Answer
$$34$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( {5x + 8y} \right)}dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 3,\,\,\,\,\,\,\,0 \leqslant y \leqslant x - 1 \cr
& {\text{replacing the limits for the region }}R \cr
& \int_1^3 {\int_0^{x - 1} {\left( {5x + 8y} \right)dy} dx} \cr
& = \int_1^3 {\left[ {\int_0^{x - 1} {\left( {5x + 8y} \right)dy} } \right]dx} \cr
& {\text{solve the inner integral by treating }}x{\text{ as a constant}} \cr
& \int_0^{x - 1} {\left( {5x + 8y} \right)dy} \cr
& {\text{integrate using the power rule}} \cr
& = \left[ {5xy + 4{y^2}} \right]_0^{x - 1} \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left[ {5x\left( {x - 1} \right) + 4{{\left( {x - 1} \right)}^2}} \right] - \left[ {5x\left( 0 \right) + 4{{\left( 0 \right)}^2}} \right] \cr
& = 5{x^2} - 5x + 4\left( {{x^2} - 2x + 1} \right) \cr
& = 5{x^2} - 5x + 4{x^2} - 8x + 4 \cr
& = 9{x^2} - 13x + 4 \cr
& {\text{then}} \cr
& \cr
& \int_1^3 {\left[ {\int_0^{x - 1} {\left( {5x + 8y} \right)dy} } \right]dx} = \int_1^3 {\left( {9{x^2} - 13x + 4} \right)dx} \cr
& {\text{integrating}} \cr
& = \left( {3{x^3} - \frac{{13{x^2}}}{2} + 4x} \right)_1^3 \cr
& = \left( {3{{\left( 3 \right)}^3} - \frac{{13{{\left( 3 \right)}^2}}}{2} + 4\left( 3 \right)} \right) - \left( {3{{\left( 1 \right)}^3} - \frac{{13{{\left( 1 \right)}^2}}}{2} + 4\left( 1 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{69}}{2} - \frac{1}{2} \cr
& = 34 \cr} $$
