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

Given f(x) is 2y=3x+12$\Rightarrow y=\frac{3x+12}{2}$-----(1)
and g(x) is 4y=3x^2$\Rightarrow y=\frac{3x^2}{4}$--------(2)
The required area to be found is enclosed between the straight line and the parabola.
To find the limits,we will have to find the points of intersection of the straight line and the parabola.
By equating equ(1) and equ(2) we get
$\frac{3x+12}{2}=\frac{3x^2}{4}$
$\Rightarrow4(3x+12)=2(3x^2)$ [dividing by 2 on both sides]
$\Rightarrow2(3x+12)=3x^2$
$\Rightarrow6x+24=3x^2$
$\Rightarrow3x^2-6x-24=0$
dividing through out by 3 we get
$x^2-2x-8=0.$
On factorising we get (x-4)(x+2=0 or x=4,-2;hence y=3,12 respectively.
The points of intersection are (-2,3) and (4,12)
Hence the required area is
$A=\int_a^b[f(x)-g(x)]dx$
Here a=-2;b=4,$f(x)=\frac{3x+12}{12}$ and $g(x)=\frac{3x^2}{4}.$
$A=\int_{-2}^4\begin{bmatrix}\frac{3x+12}{2}-\frac{3x^2}{4}\end{bmatrix}dx$
This can be splitter as $\int_{-2}^4\bigg(\frac{3x+12}{2}\bigg)dx-\int_{-2}^4\bigg(\frac{3x^2}{4}\bigg)dx$
On integrating we get,
$A=\frac{1}{2}\begin{bmatrix}\frac{3x^2}{2}+\frac{12x}{2}\end{bmatrix}_{-2}^4-\frac{3}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_{-2}^4$
On applying the limits we get
$A=\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8]$
$\;=\frac{1}{2}(90)-\frac{1}{4}(72)=27sq.units$
Hence the required area is 27sq.units.