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Question Number 141713 by rs4089 last updated on 22/May/21

Answered by Dwaipayan Shikari last updated on 22/May/21

$$1-x=u$$$${\int }_{-1}^1{\int }_0^u{x}^{1/3}{y}^{-1/2}{(u-y)}^{1/2}dydxy=uk$$$$={\int }_{-1}^1{\int }_0^1{x}^{1/3}{uk}^{-1/2}{(1-k)}^{1/2}dkdx$$$$=\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right){\int }_{-1}^1{x}^{1/3}(1-x)dx=\frac{\pi }{2}{[\frac{3}{4}{x}^{4/3}-\frac{3}{7}{x}^{7/3}]}_{-1}^1$$$$=-\frac{3\pi }{7}$$

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