Here is what I got for this question :
$ \int_1^2 \int_1^3 xy^3e^{x^2y^2} dx dy$
$= \frac{1}{2} \int_1^2 y\int_1^3 2xy^2e^{x^2y^2} dx dy$
$= \frac{1}{2} \int_1^2 y\int_1^3 2xy^2e^{x^2y^2} dx dy$
$= \frac{1}{2} \int_1^2 y\int_a^b e^{u} du dy$
$= \frac{1}{2} \int_1^2 y \cdot [e^{x^2y^2}] |_1^2 dy$
$= \frac{1}{2} \int_1^2 y \cdot [[e^{4y^2}] - [e^{y^2}] |_1^2 dy$
$= \frac{1}{2} \int_1^2 y \cdot e^{3y^2} dy$
$= \frac{1}{2} \frac{1}{6}\int_1^2 6y \cdot e^{3y^2} dy$
$= \frac{1}{2} \frac{1}{6}\int_a^b e^{u} du $
$= \frac{1}{2} \frac{1}{6} [e^{3y^2}] |_1^2 dy $
$= \frac{1}{2} \frac{1}{6} ([e^{12}] - [e^{3}])$
$ = \frac{e^12 - e^3}{12}$
However the answer in the back says
$ \frac{1}{36}(e^{36}-e^9) - \frac{1}{4}(e^4 - e)$
Where did I go wrong while computing this integral ?
$e^{4y^2}-e^{y^2}\ne e^{3y^2}$ Also what happened to the upper limit of $3$? Magically changed to $2$