The area cut off the parabola $4y=3x^2$ by the straight line $2y=3x+12$ is:

1. Find intersection points:From line: $2y=3x+12⟹4y=6x+24$.Equate to parabola: $3x^2=6x+24$.$3x^2-6x-24=0⟹x^2-2x-8=0$.$(x-4)(x+2)=0⟹x=4,-2$.2. Setup integral for area:Area $={\int }_{-2}^4(y_{\text{line}}-y_{\text{parabola}})dx$ Area $={\int }_{-2}^4(\frac{3x+12}{2}-\frac{3x^2}{4})dx$ Area $=[\frac{3x^2}{4}+6x-\frac{x^3}{4}{]}_{-2}^4$ Upper limit (4): $\frac{3(16)}{4}+6(4)-\frac{64}{4}=12+24-16=20$.Lower limit (-2): $\frac{3(4)}{4}+6(-2)-\frac{-8}{4}=3-12+2=-7$.Area $=20-(-7)=27\text{sq. units}$.