Question

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

Answer 1

The area enclosed between the parabola, $4y=3x^2$ and the line, $2y=3x+12$ is represented by the shaded are $OBAO$
The points of intersection of the given curves are $A(-2,3)$ and $(4,12)$.
We draw $AC$ and $BD$ perpendicular to $x$-axis.

$\therefore AreaOBAO=AreaCDBA-(AreaODBO+AreaOACO)$

$={\int }_{-2}^4\frac{1}{2}(3x+12)dx-{\int }_{-2}^4\frac{3x^2}{4}dx$

$=\frac{1}{2}{[\frac{3x^2}{2}+12x]}_{-2}^4-\frac{3}{4}{\left[\frac{x^3}{3}\right]}_{-2}^4$

$=\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8]$

$=\frac{1}{2}\left[90\right]-\frac{1}{4}\left[72\right]$

$=45-18$

$=27$ sq. units