Question

Find the area of the region $\left\{\right(x,y)|0\le y\le x^2,0\le y\le x+2,0\le x\le 3\}$.

Answer 1

$\left\{\right(x,y)|0\le y\le x^2,0\le y\le x+2,0\le x\le 3\}$
Here $0\le y\le x^2$,
$0\le y\le x+2$,
And $0\le x\le 3$ To find area of these inequalities, for that we conver in to equality.
$y=x^2,y=x+2$ and $x=3$
$y=x^2$ is the parabola with vertex origin, which intersects line $y=x+2$
$\therefore$ We solve equations $y=x^2$ and $y=x+2$
$\therefore x^2=x+2$
$\therefore x^2-x-2=0$
$\therefore (x-2)(x+1)=0$
$\therefore x=2$ and $x=-1$
$\therefore$ For $x=2$ we have $y=4$
$\therefore$ Point of intersection $P(2,4)$
And for $x=-1$ we have $y=1$
$\therefore$ Point of intersection will be $M(1,1)$
Required are of the bounded region $=$ Area of the region OPSO$+$Area of the region of SPQRS
Required area $=\stackrel{2}{\underset{0}{\int }}{x}^{2}dx+\stackrel{2}{\underset{0}{\int }}(x+2)dx$
($\because$ Region OPSO is bounded by $y=x^2$, $x=0$ and $x=2$ and X-axis and region SPQRS is bounded by line y$=x+2,x=2,x=3$, and X-axis)
$\therefore$ Required area $={\left(\frac{x^3}{3}\right)}_0^2+{(\frac{x^2}{2}+2x)}_2^5$
$=(\frac{8}{3}-0)+((\frac{9}{2}+6)-(2+4\left)\right)$
$=\frac{8}{3}+(\frac{21}{2}-6)$
$=\frac{8}{3}+\frac{9}{2}$
$=\frac{16+27}{6}$
$=\frac{43}{6}$
Required area $=\frac{43}{6}$ sq. unit.