(n)=−e^(−u) u^n −ne^(−u) u^(n−1) −n(n−1)e^(−u) u^(n−2) +n(n−1)(n−2)I(n−3).  I(n)=−e^(−u) (((n!)/((n−0)!))u^(n−0) +((n!)/((n−1)!))u^(n−1) +((n!)/((n−2)!))u^(n−2) +((n!)/((n−3)!))u^(n−3) +...+((n!)/(2!))u^2 )+((n!)/(1!))I(1)  I(1)=∫e^(−u) udu=−e^(−u) u+∫e^(−u) du=−e^(−u) u−e^(−u) +C.  I(n)=−e^(−u) (((n!)/((n−0)!))u^(n−0) +((n!)/((n−1)!))u^(n−1) +((n!)/((n−3)!))u^(n−3) +...+((n!)/(2!))u^2 +((n!)/(1!))u^1 +((n!)/(0!))u^0 )+D  I(n)=D−e^(−u) n!(Σ_(r=0) ^n (u^(n−r) /((n−r)!)))

$${Define}\:{I}\left({n}\right)=\int{e}^{−{u}} {u}^{{n}} {du}\:\left({n}\geqslant\mathrm{0}\right). \\ $$$$\therefore\:{Integrating}\:{by}\:{parts}\:{gives} \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} −\int−{ne}^{−{u}} {u}^{{n}−\mathrm{1}} {du} \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} +{n}\int{e}^{−{u}} {u}^{{n}−\mathrm{1}} {du} \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} +{nI}\left({n}−\mathrm{1}\right) \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} −{ne}^{−{u}} {u}^{{n}−\mathrm{1}} +{n}\left({n}−\mathrm{1}\right){I}\left({n}−\mathrm{2}\right) \\ $$$${I}\left({n}\right)=−{e}^{−{u}} {u}^{{n}} −{ne}^{−{u}} {u}^{{n}−\mathrm{1}} −{n}\left({n}−\mathrm{1}\right){e}^{−{u}} {u}^{{n}−\mathrm{2}} +{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right){I}\left({n}−\mathrm{3}\right). \\ $$$${I}\left({n}\right)=−{e}^{−{u}} \left(\frac{{n}!}{\left({n}−\mathrm{0}\right)!}{u}^{{n}−\mathrm{0}} +\frac{{n}!}{\left({n}−\mathrm{1}\right)!}{u}^{{n}−\mathrm{1}} +\frac{{n}!}{\left({n}−\mathrm{2}\right)!}{u}^{{n}−\mathrm{2}} +\frac{{n}!}{\left({n}−\mathrm{3}\right)!}{u}^{{n}−\mathrm{3}} +...+\frac{{n}!}{\mathrm{2}!}{u}^{\mathrm{2}} \right)+\frac{{n}!}{\mathrm{1}!}{I}\left(\mathrm{1}\right) \\ $$$${I}\left(\mathrm{1}\right)=\int{e}^{−{u}} {udu}=−{e}^{−{u}} {u}+\int{e}^{−{u}} {du}=−{e}^{−{u}} {u}−{e}^{−{u}} +{C}. \\ $$$${I}\left({n}\right)=−{e}^{−{u}} \left(\frac{{n}!}{\left({n}−\mathrm{0}\right)!}{u}^{{n}−\mathrm{0}} +\frac{{n}!}{\left({n}−\mathrm{1}\right)!}{u}^{{n}−\mathrm{1}} +\frac{{n}!}{\left({n}−\mathrm{3}\right)!}{u}^{{n}−\mathrm{3}} +...+\frac{{n}!}{\mathrm{2}!}{u}^{\mathrm{2}} +\frac{{n}!}{\mathrm{1}!}{u}^{\mathrm{1}} +\frac{{n}!}{\mathrm{0}!}{u}^{\mathrm{0}} \right)+{D} \\ $$$${I}\left({n}\right)={D}−{e}^{−{u}} {n}!\left(\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{u}^{{n}−{r}} }{\left({n}−{r}\right)!}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by FilupSmith last updated on 11/Jun/16

is a D constant?

$$\mathrm{is}\:\mathrm{a}\:{D}\:\mathrm{constant}? \\ $$

Commented by Yozzii last updated on 11/Jun/16

D=n!C where n=constant  and C=constant. You get that  after substituting the expression for I(1).

$${D}={n}!{C}\:{where}\:{n}={constant} \\ $$$${and}\:{C}={constant}.\:{You}\:{get}\:{that} \\ $$$${after}\:{substituting}\:{the}\:{expression}\:{for}\:{I}\left(\mathrm{1}\right). \\ $$$$ \\ $$

Commented by FilupSmith last updated on 11/Jun/16

I see, thank you!

$$\mathrm{I}\:\mathrm{see},\:\mathrm{thank}\:\mathrm{you}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com