Question

Find the area enclosed by the parabola $4y=3x^2$ and the line 2y = 3x+12. ?

Answer 1

Given the curve (represents an upward parabola with vertex (0, 0))

$4y=3x^2$ ...(i)

and equation of line $2y=3x+12$ ...(ii)

For intersection point from Eqs. (i) and (ii), we get

$2(3x+12)=3x^2\Rightarrow 3x^2-6x-24=0\Rightarrow x^2-2x-8=0$

$\Rightarrow (x-4)(x+2)=0\Rightarrow x=4,-2$
When $x=4,y=\frac{3\times 4+12}{2}=12$
When $x=-2,y=\frac{3\times (-2)+12}{2}=3$
Thus, intersection points are (-2, 3) and (4, 12).
Required area (shown in shaded region)
= (area under the line between x = - 2 and x = 4)
= (Area under the parabola between x = - 2 and x = 4)

$={\int }_{-2}^4(\frac{3x+12}{2}-\frac{3x^2}{4})dx=\left[\frac{1}{2}\right(\frac{3x^2}{2}+12x)-\frac{3x^3}{4\times 3}{]}_{-2}^4=[\frac{3x^2}{4}+6x-\frac{x^3}{4}{]}_{-2}^4=\frac{3\times 4^2}{4}+6\times 4-\frac{4^3}{4}-[\frac{3\times 4}{4}-12+\frac{8}{4}]=12+24-16-3+12-2=27squnit$