Answer
$$4$$
Work Step by Step
$$\eqalign{
& \int_1^3 {\int_0^{\pi /2} {x\sin y} } dydx \cr
& = \int_1^3 {\left[ {\int_0^{\pi /2} {x\sin y} dy} \right]} dx \cr
& = \int_1^3 {x\left[ {\int_0^{\pi /2} {\sin y} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& \int_0^{\pi /2} {\sin y} dy = \left[ { - \cos y} \right]_0^{\pi /2} \cr
& = \left[ {\cos y} \right]_{\pi /2}^0 \cr
& {\text{evaluating the limits}} \cr
& = \cos 0 - cos\pi /2 \cr
& {\text{simplifying}} \cr
& = 1 \cr
& \cr
& = \int_1^3 {x\left[ {\int_0^{\pi /2} {\sin y} dy} \right]} dx = \int_1^3 {x\left( 1 \right)} dx \cr
& = \int_1^3 x dx \cr
& {\text{integrating}} \cr
& = \left( {\frac{{{x^2}}}{2}} \right)_1^3 \cr
& {\text{evaluate}} \cr
& = \frac{{{{\left( 3 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^2}}}{2} \cr
& = \frac{9}{2} - \frac{1}{2} \cr
& = 4 \cr} $$
