Originally Posted by
Amer $\displaystyle \int_{1<y<2}\int_{y<x<y^3}-e^{\frac{x}{y}}dxdy$
that is your integral right
it is the same as
$\displaystyle \int_{1}^{2}\int_{y}^{y^3}-e^{\frac{x}{y}} dx dy $
first we integrate with respect to x and y constant
$\displaystyle \int_{1}^{2}-ye^{\left(\dfrac{x}{y}\right)}\mid_{y}^{y^3}$
sub now
$\displaystyle \int_{1}^{2}-y(e^{\left(\frac{y}{y}\right)}-e^{\left(\frac{y^3}{y}\right)})dy$
$\displaystyle \int_{1}^{2}-y(e^{1}-e^{y^2})dy$
$\displaystyle \int_{1}^{2}-y(e) dy+ \int_{1}^{2}ye^{y^2} dy $
the first one is easy the second let u=y^2 and you continue