Question

Find the area cut off from the parabola $4y=3x^2$ with the line $2y=3x+12$.

Answer 1

$4y=3x^2$ and $2y=3x+12$
$4y=3x^2$
$\Rightarrow 2y=\frac{3}{2}{x}^2$
$\Rightarrow 3x+12=\frac{3}{2}{x}^2$
$\Rightarrow 6x+24=3x^2$
$\Rightarrow 2x+8=x^2$
$\Rightarrow x^2-2x-8=0$
$x=\frac{2\pm \sqrt{4+32}}{2}=\frac{2\pm 6}{2}$
$\therefore x=4,-2$
and $y=12,3$
$\therefore$ the line $2y=3x+12$ intersects the parabola $4y=3x^2$ at point $(-2,3)$ and $(4,12)$
$\therefore$ the area enclosed between the parabola and line is shown in the figure as the shaded portion.
$Area={\int }_a^b\left[f\right(x)-g(x\left)\right]dx$
$a=-2b=4f\left(x\right)=\frac{3x+12}{2}g\left(x\right)=\frac{3x^2}{4}$
$\therefore A={\int }_{-2}^4[\left[\frac{3x+12}{2}\right]-\left(\frac{3x^2}{4}\right)]dx={\int }_{-2}^4\frac{3x+12}{2}dx-{\int }_{-2}^4\frac{3x^2}{4}dx$
On integrating we get
$A=\frac{1}{2}{[\frac{3x^2}{2}+\frac{12x}{4}]}_2^4-\frac{3}{4}{\left[\frac{x^3}{3}\right]}_{-2}^4$
$=\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8]$
$=45-18=27squnits$.