Answer
$\frac{3}{4}{e^4} - \frac{3}{8}$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{e^{2y/x}}}dydx \cr
& R{\text{ is bounded by }}{x^2},{\text{ }}y = 0{\text{ and }}x = 2. \cr
& {\text{The region }}R{\text{ of integration is given by}} \cr
& {\text{See graph below}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 2,{\text{ 0}} \leqslant y \leqslant {x^2}} \right\} \cr
& {\text{Therefore}}{\text{,}} \cr
& \iint\limits_R {{e^{2y/x}}}dydx = \int_0^2 {\int_0^{{x^2}} {{e^{2y/x}}} dy} dx \cr
& {\text{Integrating}} \cr
& = \int_0^2 {\left[ {\frac{x}{2}{e^{2y/x}}} \right]_0^{{x^2}}} dx \cr
& = \int_0^2 {\left[ {\frac{x}{2}{e^{2\left( {{x^2}} \right)/x}} - \frac{x}{2}{e^{2\left( 0 \right)/x}}} \right]} dx \cr
& = \int_0^2 {\left( {\frac{x}{2}{e^{2x}} - \frac{x}{2}} \right)} dx \cr
& = \frac{1}{2}\left[ {\frac{1}{2}x{e^{2x}} - \frac{1}{4}{e^{2x}} - \frac{{{x^2}}}{4}} \right]_0^2 \cr
& {\text{Evaluating}} \cr
& = \frac{1}{2}\left[ {\frac{1}{2}\left( 2 \right){e^{2\left( 2 \right)}} - \frac{1}{4}{e^{2\left( 2 \right)}} - \frac{{{{\left( 2 \right)}^2}}}{4}} \right] - \frac{1}{2}\left[ {\frac{1}{2}\left( 0 \right){e^{2\left( 0 \right)}} - \frac{1}{4}{e^{2\left( 0 \right)}} - \frac{{{{\left( 0 \right)}^2}}}{4}} \right] \cr
& = \frac{1}{2}\left[ {{e^4} - \frac{1}{4}{e^4} - 1} \right] - \frac{1}{2}\left[ { - \frac{1}{4}} \right] \cr
& = \frac{3}{4}{e^4} - \frac{3}{8} \cr} $$
