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

Given, y=√x, 2y-x+3=0
from above we have, 2√x-x+3=0
2√x=x-3
4x = x²-6x+9
x²-10x+9=0
(x-9)(x-1)=0
x=9, x=1
When x=1, y=√1=1
When x=9, y=√9=3
Area = ∫(√x)dx from 0 to 1 + ∫( (x+3)/2 )dx from 1 to 9
= [ (2/3)x^(3/2) ] from 0 to 1 + [ (x²/4) + (3x/2) ] from 1 to 9
= (2/3)(1)^(3/2) - 0 + (81/4 + 27/2) - (1/4 + 3/2)
= 2/3 + 81/4 + 27/2 - 1/4 - 3/2
= 2/3 + 80/4 + 24/2
= 2/3 + 20 + 12
= 2/3 + 32
= (2+96)/3 = 98/3
However, the area should be calculated as follows:
Area = ∫₀¹ √x dx + ∫₁⁹ [(x+3)/2] dx - ∫₀⁹ √x dx
= [2x^(3/2)/3]₀¹ + [(x²/4) + (3x/2)]₁⁹ - [2x^(3/2)/3]₀⁹
= 2/3 + (81/4 + 27/2 - 1/4 - 3/2) - (2(27)/3)
= 2/3 + 12 + 20 - 18
= 2/3 + 14
= 44/3 ≈ 14.67
Let's reconsider the area:
Area = ∫₀¹ √x dx + ∫₁⁹ ( (x+3)/2 - √x ) dx
= [2x^(3/2)/3]₀¹ + [(x²/4) + (3x/2) - (2x^(3/2)/3)]₁⁹
= 2/3 + (81/4 + 27/2 - 18) - (1/4 + 3/2 - 2/3)
= 2/3 + (81/4 + 54/4 - 72/4) - (3/4 + 9/6 - 4/6)
= 2/3 + 63/4 - 5/6 = 2/3 + 63/4 - 5/6
= 8/12 + 189/12 - 10/12 = 187/12 ≈ 15.58
The given options do not match the calculated area.