Answer
$$\frac{7}{{24}}$$
Work Step by Step
$$\eqalign{
& \int_1^{3/2} {\int_y^{3 - y} y dxdy} \cr
& = \int_1^{3/2} {\left[ {\int_y^{3 - y} y dx} \right]dy} \cr
& {\text{solve the inner integral and treat }}y{\text{ as a constant}} \cr
& \int_y^{3 - y} y dx = y\int_y^{3 - y} {dx} \cr
& {\text{using the power rule }} \cr
& = y\left( x \right)_y^{3 - y} \cr
& {\text{evaluating the limits in the variable }}x \cr
& = y\left( {3 - y - y} \right) \cr
& {\text{simplifying}} \cr
& = 3y - 2{y^2} \cr
& {\text{then}} \cr
& \int_1^{3/2} {\left[ {\int_y^{3 - y} y dx} \right]dy} = \int_1^{3/2} {\left( {3y - 2{y^2}} \right)dy} \cr
& {\text{integrating}} \cr
& = \left[ {\frac{{3{y^2}}}{2} - \frac{{2{y^3}}}{3}} \right]_1^{3/2} \cr
& = \left( {\frac{{3{{\left( {3/2} \right)}^2}}}{2} - \frac{{2{{\left( {3/2} \right)}^3}}}{3}} \right) - \left( {\frac{{3{{\left( 1 \right)}^2}}}{2} - \frac{{2{{\left( 1 \right)}^3}}}{3}} \right) \cr
& = \frac{9}{8} - \frac{5}{6} \cr
& = \frac{7}{{24}} \cr} $$
