Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^x}\sin y{\rm{d}}A \simeq 1.87131$
Work Step by Step
We have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^x}\sin y{\rm{d}}A$ and ${\cal R} = \left[ {0,2} \right] \times \left[ {0,\frac{\pi }{4}} \right]$.
Write
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^x}\sin y{\rm{d}}A = \mathop \smallint \limits_{y = 0}^{\pi /4} \left( {\mathop \smallint \limits_{x = 0}^2 {{\rm{e}}^x}{\rm{d}}x} \right)\sin y{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^{\pi /4} \left( {{{\rm{e}}^x}|_0^2} \right)\sin y{\rm{d}}y$
$ = \left( {{{\rm{e}}^2} - 1} \right)\mathop \smallint \limits_{y = 0}^{\pi /4} \sin y{\rm{d}}y$
$ = \left( {{{\rm{e}}^2} - 1} \right)\left( { - \cos y|_0^{\pi /4}} \right)$
$ = \left( {{{\rm{e}}^2} - 1} \right)\left( { - \frac{1}{2}\sqrt 2 + 1} \right)$
$ \simeq 1.87131$
