The area (in sq. units) bounded by the curves | vedclass

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(D) The given curves are $y=\sqrt{x}$ (or $x=y^2$) and $2y-x+3=0$ (or $x=2y+3$).To find the intersection points,set $y^2 = 2y+3$,which gives $y^2-2y-3

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The given curves are $y=\sqrt{x}$ (or $x=y^2$) and $2y-x+3=0$ (or $x=2y+3$).To find the intersection points,set $y^2 = 2y+3$,which gives $y^2-2y-3=0$,so $(y-3)(y+1)=0$. Since the region is in the first quadrant,$y=3$. Thus,$x=9$.The intersection with the $X$-axis for $2y-x+3=0$ is at $y=0$,which gives $x=3$.The area bounded by the curves and the $X$-axis in the first quadrant is given by integrating with respect to $y$ from $y=0$ to $y=3$:$	ext{Area} = \int_0^3 (x_{line} - x_{curve}) dy = \int_0^3 ((2y+3) - y^2) dy$$= [y^2 + 3y - \frac{y^3}{3}]_0^3$$= (9 + 9 - \frac{27}{3}) - 0$$= 18 - 9 = 9 	ext{ sq. units.}$

The area (in sq. units) bounded by the curves $y=\sqrt{x}$,$2y-x+3=0$,$X$-axis and lying in the first quadrant is

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