Question Number 141713 by rs4089 last updated on 22/May/21
Answered by Dwaipayan Shikari last updated on 22/May/21
![1−x=u ∫_(−1) ^1 ∫_0 ^u x^(1/3) y^(−1/2) (u−y)^(1/2) dydx y=uk =∫_(−1) ^1 ∫_0 ^1 x^(1/3) uk^(−1/2) (1−k)^(1/2) dkdx =Γ((1/2))Γ((3/2))∫_(−1) ^1 x^(1/3) (1−x)dx=(π/2)[(3/4)x^(4/3) −(3/7)x^(7/3) ]_(−1) ^1 =−((3π)/7)](https://www.tinkutara.com/question/Q141722.png)
$$\mathrm{1}−{x}={u} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{{u}} {x}^{\mathrm{1}/\mathrm{3}} {y}^{−\mathrm{1}/\mathrm{2}} \left({u}−{y}\right)^{\mathrm{1}/\mathrm{2}} {dydx}\:\:\:\:{y}={uk} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{1}/\mathrm{3}} {uk}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}−{k}\right)^{\mathrm{1}/\mathrm{2}} {dkdx} \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\int_{−\mathrm{1}} ^{\mathrm{1}} {x}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}−{x}\right){dx}=\frac{\pi}{\mathrm{2}}\left[\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{4}/\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{7}}{x}^{\mathrm{7}/\mathrm{3}} \right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{3}\pi}{\mathrm{7}} \\ $$