Question Number 170532 by MikeH last updated on 26/May/22
$$\mathrm{Let}\:{I}_{{n}} \:=\int{x}^{{n}} {e}^{−{x}} {dx},\:{n}\:=\:\mathrm{0},\mathrm{1},\mathrm{2},… \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Show}\:\mathrm{that}\:{I}_{{n}} \:=\:−{x}^{{n}} {e}^{−{x}} +{nI}_{{n}−\mathrm{1}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{x}} {dx}\:=\:{n}! \\ $$
Answered by FelipeLz last updated on 26/May/22
![(i) I_n = ∫x^n e^(−x) dx x^n = u ⇒ u′ = (d/dx)[x^n ] = nx^(n−1) e^(−x) = v^′ ⇒ v = ∫e^(−x) dx = −e^(−x) ∫uv′dx = uv−∫u′vdx I_n = x^n (−e^(−x) )−∫nx^(n−1) (−e^(−x) )dx I_n = −x^n e^(−x) +n∫x^(n−1) e^(−x) dx I_n = −x^n e^(−x) +nI_(n−1) (ii) ∫x^n e^(−x) dx = −x^n e^(−x) +n∫x^(n−1) e^(−x) dx ∫x^n e^(−x) dx = −x^n e^(−x) −nx^(n−1) e^(−x) +n(n−1)∫x^(n−2) e^(−x) dx ∫x^n e^(−x) dx = −x^n e^(−x) −nx^(n−1) e^(−x) −n(n−1)x^(n−2) e^(−x) +n(n−1)(n−2)∫x^(n−3) e^(−x) dx ∫x^n e^(−x) dx = −x^n e^(−x) −nx^(n−1) e^(−x) −n(n−1)x^(n−2) e^(−x) −n(n−1)(n−2)x^(n−3) e^(−x) +n(n−1)(n−2)(n−3)∫x^(n−4) e^(−x) dx ⋮ ∫x^n e^(−x) dx = −e^(−x) Σ_(k=0) ^n ((n!)/((n−k)!))x^(n−k) ∫_0 ^∞ x^n e^(−x) dx = −Σ_(k=0) ^n [((n!)/((n−k)!))∙lim_(x→∞) ((x^(n−k) /e^x ))]+Σ_(k=0) ^(n−1) [((n!)/((n−k)!))∙(0^(n−k) /e^0 )]+((n!)/e^0 ) ∫_0 ^∞ x^n e^(−x) dx = −Σ_(k=0) ^n [((n!)/((n−k)!))∙0]+Σ_(k=0) ^(n−1) [((n!)/((n−k)!))∙0]+((n!)/1) = n!](https://www.tinkutara.com/question/Q170540.png)
$$\left({i}\right) \\ $$$$\:\:\:\:\:\:\:{I}_{{n}} \:=\:\int{x}^{{n}} {e}^{−{x}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{n}} \:=\:{u}\:\Rightarrow\:{u}'\:=\:\frac{{d}}{{dx}}\left[{x}^{{n}} \right]\:=\:{nx}^{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{−{x}} \:=\:{v}^{'} \:\Rightarrow\:{v}\:=\:\int{e}^{−{x}} {dx}\:=\:−{e}^{−{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\int{uv}'{dx}\:=\:{uv}−\int{u}'{vdx} \\ $$$$\:\:\:\:\:\:\:{I}_{{n}} \:=\:{x}^{{n}} \left(−{e}^{−{x}} \right)−\int{nx}^{{n}−\mathrm{1}} \left(−{e}^{−{x}} \right){dx} \\ $$$$\:\:\:\:\:\:\:{I}_{{n}} \:=\:−{x}^{{n}} {e}^{−{x}} +{n}\int{x}^{{n}−\mathrm{1}} {e}^{−{x}} {dx} \\ $$$$\:\:\:\:\:\:\:{I}_{{n}} \:=\:−{x}^{{n}} {e}^{−{x}} +{nI}_{{n}−\mathrm{1}} \\ $$$$\left({ii}\right) \\ $$$$\:\:\:\:\:\:\int{x}^{{n}} {e}^{−{x}} {dx}\:=\:−{x}^{{n}} {e}^{−{x}} +{n}\int{x}^{{n}−\mathrm{1}} {e}^{−{x}} {dx} \\ $$$$\:\:\:\:\:\:\int{x}^{{n}} {e}^{−{x}} {dx}\:=\:−{x}^{{n}} {e}^{−{x}} −{nx}^{{n}−\mathrm{1}} {e}^{−{x}} +{n}\left({n}−\mathrm{1}\right)\int{x}^{{n}−\mathrm{2}} {e}^{−{x}} {dx} \\ $$$$\:\:\:\:\:\:\int{x}^{{n}} {e}^{−{x}} {dx}\:=\:−{x}^{{n}} {e}^{−{x}} −{nx}^{{n}−\mathrm{1}} {e}^{−{x}} −{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} {e}^{−{x}} +{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\int{x}^{{n}−\mathrm{3}} {e}^{−{x}} {dx} \\ $$$$\:\:\:\:\:\:\int{x}^{{n}} {e}^{−{x}} {dx}\:=\:−{x}^{{n}} {e}^{−{x}} −{nx}^{{n}−\mathrm{1}} {e}^{−{x}} −{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} {e}^{−{x}} −{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right){x}^{{n}−\mathrm{3}} {e}^{−{x}} +{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)\int{x}^{{n}−\mathrm{4}} {e}^{−{x}} {dx} \\ $$$$\:\:\:\:\:\:\vdots \\ $$$$\:\:\:\:\:\:\int{x}^{{n}} {e}^{−{x}} {dx}\:=\:−{e}^{−{x}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{\left({n}−{k}\right)!}{x}^{{n}−{k}} \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{x}} {dx}\:=\:−\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left[\frac{{n}!}{\left({n}−{k}\right)!}\centerdot\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}^{{n}−{k}} }{{e}^{{x}} }\right)\right]+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\frac{{n}!}{\left({n}−{k}\right)!}\centerdot\frac{\mathrm{0}^{{n}−{k}} }{{e}^{\mathrm{0}} }\right]+\frac{{n}!}{{e}^{\mathrm{0}} }\: \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{x}} {dx}\:=\:−\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left[\frac{{n}!}{\left({n}−{k}\right)!}\centerdot\mathrm{0}\right]+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\frac{{n}!}{\left({n}−{k}\right)!}\centerdot\mathrm{0}\right]+\frac{{n}!}{\mathrm{1}}\:=\:{n}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$