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I have to evaluate this integral:

$$ \int_0^4 \int_\sqrt{y}^2 y^2 {e}^{x^7} \operatorname d\!x \operatorname d\!y\, $$

I have no idea what to do with $\;{e}^{x^7}$.

I have even tried $\int{e}^{x^7} dx$ with WolframAlpha, but it gives me something with $\;\Gamma\;$ and I don't know what to do with that.

I tried posing $\;u = x^7\;$ and doing another change of variables. I got $\;445 {e}^{128}/9408\;$, but I'm not really sure about it.

If anyone could at least point me in the right direction, it would be awesome! Thanks.

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2 Answers 2

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The interchange of the order of integration is justified by Fubini's theorem:

$$ \int_0^4 \int_\sqrt{y}^2 y^2 {e}^{x^7} dxdy=\int_0^2\int_0^{x^2}y^2 {e}^{x^7} dydx=\frac 1 3\int_0^2 x^6{e}^{x^7} dx=\frac1{21}{e}^{x^7} \bigg|_0^2=\frac {{e}^{2^7}-1}{21} $$

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  • $\begingroup$ That's pretty close to $\;445 {e}^{128}/9408\;$ when you evaluate them. Does this mean anything? $\endgroup$ Feb 15, 2014 at 21:36
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    $\begingroup$ yes: if your result is not $=$ to it, it must be wrong ;) $\endgroup$ Feb 15, 2014 at 21:53
  • $\begingroup$ Ha! Yeah, that's a good point. $\endgroup$ Feb 15, 2014 at 21:58
  • $\begingroup$ I just did it; I got it! Thanks a lot! $\endgroup$ Feb 15, 2014 at 22:11
  • $\begingroup$ You're welcome. $\endgroup$
    – user63181
    Feb 15, 2014 at 22:16
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Change the order of integration; this leads to

$$\int_0^2 \int_0^{x^2} y^2 e^{x^7} dy dx = \frac 1 3\int_0^2 x^6 e^{x^7} dx$$

which is an easy integral.

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