Question

Find the area enclosed between the parabola $4y=3x^2$ and the straight line $3x-2y+12=0$

Answer 1

Here, $3x-2y+12=0$ $\Rightarrow$ $y=\frac{3x+12}{2}$

To find the intersection of parabola and the line, we substitute $y=\frac{3x^2}{4}$ in the line equation.

$\therefore 3x-2\times \frac{3x^2}{4}+12=0$

$\Rightarrow 3x-\frac{3x^2}{2}+12=0$

$\Rightarrow 6x-3x^2+24=0or{x}^2-2x-8=0$

$\Rightarrow x=-2,4$

The area enclosed by the curves is given by

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

$=\frac{1}{4}{\int }_{-2}^4(6x+24-3x^2)dx$

$=\frac{1}{4}[3x^2+24x-x^3{]}_{-2}^4$

$=\frac{1}{4}\left[\right(3\times 16+24\times 4-64)-(3\times 4-48+8\left)\right]$

$=\frac{1}{4}\times [48+96-64-12+48-8]$

$=\frac{1}{4}\times 108$

$=27$ sq units