Answer
$5y +\displaystyle \frac{25}{2}e^{y},\qquad\frac{1}{2}+ex -x$
Work Step by Step
Working on $\displaystyle \int f(x,y)dx$, treat y as a constant.
$\displaystyle \int_{0}^{5} (y +xe^{y})dx=[xy +\displaystyle \frac{x^{2}}{2}e^{y}]_{x=0}^{x=5}$
$=(5y +\displaystyle \frac{25}{2}e^{y})-(0+0)$
$=5y +\displaystyle \frac{25}{2}e^{y}$,
Working on $\displaystyle \int f(x,y)dy$, treat $x$ as a constant.
$\displaystyle \int_{0}^{1} (y +xe^{y})dy =[\displaystyle \frac{y^{2}}{2}+xe^{y}]_{y=0}^{y=1}$
$=(\displaystyle \frac{1}{2}+xe^{1})-(0+xe^{0})$
$=\displaystyle \frac{1}{2}+ex -x$
