Menu Close

Given-that-I-n-0-1-x-1-x-n-dx-obtain-a-reduction-formulae-for-I-n-in-terms-of-I-n-2-Hence-evaluate-0-1-x-1-x-5-dx-

Tinku Tara 0 Comments
Pin




Question Number 122166 by physicstutes last updated on 14/Nov/20
Given that I_n = ∫_0 ^1 x(1−x)^n dx obtain a reduction formulae for I_(n ) in terms of I_(n−2) Hence evaluate ∫_0 ^1 x(1−x)^5 dx.
$$\mathrm{Given}\:\mathrm{that}\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\left(\mathrm{1}−{x}\right)^{{n}} {dx} \\ $$$$\mathrm{obtain}\:\mathrm{a}\:\mathrm{reduction}\:\mathrm{formulae}\:\mathrm{for}\:{I}_{{n}\:} \:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:{I}_{{n}−\mathrm{2}} \:\mathrm{Hence}\:\mathrm{evaluate}\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\left(\mathrm{1}−{x}\right)^{\mathrm{5}} {dx}. \\ $$
Answered by liberty last updated on 14/Nov/20
∫_0 ^1 x(1−x)^5 dx = −((x(1−x)^6 )/6)]_0 ^1 + ∫_0 ^1 (((1−x)^6 )/6) dx = −(1/(42)) [ (1−x)]_0 ^1 = (1/(42))
$$\left.\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{5}} \:\mathrm{dx}\:=\:−\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{6}} }{\mathrm{6}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:+\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{6}} }{\mathrm{6}}\:\mathrm{dx} \\ $$$$\:=\:−\frac{\mathrm{1}}{\mathrm{42}}\:\left[\:\left(\mathrm{1}−\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{42}} \\ $$
Commented by physicstutes last updated on 14/Nov/20
from your method i deduce that: I_n = −[((x(1−x)^(n+1) )/(n+1))]_0 ^1 + ∫_0 ^1 (((1−x)^(n+1) )/(n+1)) dx I_n = (1/(n+1)) ∫_0 ^1 (1−x)^(n+1) dx Thanks
$$\mathrm{from}\:\mathrm{your}\:\mathrm{method}\:\mathrm{i}\:\mathrm{deduce}\:\mathrm{that}: \\ $$$${I}_{{n}} \:=\:−\left[\frac{{x}\left(\mathrm{1}−{x}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\left(\mathrm{1}−{x}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:{dx} \\ $$$${I}_{{n}} \:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{1}−{x}\right)^{{n}+\mathrm{1}} {dx} \\ $$$$\:\mathrm{Thanks} \\ $$
Commented by Dwaipayan Shikari last updated on 14/Nov/20
∫_0 ^1 x(1−x)^n dx=∫_0 ^1 (1−x)x^n dx =(1/(n+1))−(1/(n+2)) here n=5 so, (1/(42))
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}\right)^{{n}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right){x}^{{n}} {dx}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$${here}\:{n}=\mathrm{5}\:\:{so},\:\frac{\mathrm{1}}{\mathrm{42}} \\ $$
Answered by Dwaipayan Shikari last updated on 14/Nov/20
∫_0 ^1 x(1−x)^n dx =β(2,n+1)=((Γ(2)Γ(n+1))/(Γ(n+3)))=((Γ(n+1))/(Γ(n+3))) So ∫_0 ^1 x(1−x)^5 dt=((Γ(5+1))/(Γ(5+3)))=((5!)/(7!))=(1/(42))
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}\right)^{{n}} {dx} \\ $$$$=\beta\left(\mathrm{2},{n}+\mathrm{1}\right)=\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\mathrm{3}\right)}=\frac{\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\mathrm{3}\right)} \\ $$$${So} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}\right)^{\mathrm{5}} {dt}=\frac{\Gamma\left(\mathrm{5}+\mathrm{1}\right)}{\Gamma\left(\mathrm{5}+\mathrm{3}\right)}=\frac{\mathrm{5}!}{\mathrm{7}!}=\frac{\mathrm{1}}{\mathrm{42}} \\ $$
Commented by physicstutes last updated on 14/Nov/20
Magnifcent!
$$\mathrm{Magnifcent}! \\ $$
Answered by mathmax by abdo last updated on 14/Nov/20
I_n =∫_0 ^1 x(1−x)^n dx we know B(p,q) =∫_0 ^1 x^(p−1) (1−x)^(q−1) dx ⇒ B(p,q) =((Γ(p).Γ(q))/(Γ(p+q))) ⇒I_n =∫_0 ^1 x^(2−1) (1−x)^(n+1−1) dx =B(2,n+1) =((Γ(2).Γ(n+1))/(Γ(2+n+1))) =((1! n!)/((n+2)!)) =((n!)/((n+2)(n+1)n!)) =(1/((n+1)(n+2))) (I_n /I_(n−2) )=(1/((n+1)(n+2))) .(n−1)n =((n(n−1))/((n+1)(n+2))) ⇒ I_n =((n(n−1))/((n+1)(n+2)))I_(n−2) (n≥2) ∫_0 ^1 x(1−x)^5 dx =I_5 =(1/(6.7)) =(1/(42))
$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}} \mathrm{dx}\:\mathrm{we}\:\mathrm{know}\:\:\:\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{p}−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{q}−\mathrm{1}} \mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)\:=\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)}\:\Rightarrow\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2}−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}+\mathrm{1}−\mathrm{1}} \mathrm{dx}\:=\mathrm{B}\left(\mathrm{2},\mathrm{n}+\mathrm{1}\right) \\ $$$$=\frac{\Gamma\left(\mathrm{2}\right).\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}+\mathrm{n}+\mathrm{1}\right)}\:=\frac{\mathrm{1}!\:\mathrm{n}!}{\left(\mathrm{n}+\mathrm{2}\right)!}\:=\frac{\mathrm{n}!}{\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)\mathrm{n}!}\:=\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\: \\ $$$$\frac{\mathrm{I}_{\mathrm{n}} }{\mathrm{I}_{\mathrm{n}−\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:.\left(\mathrm{n}−\mathrm{1}\right)\mathrm{n}\:=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} =\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\mathrm{I}_{\mathrm{n}−\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{n}\geqslant\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{5}} \mathrm{dx}\:=\mathrm{I}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{6}.\mathrm{7}}\:=\frac{\mathrm{1}}{\mathrm{42}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *