- #1
springo
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Homework Statement
[tex]\int_{0}^{3}\sqrt[3]{\frac{3-x}{x^2}}\: \mathrm{d}x[/tex]
2. The attempt at a solution
I'm pretty sure I have to use Euler's Beta function, so I tried to change the limits to 0 and 1 by setting x = 3·u (so dx = 3·du). However there must be some mistake when I did it because I checked and I did it wrong:
[tex]\int_{0}^{3}\sqrt[3]{\frac{3-x}{x^2}}\: \mathrm{d}x=\int_{0}^{1}\sqrt[3]{\frac{3-3u}{(3u)^2}}\: 3\cdot\mathrm{d}u=3^{\frac{2}{3}}\cdot\beta\left(\frac{1}{3},\frac{4}{3}\right)[/tex]
Thanks a lot for your help.
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