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Question Number 127521 by mohammad17 last updated on 30/Dec/20

$${\underset{0}{\int }}^2\sqrt[3]{4x-x^3}dx$$

Answered by Dwaipayan Shikari last updated on 30/Dec/20

$${\int }_0^2\sqrt[3]{x}\left(\sqrt[3]{4-x^2}\right)dx{x}^2=4u\Rightarrow x=2\frac{du}{dx}$$$$=2{\int }_0^1{x}^{-\frac{2}{3}}{(4-4u)}^{\frac{1}{3}}du=2{\int }_0^1{u}^{-\frac{1}{3}}{(1-u)}^{\frac{1}{3}}du=2\beta (\frac{2}{3},\frac{4}{3})$$$$=2\frac{\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)}{3.\mathrm{\Gamma }\left(2\right)}=\frac{2\pi }{3sin\left(\frac{\pi }{3}\right)}=\frac{4\pi }{3\sqrt{3}}$$

Answered by mathmax by abdo last updated on 01/Jan/21

$$\mathrm{I}={\int }_0^2{}^3\sqrt{4\mathrm{x}-{\mathrm{x}}^3}\mathrm{dx}\Rightarrow \mathrm{I}=-{\int }_0^2\mathrm{x}\left({}^3\sqrt{-\frac{4}{{\mathrm{x}}^2}+1}\right)\mathrm{dx}$$$$\mathrm{changement}\frac{2}{\mathrm{x}}=\frac{1}{\mathrm{t}}\Rightarrow \mathrm{x}=2\mathrm{t}\Rightarrow \mathrm{I}=-{\int }_0^1\left(2\mathrm{t}\right){(-\frac{1}{{\mathrm{t}}^2}+1)}^{\frac{1}{3}}\left(2\mathrm{dt}\right)$$$$=-4{\int }_0^1\mathrm{t}{\left(\frac{-1+{\mathrm{t}}^2}{{\mathrm{t}}^2}\right)}^{\frac{1}{3}}\mathrm{dt}=-4{\int }_0^1\frac{\mathrm{t}}{{\mathrm{t}}^{\frac{2}{3}}}{({\mathrm{t}}^2-1)}^{\frac{1}{3}}\mathrm{dt}$$$$=4{\int }_0^1{\mathrm{t}}^{\frac{1}{3}}{(1-{\mathrm{t}}^2)}^{\frac{1}{3}}\mathrm{dt}{=}_{\mathrm{t}=\sqrt{\mathrm{u}}}4{\int }_0^1{\mathrm{u}}^{\frac{1}{6}}{(1-\mathrm{u})}^{\frac{1}{3}}\frac{\mathrm{du}}{2\sqrt{\mathrm{u}}}$$$$=2{\int }_0^1{\mathrm{u}}^{\frac{1}{6}-\frac{1}{2}}{(1-\mathrm{u})}^{\frac{1}{3}}\mathrm{du}=2{\int }_0^1{\mathrm{u}}^{-\frac{1}{3}}{(1-\mathrm{u})}^{\frac{1}{3}}\mathrm{du}=2\mathrm{B}(\frac{2}{3},\frac{4}{3})$$$$=2\frac{\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{4}{3}\right)}{\mathrm{\Gamma }(\frac{2}{3}+\frac{4}{3})}=\frac{2}{\mathrm{\Gamma }\left(2\right)}\times \mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{4}{3}\right)$$

Commented by mathmax by abdo last updated on 01/Jan/21

$$\mathrm{I}=2\mathrm{\Gamma }\left(\frac{2}{3}\right).\mathrm{\Gamma }(\frac{1}{3}+1)=2\mathrm{\Gamma }\left(\frac{2}{3}\right)\times \frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{3}\right)$$$$=\frac{2}{3}\mathrm{\Gamma }\left(\frac{1}{3}\right).\mathrm{\Gamma }(1-\frac{1}{3})=\frac{2}{3}\times \frac{\pi }{\sin \left(\frac{\pi }{3}\right)}=\frac{2\pi }{3\times \frac{\sqrt{3}}{2}}=\frac{4\pi }{3\sqrt{3}}$$

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