Question

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

Answer 1

Given equation of parabola $4y$ = $3x^2$
$\Rightarrow y=\frac{3x^2}{4}\dots$ $\left(i\right)$
and the line $3x-2y+12=0$
$\Rightarrow y=\frac{3x+12}{2}\dots$ $\left(ii\right)$
The lineinteresect the parabola at $(-2,3)$ and $(4,12)$.
Hence, the required area will be the shaded region.



Required area = ${\int }_{-2}^4\frac{3x+12}{2}dx-{\int }_{-2}^4\frac{3x^2}{4}dx$
= ${[\frac{3}{4}{x}^2+6x-\frac{x^3}{4}]}_{-2}^4$
$\text{= (12 + 24 - 16) - (3 - 12 + 2) = 20 + 7 = 27 square units.}$