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 are required to compute the area enclosed between the right-opening parabola $y^2 = x$ and the straight line $x + y = 2$, restricted entirely within the first quadrant ($x \ge 0, y \ge 0$).

Step 2: Key Formula or Approach:
First, we find the intersection points of the curves by substituting $x = 2 - y$ into the parabola's equation. To find the enclosed region's area, it is much easier to integrate along the $y$-axis from the lower boundary to the upper boundary using: $$\text{Area} = \int_{y_1}^{y_2} (x_{\text{line}} - x_{\text{parabola}}) \, dy$$ Alternatively, if integrating with respect to $x$, the region must be split where the upper bounding curve function changes. Let's use integration with respect to $y$ as it handles the entire boundary cleanly in a single integral block.

Step 3: Detailed Explanation:
Let's find the intersection points by substituting $x = 2 - y$ into $y^2 = x$: $$y^2 = 2 - y \implies y^2 + y - 2 = 0$$ Factoring the quadratic equation: $$(y + 2)(y - 1) = 0 \implies y = 1 \quad \text{or} \quad y = -2$$ Since we are strictly bounded within the first quadrant, $y$ ranges from $0$ to $1$. Let's look at the functions expressing $x$ in terms of $y$: $$\text{From the line: } x_1 = 2 - y$$ $$\text{From the parabola: } x_2 = y^2$$ For the interval $y \in [0, 1]$, the line lies to the right of the parabola ($2 - y \ge y^2$). Therefore, the area is given by: $$\text{Area} = \int_{0}^{1} \left((2 - y) - y^2\right) \, dy$$ $$\text{Area} = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{0}^{1}$$ Evaluating this expression at the upper limit $1$ and lower limit $0$: $$\text{Area} = \left( 2(1) - \frac{1^2}{2} - \frac{1^3}{3} \right) - 0$$ $$\text{Area} = 2 - \frac{1}{2} - \frac{1}{3}$$ Find a common denominator, which is 6: $$\text{Area} = \frac{12}{6} - \frac{3}{6} - \frac{2}{6} = \frac{12 - 3 - 2}{6} = \frac{7}{6} \text{ sq. units}$$

Step 4: Final Answer:
The bounded area in the first quadrant is $\frac{7}{6}$ square units, which corresponds to option (A).