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Question Number 50416 by Abdo msup. last updated on 16/Dec/18

calculate ∫_0 ^1 ^3 (√(x^2 (1−x^3 )))dx

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)}{dx} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

∫_0 ^1 {x^2 (1−x^3 )}^(1/3) dx t=x^3 dt=3x^2 dx (dt/(3(t)^(2/3) ))=dx ∫_0 ^1 (({t^(2/3) (1−t)}^(1/3) ×(dt/(3t^(2/3) )))/1) (1/3)∫_0 ^1 t^((2/9)−(2/3)) (1−t)^(1/3) dt (1/3)∫_0 ^1 t^((2−6)/9) (1−t)^(1/3) dt wait now use beta function... (1/3)∫t^((5/9)−1) (1−t)^((4/3)−1) dt formula ∫_0 ^1 x^(m−1) (1−x)^(n−1) dx=β(m,n) =((⌈(m)⌈(n))/(⌈(m+n))) answer is =(1/3)×((⌈((5/9))⌈((4/3)))/(⌈((5/9)+(4/3)))) =(1/3)×((⌈((5/9))⌈((4/3)))/(⌈(((17)/9)))) wait busy...

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{3}} \right)\right\}^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\ $$$${t}={x}^{\mathrm{3}} \:\:\:{dt}=\mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$$\frac{{dt}}{\mathrm{3}\left({t}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }={dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left\{{t}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}−{t}\right)\right\}^{\frac{\mathrm{1}}{\mathrm{3}}} ×\frac{{dt}}{\mathrm{3}{t}^{\frac{\mathrm{2}}{\mathrm{3}}} }}{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{2}}{\mathrm{9}}−\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{2}−\mathrm{6}}{\mathrm{9}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dt} \\ $$$${wait}\:{now}\:{use}\:{beta}\:{function}... \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int{t}^{\frac{\mathrm{5}}{\mathrm{9}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}} {dt} \\ $$$${formula}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{m}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} {dx}=\beta\left({m},{n}\right) \\ $$$$=\frac{\lceil\left({m}\right)\lceil\left({n}\right)}{\lceil\left({m}+{n}\right)} \\ $$$${answer}\:{is}\:=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\lceil\left(\frac{\mathrm{5}}{\mathrm{9}}\right)\lceil\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}{\lceil\left(\frac{\mathrm{5}}{\mathrm{9}}+\frac{\mathrm{4}}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\lceil\left(\frac{\mathrm{5}}{\mathrm{9}}\right)\lceil\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}{\lceil\left(\frac{\mathrm{17}}{\mathrm{9}}\right)} \\ $$$${wait}\:{busy}... \\ $$$$ \\ $$

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