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 168303 by Mathspace last updated on 07/Apr/22

calculate ∫_0 ^1 x(√(1−x^6 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }{dx} \\ $$

Answered by peter frank last updated on 07/Apr/22

check Qn 168231

$$\mathrm{check}\:\mathrm{Qn}\:\mathrm{168231} \\ $$

Answered by Mathspace last updated on 08/Apr/22

this integral is resoluble  we do the changement x=t^(1/6)  ⇒  ∫_0 ^1 x(√(1−x^6 ))dx=(1/6)∫_0 ^1 t^(1/6) (1−t)^(1/2)   t^((1/6)−1) dt  =(1/6)∫_0 ^1  t^((1/3)−1) (1−t)^((3/2)−1)  dt  we know B(p,q)=∫_0 ^1 t^(p−1) (1−t)^(q−1) dt  =((Γ(p).Γ(q))/(Γ(p+q))) ⇒  I=(1/6)B((1/3),(3/2))=(1/6)×((Γ((1/3)).Γ((3/2)))/(Γ((1/3)+(3/2))))  Γ((3/2))=Γ((1/2)+1)=(1/2)Γ((1/2))=((√π)/2)  ⇒I=((√π)/(12)).((Γ((1/3)))/(Γ(((11)/6))))

$${this}\:{integral}\:{is}\:{resoluble} \\ $$$${we}\:{do}\:{the}\:{changement}\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }{dx}=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{t}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \:{dt} \\ $$$${we}\:{know}\:{B}\left({p},{q}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{p}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{q}−\mathrm{1}} {dt} \\ $$$$=\frac{\Gamma\left({p}\right).\Gamma\left({q}\right)}{\Gamma\left({p}+{q}\right)}\:\Rightarrow \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{6}}{B}\left(\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{6}}×\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\Rightarrow{I}=\frac{\sqrt{\pi}}{\mathrm{12}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{11}}{\mathrm{6}}\right)} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com