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5050-0-1-1-x-50-100-dx-0-1-1-x-50-101-dx-




Question Number 97531 by 281981 last updated on 08/Jun/20
5050((∫_0 ^1 (1−x^(50) )^(100) dx)/(∫_0 ^1 (1−x^(50) )^(101) dx))=
$$\mathrm{5050}\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{50}} \right)^{\mathrm{100}} {dx}}{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{50}} \right)^{\mathrm{101}} {dx}}= \\ $$
Answered by smridha last updated on 08/Jun/20
let (1−x^(50) )=m so x=(1−m)^(1/(50))   now,  5050(((1/(50))∫_0 ^1 m^((101−1)) (1−m)^((1/(50))−1) dm)/((1/(50))∫_0 ^1 (m)^((102−1)) (1−m)^((1/(50))−1) dm))  =5050(((Γ(101)Γ((1/(50))))/(𝚪(101+(1/(50)))))/((Γ(102)Γ((1/(50))))/(Γ(102+(1/(50))))))=5050(((101+(1/(50))))/(101))  =5051
$$\boldsymbol{{let}}\:\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{50}} \right)=\boldsymbol{{m}}\:{so}\:{x}=\left(\mathrm{1}−{m}\right)^{\frac{\mathrm{1}}{\mathrm{50}}} \\ $$$${now}, \\ $$$$\mathrm{5050}\frac{\frac{\mathrm{1}}{\mathrm{50}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{m}}^{\left(\mathrm{101}−\mathrm{1}\right)} \left(\mathrm{1}−\boldsymbol{{m}}\right)^{\frac{\mathrm{1}}{\mathrm{50}}−\mathrm{1}} \boldsymbol{{dm}}}{\frac{\mathrm{1}}{\mathrm{50}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({m}\right)^{\left(\mathrm{102}−\mathrm{1}\right)} \left(\mathrm{1}−{m}\right)^{\frac{\mathrm{1}}{\mathrm{50}}−\mathrm{1}} \boldsymbol{{dm}}} \\ $$$$=\mathrm{5050}\frac{\frac{\Gamma\left(\mathrm{101}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{50}}\right)}{\boldsymbol{\Gamma}\left(\mathrm{101}+\frac{\mathrm{1}}{\mathrm{50}}\right)}}{\frac{\Gamma\left(\mathrm{102}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{50}}\right)}{\Gamma\left(\mathrm{102}+\frac{\mathrm{1}}{\mathrm{50}}\right)}}=\mathrm{5050}\frac{\left(\mathrm{101}+\frac{\mathrm{1}}{\mathrm{50}}\right)}{\mathrm{101}} \\ $$$$=\mathrm{5051} \\ $$
Commented by 281981 last updated on 10/Jun/20
explain 3rd step briefly sir.tnq
$${explain}\:\mathrm{3}{rd}\:{step}\:{briefly}\:{sir}.{tnq} \\ $$
Commented by smridha last updated on 10/Jun/20
 I just used 𝚪(n+1)=n𝚪(n)
$$\:\boldsymbol{{I}}\:\boldsymbol{{just}}\:\boldsymbol{{used}}\:\boldsymbol{\Gamma}\left(\boldsymbol{{n}}+\mathrm{1}\right)=\boldsymbol{{n}\Gamma}\left({n}\right) \\ $$
Commented by smridha last updated on 10/Jun/20
Γ(102+(1/(50)))=(101+(1/(50)))𝚪(101+(1/(50)))
$$\Gamma\left(\mathrm{102}+\frac{\mathrm{1}}{\mathrm{50}}\right)=\left(\mathrm{101}+\frac{\mathrm{1}}{\mathrm{50}}\right)\boldsymbol{\Gamma}\left(\mathrm{101}+\frac{\mathrm{1}}{\mathrm{50}}\right) \\ $$

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