Answer
$$\frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {xy\sin {x^2}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \sqrt {\pi /2} ,\,\,\,\,0 \leqslant y \leqslant 1} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^1 {\int_0^{\sqrt {\pi /2} } {xy\sin {x^2}} } dxdy \cr
& = \int_0^1 {\left[ {\int_0^{\sqrt {\pi /2} } {xy\sin {x^2}} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = \frac{y}{2}\int_0^{\sqrt {\pi /2} } {\sin {x^2}\left( {2x} \right)} dx \cr
& {\text{integrating}} \cr
& = \frac{y}{2}\left( { - \cos {x^2}} \right)_0^{\sqrt {\pi /2} } \cr
& {\text{evaluating the limits for the variable }}x \cr
& = - \frac{y}{2}\left( {\cos {{\left( {\sqrt {\pi /2} } \right)}^2} - \cos {{\left( 0 \right)}^2}} \right) \cr
& {\text{simplifying}} \cr
& = - \frac{y}{2}\left( {0 - 1} \right) \cr
& = \frac{y}{2} \cr
& \cr
& = \int_0^1 {\left[ {\int_0^{\sqrt {\pi /2} } {xy\sin {x^2}} dx} \right]} dy = \int_0^1 {\frac{y}{2}} dy \cr
& = \frac{1}{2}\int_0^1 y dy \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {\frac{{{y^2}}}{2}} \right)_0^1 \cr
& {\text{evaluate}} \cr
& = \frac{1}{4}\left( {{{\left( 1 \right)}^2} - {{\left( 0 \right)}^2}} \right) \cr
& = \frac{1}{4} \cr} $$
