Answer
$$15$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {4{x^3}\cos y}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):1 \leqslant x \leqslant 2,\,\,\,\,0 \leqslant y \leqslant \pi /2} \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& \int_1^2 {\int_0^{\pi /2} {4{x^3}\cos y} dy} dx \cr
& = \int_1^2 {\left[ {\int_0^{\pi /2} {4{x^3}\cos y} dy} \right]} dx \cr
& = \int_1^2 {4{x^3}\left[ {\int_0^{\pi /2} {\cos y} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& \int_0^{\pi /2} {\cos y} dy \cr
& = \left[ {\sin y} \right]_0^{\pi /2} \cr
& {\text{evaluating the limits for the variable }}y \cr
& = \sin \left( {\frac{\pi }{2}} \right) - \sin \left( 0 \right) \cr
& = 1 \cr
& {\text{simplifying}} \cr
& \int_1^2 {4{x^3}\left[ {\int_0^{\pi /2} {\cos y} dy} \right]} dx = \int_1^2 {4{x^3}\left( 1 \right)} dx \cr
& = \int_1^2 {4{x^3}} dx \cr
& {\text{integrating}} \cr
& = \left( {{x^4}} \right)_1^2 \cr
& {\text{evaluate}} \cr
& = {\left( 2 \right)^4} - {\left( 1 \right)^4} \cr
& = 16 - 1 \cr
& = 15 \cr} $$
