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

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.

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

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

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

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.