Question

Using intergration find the area of the region included between the parabola $4y={3x}^2$ and the line 3x=2y+12=0.

Answer 1

Given equation of line is, $3x-2y+120$

$\Rightarrow y=\frac{3x+12}{2}$ ………….$\left(1\right)$

Equation of parabola is, $3x^2=4y$ …………..$\left(2\right)$

Substitute value of y from equation $\left(1\right)$ in equation $\left(2\right)$,

$3x^2=4\left(\frac{3x+12}{2}\right)$

$3x^2=6x+24$

$\Rightarrow x^2=2x+28$

$\Rightarrow (x-4)(x+2)=0$

$\Rightarrow x=4$ and $x=-2$

For $x=4$, $y=\frac{3\times 4+12}{2}=12\Rightarrow y=12$

For $x=-2$, $y=\frac{3\times (-2)+12}{2}=3\Rightarrow y=3$

Points of intersection are $(4,12)$ and $(-2,3)$

Draw graph of line $3x-4y+12=0$ & parabola $3x^2=4y$,

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

$=\frac{3}{4}{\int }_{-2}^4(2x+8-x^2)dx$

$=\frac{3}{4}{[x^2+8x-\frac{x^3}{3}]}_{-2}^4$

$=\frac{3}{4}[(16+32-\frac{64}{3})-(4-16+\frac{8}{3})]$

Required area$=27$ sq.units.