Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 108118 by mathdave last updated on 14/Aug/20

Answered by abdomathmax last updated on 14/Aug/20

$$\mathrm{I}={\int }_0^1{(1-{\mathrm{x}}^{18})}^{\frac{1}{20}}\mathrm{dx}-{\int }_0^1{(1-{\mathrm{x}}^{20})}^{\frac{1}{18}}\mathrm{dx}=\mathrm{H}-\mathrm{K}$$$$\mathrm{J}={\int }_0^1{(1-{\mathrm{x}}^{18})}^{\frac{1}{20}}\mathrm{dx}{=}_{{\mathrm{x}}^{18}={\sin }^2\mathrm{t}\to \mathrm{x}={\sin }^{\frac{1}{9}}\mathrm{t}}\frac{1}{9}{\int }_0^{\frac{\pi }{2}}{\cos }^{\frac{1}{10}}{\mathrm{tcostsin}}^{\frac{1}{9}-1}\mathrm{t}\mathrm{dt}$$$$=\frac{1}{9}{\int }_0^{\frac{\pi }{2}}{\cos }^{\frac{1}{10}+1}\mathrm{t}{\sin }^{\frac{1}{9}-1}\mathrm{t}\mathrm{dt}\mathrm{we}\mathrm{know}$$$$2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{p}-1}\mathrm{t}{\sin }^{2\mathrm{q}-1}\mathrm{t}\mathrm{dt}=\mathrm{B}(\mathrm{p},\mathrm{q})=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}$$$$2\mathrm{p}-1=\frac{1}{10}+1\Rightarrow 2\mathrm{p}=\frac{1}{10}+2\Rightarrow \mathrm{p}=\frac{1}{20}+1=\frac{21}{20}$$$$2\mathrm{q}-1=\frac{1}{9}-1\Rightarrow \mathrm{q}=\frac{1}{18}\Rightarrow$$$$\mathrm{H}=\frac{1}{18}\mathrm{B}(\frac{21}{20},\frac{1}{18})=\frac{1}{18}\frac{\mathrm{\Gamma }\left(\frac{21}{20}\right).\mathrm{\Gamma }\left(\frac{1}{18}\right)}{\mathrm{\Gamma }(\frac{21}{20}+\frac{1}{18})}$$$$\mathrm{K}={\int }_0^1{(1-{\mathrm{x}}^{20})}^{\frac{1}{18}}\mathrm{dx}{=}_{{\mathrm{x}}^{20}={\sin }^2\mathrm{t}\to \mathrm{x}={\sin }^{\frac{1}{10}}\mathrm{t}}\frac{1}{10}{\int }_0^{\frac{\pi }{2}}{\cos }^{\frac{1}{9}}\mathrm{t}\mathrm{cost}{\sin }^{\frac{1}{10}-1}\mathrm{t}\mathrm{dt}$$$$=\frac{1}{10}{\int }_0^{\frac{\pi }{2}}{\cos }^{\frac{1}{9}+1}\mathrm{t}{\sin }^{\frac{1}{10}-1}\mathrm{t}\mathrm{dt}$$$$2\mathrm{p}-1=\frac{1}{9}+1\Rightarrow 2\mathrm{p}=\frac{1}{9}+2\Rightarrow \mathrm{p}=\frac{1}{18}+1=\frac{19}{18}$$$$2\mathrm{q}-1=\frac{1}{10}-1\Rightarrow \mathrm{q}=\frac{1}{20}\Rightarrow$$$$\mathrm{K}=\frac{1}{20}\mathrm{B}(\frac{19}{18},\frac{1}{20})=\frac{1}{20}\frac{\mathrm{\Gamma }\left(\frac{19}{18}\right).\mathrm{\Gamma }\left(\frac{1}{20}\right)}{\mathrm{\Gamma }(\frac{19}{18}+\frac{1}{20})}$$$$\mathrm{I}=\mathrm{H}-\mathrm{K}$$

Commented by Her_Majesty last updated on 14/Aug/20

$$\mathrm{\Gamma }\left(\frac{21}{20}\right)=\frac{1}{20}\mathrm{\Gamma }\left(\frac{1}{20}\right)$$$$\mathrm{\Gamma }\left(\frac{19}{18}\right)=\frac{1}{18}\mathrm{\Gamma }\left(\frac{1}{18}\right)$$$$\frac{21}{20}+\frac{1}{18}=\frac{19}{18}+\frac{1}{20}$$$$\Rightarrow H=K$$$$\Rightarrow I=0$$

Commented by abdomathmax last updated on 15/Aug/20

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{completing}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com