TITLE: Integral $\int_0^4 \int_\sqrt{y}^2 y^2 {e}^{x^7} \operatorname d\!x \operatorname d\!y\,$
QUESTION [5 upvotes]: 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.

REPLY [9 votes]: The interchange of the order of integration is justified by Fubini's theorem:
$$
\int_0^4 \int_\sqrt{y}^2 y^2 {e}^{x^7} dxdy=\int_0^2\int_0^{x^2}y^2 {e}^{x^7} dydx=\frac 1 3\int_0^2 x^6{e}^{x^7} dx=\frac1{21}{e}^{x^7} \bigg|_0^2=\frac {{e}^{2^7}-1}{21} 
$$

REPLY [5 votes]: 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.