Question

Sketch the graphs of the curves $y^2=x$ and $y^2=4-3x$ and find the area enclosed between them

Answer 1

$y^2=-3x+4$
$=-3(x-\frac{4}{3})$
The vertex $L$ of this parabola is $(\frac{4}{3},0)$. It cuts the y-axis at $A(0,2)$ and $B(0,-2)$.
The points of intersection of these two parabolas are given by the equation
$x=-3x+4\Rightarrow x=1$
Then $y^2=1\Rightarrow y=\pm 1$
Thus, the points of intersection are $P(1,1)$ and $Q(1,-1)$. Let $PQ$ cut the x-axis at $R$
$\therefore$ Total area of $POQLP=2$ area of $OPLRO$
$=2[{\int }_0^1\sqrt{x}dx+{\int }_1^{4/3}\sqrt{4-3x}dx]$
$=2[{\left(\frac{x^{3/2}}{3/2}\right)}_0^1+{\left(\frac{2(4-3x{)}^{3/2}}{(-3)\times 3}\right)}_1^{4/3}]$
$=2[(\frac{2}{3}-0)-\frac{2}{9}(0-1\left)\right]$
$=2[\frac{2}{3}+\frac{2}{9}]=2\left[\frac{6+2}{9}\right]$
$=2\left(\frac{8}{9}\right)=\frac{16}{9}sq.units$