The area (in square units) bounded by the curves `y=sqrt(x),2y-x+3=0,` x-axis, and lying in the first quadrant is

The area (in square units) bounded by the curves : y=sqrtx, 2y-x +3 =0 x-axis , and lying in the first quadrant is :

The area (in sqaure units) bounded by the curves: y=sqrtx , 2y-x+3=0, x-axis, and lying in the first quadrant is:

The area (in square units) bounded by the curves y=sqrtx,2y-x+3=0 , x-axis, and lying in the fir stquadrant is :

The area bounded by the curves y=sqrt(x),2y-x+3=0, X-axis and lying in the first quadrant is

The area bounded by the curves y=sqrt(x),2y-x+3=0 , x-axis and lying in the first quadrant is

The area bounded by the curves y=sqrt(x),2y+3=x , and x-axis in the 1st quadrant is

Area ( in square unit ) bounded by the curve y= sqrt(x) , the straight line x=2y+3 in first quadrant and x- axis is-

Find the area of the region bounded by the curve y = sqrt ( 36 - x ^(2)), the X- axis lying in the first quadrant and the lines X= 0, X = 6 ?

Find the area of the region bounded by the curves y= sqrt(4-x^2) ,y= sqrt(3x) and X- Axis in the first quadrant .