The area bounded by the curves y = \(\sqrt{x}\), 2y - x + 3 = 0, x-axis and lying in the first quadrant is :
Answer & explanation
Correct answer: option 4
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