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Compute: $\int_{0}^{1}\int_{x}^{1} e^{x/y}$

I just don't know how to compute this integral. I tried u = x/y, but that didn't really lead me anywhere. It was suggested by fellows on a IRC that I graph this, but I didn't really understand. Could anyone expand on this?

Ozera
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2 Answers2

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Hint: By changing the order of integration:

$$\int_0^1 \int_x^1 e^{x/y} dy dx = \int_0^1 \int_0^y e^{x/y} dx dy$$

Now the inside integral can be evaluated by

$$\int_0^y e^{x/y} dx = y e^{x/y} |_0^y = y e^1 - y e^0$$

  • I just learned how to compute Double integrals today, sorry, but how/why are you able to switch the order of integration? – Ozera Aug 22 '13 at 23:38
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    @Ozera Draw a picture of the region that you're integrating over, and check that both integrals describe the same region. Then we can reverse the order using Fubini's Theorem. –  Aug 22 '13 at 23:42
  • I drew the graph, but why is the bounds not: $\int_{0}^{1}\int_{y}^{0}$ ? Is it because since $x=1$ is the farthest line out, to find the difference we would subtract it by $x=y$ ? I'm not sure if that same argument works for when choosing bounds in a dy case though... – Ozera Aug 22 '13 at 23:54
  • @Ozera Did you draw the region and shade in the interval that you're integrating over? – rurouniwallace Aug 23 '13 at 00:02
  • @Ataraxia Sorry, I drew but did not shade and consequently misunderstood my graph. I understand why the bounds are the way they are now. Thank you all. – Ozera Aug 23 '13 at 00:03
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$$ \begin{align} \int_0^1\int_x^1e^{x/y}\,\mathrm{d}y\,\mathrm{d}x &=\int_0^1\int_x^1e^t\frac{x}{t^2}\,\mathrm{d}t\,\mathrm{d}x&&y=\frac xt\\ &=\int_0^1\int_0^te^t\frac{x}{t^2}\,\mathrm{d}x\,\mathrm{d}t&&\text{swap order}\\ &=\int_0^1e^t\frac{\frac12t^2}{t^2}\,\mathrm{d}t&&\text{inner integral}\\[4pt] &=\frac12(e-1)&&\text{outer integral} \end{align} $$

robjohn
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