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If $$f(x) = \int_x^2{\frac{dy}{\sqrt{1+y^3}}}$$

then find the value of $$\int_0^2{xf(x)}dx$$

I have no idea how to solve this question. Please help.

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    $\begingroup$ Integrate by part would be work. $\endgroup$ – C.F.G Oct 6 '17 at 10:47
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The integration domain can be equivalently written as

$$ \Omega = \{(x,y): x<y<2 ~~\mbox{and}~~~ 0 < x < 2 \} $$ or

$$ \Omega = \{(x,y): 0<x<y ~~\mbox{and}~~~ 0 < y < 2 \} $$

Such that

\begin{eqnarray} \int_0^2{\rm d}x\int_{x}^2{\rm d}y ~\frac{x}{\sqrt{1 + y^3}} &=& \int_0^2{\rm d}y\int_{0}^y{\rm d}x ~\frac{x}{\sqrt{1 + y^3}} \\ &=& \int_0^2{\rm d}y ~\frac{y^2}{2}\frac{1}{\sqrt{1 + y^3}} \\ &=& \frac{1}{2} \int_0^2{\rm d}y \frac{y^2}{\sqrt{1 + y^3}} \\ &=&\frac{1}{2}\times\frac{4}{3} = \frac{2}{3} \end{eqnarray}

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  • $\begingroup$ Thanks @caverac $\endgroup$ – MathsLearner Oct 6 '17 at 11:08
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Using integration by parts

$$\int_0^2xf(x)dx=\frac{1}{2}\int_{x=0}^{x=2}f(x)d(x^2)$$ $$=\frac{1}{2}x^2f(x)\bigg|_0^2-\frac{1}{2}\int_0^2x^2f'(x)dx$$ $$=2f(2)+\frac{1}{2}\int_0^2x^2\frac{1}{\sqrt{1+x^3}}dx$$ $$=0+\frac{1}{3}\sqrt{1+x^3}\bigg|_0^2$$ $$=\frac{1}{3}(3-1)$$ $$=\frac{2}{3}$$

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  • $\begingroup$ Thanks @velut luna $\endgroup$ – MathsLearner Oct 6 '17 at 11:08

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