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

36 18 27/4 9

9

2\(\sqrt{x} = x - 3\) Squaring both sides :  4x = x2 - 6x + 9  ⇒ x = 1, 9 \(\int_{0}^{3} [(2y + 3) - y^2]dy\) = \([y^2 + 3y - \frac{y^3}{3}]_{0}^{3} = 9 + 9 - 9\) = 9