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Question Number 126577 by benjo_mathlover last updated on 22/Dec/20

$$\underset{0}{\stackrel{3}{\int }}\sqrt[3]{(3-x)x^2}dx=?$$

Answered by Dwaipayan Shikari last updated on 22/Dec/20

$${\int }_0^3{x}^{\frac{2}{3}}{(3-x)}^{\frac{1}{3}}dxx=3u$$$$=9{\int }_0^1{u}^{\frac{2}{3}}{(1-u)}^{\frac{1}{3}}du=9\beta (\frac{5}{3},\frac{4}{3})=9\frac{\mathrm{\Gamma }\left(\frac{5}{3}\right)\mathrm{\Gamma }\left(\frac{4}{3}\right)}{\mathrm{\Gamma }\left(3\right)}=\frac{9}{2}.\frac{2}{3}.\frac{1}{3}\mathrm{\Gamma }\left(\frac{2}{3}\right)\mathrm{\Gamma }\left(\frac{1}{3}\right)$$$$=\frac{\pi }{sin\left(\frac{\pi }{3}\right)}=\frac{2\pi }{\sqrt{3}}$$

Answered by liberty last updated on 22/Dec/20

$$I=\underset{1}{\stackrel{3}{\int }}x\sqrt[3]{\frac{3}{x}-1}dx;[let\frac{3}{x}-1=z^3]$$$$I=\underset{\mathrm{\infty }}{\stackrel{0}{\int }}\frac{27z^3}{{(z^3+1)}^3}dz;\left[byparts\right]$$$$I=-\frac{9}{2}z{\left(\frac{1}{{(z^3+1)}^2}\right)}_0^{\mathrm{\infty }}-\underset{0}{\stackrel{\mathrm{\infty }}{\int }}9.(-\frac{1}{2{(z^3+1)}^2})dz$$$$I=\frac{9}{2}{\int }_0^{\mathrm{\infty }}\frac{dz}{{(z^3+1)}^2}=\frac{9}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{(1+\frac{1}{z^3})}\left(\frac{dz}{z^2}\right)$$$$I={\int }_0^{\mathrm{\infty }}\frac{3z}{1+z^3}dz=-\frac{1}{2\pi i}.\underset{R\to \mathrm{\infty }}{\lim }{\int }_{C_R}\frac{3s\ln s}{s^3+1}ds$$$$I=-\frac{1}{2\pi i}.\underset{R\to \mathrm{\infty }}{\lim }{\int }_{C_R}\frac{3s\ln s}{s^3+1}ds=-\underset{k=0}{\sum ^2}\frac{(2k+1)\pi i}{3}{e}^{-(2k+1)\pi i/3}$$$$I=\frac{2\pi }{\sqrt{3}}=\underset{0}{\stackrel{3}{\int }}\sqrt[3]{x^2(3-x)}dx$$

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