\(\text{Given integral,}\)

\(\int_0^3\int_{-2}^1 (2x+3y)dxdy\)

\(\text{Now}\)

\(\int_0^3\int_{-2}^1 (2x+3y)dxdy=\int_0^3(\frac{2x^2}{2}+3xy)_{-2}^1 dy\)

\(=\int_0^3 (x^2+3xy)_{-2}^1 dy\)

\(=\int_0^3 ((1)^2+3(1)y-(-2)^2-3(-2)y)dy\)

\(=\int_0^3 (1+3y-4+6y)dy\)

\(=\int_0^3 (-3+9y)dy\)

\(=(-3y+\frac{9y^2}{2})_0^3\)

\(=(-3(3)+\frac{9(3)^2}{2}+3(0)+\frac{9(0)^2}{2})\)

\(=-9+\frac{81}{2}+0\)

\(=\frac{63}{2}\)

\(\text{Final answer:}\)

\(\int_0^3\int_{-2}^1 (2x+3y)dxdy=\frac{63}{2}\)

\(\int_0^3\int_{-2}^1 (2x+3y)dxdy\)

\(\text{Now}\)

\(\int_0^3\int_{-2}^1 (2x+3y)dxdy=\int_0^3(\frac{2x^2}{2}+3xy)_{-2}^1 dy\)

\(=\int_0^3 (x^2+3xy)_{-2}^1 dy\)

\(=\int_0^3 ((1)^2+3(1)y-(-2)^2-3(-2)y)dy\)

\(=\int_0^3 (1+3y-4+6y)dy\)

\(=\int_0^3 (-3+9y)dy\)

\(=(-3y+\frac{9y^2}{2})_0^3\)

\(=(-3(3)+\frac{9(3)^2}{2}+3(0)+\frac{9(0)^2}{2})\)

\(=-9+\frac{81}{2}+0\)

\(=\frac{63}{2}\)

\(\text{Final answer:}\)

\(\int_0^3\int_{-2}^1 (2x+3y)dxdy=\frac{63}{2}\)