The area (in sq units) bounded by the curves , , -axis and lying in the first quadrant, is
- A.9
- B.6
- C.18
- D.
Show answer and solution
Correct option: A
Solution
The line 2y - x + 3 = 0 gives x = 2y + 3. Find intersection of y = √x and the line: x = 2√x + 3, so let u = √x: u² - 2u - 3 = 0, giving u = 3 (u = -1 rejected), so intersection at (9, 3).
The line meets the x-axis (y = 0) at x = -3, which is outside the first quadrant. The curve y = √x meets the x-axis at x = 0.
Integrate with respect to y from y = 0 to y = 3. For a given y, x ranges from x = y² (curve) to x = 2y + 3 (line).
Area = ∫₀³ [(2y + 3) - y²] dy = [y² + 3y - y³/3]₀³ = (9 + 9 - 9) - 0 = 9.
The line meets the x-axis (y = 0) at x = -3, which is outside the first quadrant. The curve y = √x meets the x-axis at x = 0.
Integrate with respect to y from y = 0 to y = 3. For a given y, x ranges from x = y² (curve) to x = 2y + 3 (line).
Area = ∫₀³ [(2y + 3) - y²] dy = [y² + 3y - y³/3]₀³ = (9 + 9 - 9) - 0 = 9.
[JEE Main 2013]