Find the area enclosed by the parabola `4y=3x^2` and the line`2y = 3x + 12`.

Explanation

The points of intersection of the parabola `4y=3x^(2)` and line `2y=3x+12` are C(-2, 3) and B(4, 12).
Draw the graphs of the parabola `3x^(2)=4y` and the straight line `2y=3x+12.`

Now, required area
`=ar(OBCO)=ar(OABCDO)-ar(OABOCDO)`
`=int_(-2)^(4)y_(1)dx-int_(-2)^(4)y_(2)dx`
where `y_(1)` and `y_(2)` are the values of y along the line and parabola respectively.
`=int_(-2)^(4)(3x-12)/(2)dx-int_(-2)^(4)(3x^(2))/(4)dx`
`=(1)/(2)[(3x^(2))/(2)+12x]_(-2)^(4)-(1)/(4)[x^(3)]_(-2)^(4)`
`=(1)/(2)[(24+48)-(6-24)]-(1)/(4)[64+8]`
`=45-18=27` sq. units.