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Calculate: $$\int_0^1 \int_0^{x^3} e^\frac{y}{x} dydx$$ Obviously i need to change it to $dxdy$ thus i need to change the boundaries of the second integral but how to do that in this case?

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    $\begingroup$ I think you'll find this problem more difficult if you switch the order of integration $\endgroup$ – jameselmore Aug 20 '15 at 14:39
  • $\begingroup$ Boundaries are fine. $\endgroup$ – Anthony Aug 20 '15 at 14:40
  • $\begingroup$ Thus should i use polar coordinates in this integral or just leave it as it is? $\endgroup$ – mkropkowski Aug 20 '15 at 14:42
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    $\begingroup$ As is. It's fairly straightforward, although it looks complicated. $\endgroup$ – Daniel Fischer Aug 20 '15 at 14:45
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As presented the integral may be best suited. By integrating $y$ first makes $x$ a "constant" in that particular integral. Alternatively if $x$ is is integrated first a change of variable will need to be made, ie $x \to \frac{1}{t}$ as well as changing the integration limits.

\begin{align} \int_0^1 \, \int_0^{x^3} e^{\frac{y}{x}} \, dy \, dx &= \int_{0}^{1} \left[ x \, e^{\frac{y}{x}} \right]_{0}^{x^{3}} \, dx \\ &= \int_{0}^{1} x \, (e^{x^{2}} -1) \, dx \\ &= \int_{0}^{1} \frac{d}{dx} \left[ \frac{1}{2} \, e^{x^{2}} - \frac{x^{2}}{2} \right] \, dx \\ &= \frac{1}{2} \, \left[ e^{x^{2}} - x^{2} \right]_{0}^{1} \\ &= \frac{e - 2}{2}. \end{align}

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