## Thomas' Calculus 13th Edition

$e-1$
We reverse the order of integration to get: $I= \int_0^{1} \int_0^{3y^2} e^{y^3} dx dy$ or, $= \int_0^{1} 3y^2e^{y^3} dy$ Set $a = y^3 \implies da=3y^2 dy$ Now, $I=\int_0^{1} e^{a} da=[e^a]_0^{1}=e^1-e^0=e-1$