Question

The area bounded by the parabola $y^2 = x$ and the line $x + y = 2$ in the first quadrant is

• A. $\frac{7}{6}$ sq. units
• B. $\frac{1}{6}$ sq. units
• C. $\frac{2}{3}$ sq. units
• D. $\frac{6}{7}$ sq. units

Hint:

Whenever a parabolic boundary is written as $y^2 = x$, integrating along the $y$-axis ($\int x \, dy$) is almost always faster and less error-prone than splitting the region into multiple pieces along the $x$-axis!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We must find the area enclosed between y² = x and x + y = 2 in the first quadrant.

Step 2: Key Formula or Approach:
Integrate with respect to y: Area = ∫ (x_line – x_parabola) dy from the lower to upper intersection.

Step 3: Detailed Explanation:
Intersection: y² = 2 – y → y² + y – 2 = 0 → y = 1 (first quadrant). Area = ∫₀¹ [(2–y) – y²] dy = [2y – y²/2 – y³/3]₀¹ = 2 – 1/2 – 1/3 = 7/6.

Step 4: Final Answer:
The area is 7/6 sq. units, matching option (A).