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Q.
Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.
Application of Integrals
Solution:
The given parabola is $y=\frac{3}{4} x^{2}\quad\ldots\left(i\right)$
and the line is $3x - 2y + 12 = 0\quad\ldots\left(ii\right)$
Solving $(1)$ & $(2)$, we get
$3x-2\left(\frac{3}{4}x^{2}\right)+12=0\,$
$\Rightarrow \quad3x-\frac{3}{2}x^{2}+12=0$
$\Rightarrow \quad6x-3x^{2}+24=0\,$
$\Rightarrow \quad x^{2}-2x-8=0$
$\Rightarrow \quad\left(x-4\right)\left(x+2\right)=0\,$
$\Rightarrow \quad x=4, -2$.
Putting values of $x$ in $\left(1\right)$, we get
$y=\frac{3}{4}\left(4\right)^{2}=12$ and $y=\frac{3}{4}\left(-2\right)^{2}=3$.
Hence, the line and parabola intersects at the points $\left(-2,3\right)$ and $\left(4, 12\right)$.
$\therefore \quad$ Required area = area $(ABCDA)$
$=\int\limits_{-2}^{4}\left(\frac{3x+12}{2}\right)dx-\left(\int\limits_{-2}^{0}\frac{3}{4}x^{2}dx+\int\limits_{0}^{4}\frac{3}{4}x^{2}dx\right)$
