Question

Find the area bounded by the parabola $x^2=\left(\frac{4}{3}\right)y$ and the line $3x-2y+12=0$

Answer 1

$\begin{array}{c}f\left(x\right):-3x-2y+12=0\Rightarrow 2y=3x+12\to \left(i\right)\\ g\left(x\right):-x^2=\frac{4}{3}y\Rightarrow y=\frac{3x^2}{4}\to \left(ii\right)\end{array}$

By equating equation (i) and (ii), we get

$\begin{array}{c}\frac{3x+12}{2}=\frac{3x^2}{4}\\ \Rightarrow 6x+24=3x^2\\ \Rightarrow 3x^2-6x-24=0\\ \Rightarrow x^2-2x-8=0\\ \Rightarrow x^2+2x-4x-8=0\\ \Rightarrow x\left(x+2\right)-4\left(x+2\right)=0\\ \left(x+2\right)\left(x-4\right)=0\\ \therefore x=-2and4\\ Hence,y=\left(3and2\right)\end{array}$

$\therefore$ Point of intersection are $(-2,3)$ and $(4,12)$

Hence, the required Area is

$\begin{array}{c}A={\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]dx\\ ={\int }_{-2}^4\left[\frac{3x+12}{2}-\frac{3x^2}{4}\right]dx\\ ={\int }_{-2}^4\left[\frac{3x}{2}+6-\frac{3}{4}{x}^2\right]dx\\ ={\left[\frac{3x^2}{4}+6x-\frac{3x^3}{12}\right]}_{-2}^4\\ =\left[12+24-16-3+12-2\right]\\ =\left[48-21\right]=27sq.units.\end{array}$