Question

The area of the region enclosed by the parabola \((y - 2)^2 = x - 1\), the line \(x - 2y + 4 = 0\) and the positive coordinate axes is ______.

Answer 1

Correct Answer: 5

Rearrange the equations: The parabola:

\[ (y - 2)^2 = x - 1 \implies y - 2 = \sqrt{x - 1}. \]

- The line:

\[ x - 2y + 4 = 0 \implies y = \frac{x + 4}{2}. \]

Find points of intersection: Set the equations equal:

\[ \frac{x + 4}{2} = \sqrt{x - 1} + 2. \]

Rearranging and squaring gives:

\[ (x - 2)^2 = 0 \implies x = 2. \]

Find corresponding \( y \): Substitute \( x = 2 \):

\[ y = \frac{2 + 4}{2} = 3. \]

Identify area: Compute:

\[ A = \int_{1}^{2} \left( \frac{x + 4}{2} - (2 + \sqrt{x - 1}) \right) dx. \]

Evaluate the integral:

\[ A = \int_{1}^{2} \left( \frac{x + 4}{2} - 2 - \sqrt{x - 1} \right) dx. \]

Combine results:

\[ A = \left[ \frac{2^2 + 8 \times 2}{4} - 2(2) - \frac{2}{3}(2 - 1)^{3/2} \right] - \left[ \frac{1^2 + 8 \times 1}{4} - 2(1) - \frac{2}{3}(1 - 1)^{3/2} \right]. \]

= 9 - 18 + 15 - 1 = 5

After evaluation, we find: 5

Answer 2

The problem asks for the area of the region in the first quadrant (\(x \ge 0, y \ge 0\)) enclosed by the parabola \((y - 2)^2 = x - 1\) and the line \(x - 2y + 4 = 0\).

Concept Used:

The area of a region bounded by curves can be calculated using a definite integral. When the curves are conveniently expressed as \(x\) in terms of \(y\), it is often easier to integrate with respect to \(y\). If a region is bounded on the right by a curve \(x = x_{R}(y)\) and on the left by a curve \(x = x_{L}(y)\) from \(y = c\) to \(y = d\), the area \(A\) is given by the formula:

\[ A = \int_{c}^{d} (x_{R}(y) - x_{L}(y)) \, dy \]

To use this formula, we must first find the intersection points of the bounding curves to determine the limits of integration and to understand the layout of the region.

Step-by-Step Solution:

Step 1: Express the equations of the curves in the form \(x = f(y)\) and find their key points in the first quadrant.

The given curves are:

1. Parabola: \((y - 2)^2 = x - 1 \implies x = (y-2)^2 + 1\). This is a right-opening parabola with its vertex at \((1, 2)\).
2. Line: \(x - 2y + 4 = 0 \implies x = 2y - 4\).
3. Positive y-axis: \(x = 0\).
4. Positive x-axis: \(y = 0\).

Step 2: Find the points of intersection between the curves.

Intersection of the parabola and the line:

\[ (y-2)^2 + 1 = 2y - 4 \] \[ y^2 - 4y + 4 + 1 = 2y - 4 \] \[ y^2 - 6y + 9 = 0 \] \[ (y - 3)^2 = 0 \implies y = 3 \]

The corresponding x-coordinate is \(x = 2(3) - 4 = 2\). The line is tangent to the parabola at the point \((2, 3)\).

Intersection of the line with the y-axis (\(x=0\)):

\[ 0 = 2y - 4 \implies y = 2 \]

The intersection point is \((0, 2)\).

Intersection of the parabola with the x-axis (\(y=0\)):

\[ x = (0-2)^2 + 1 = 5 \]

The intersection point is \((5, 0)\).

Step 3: Visualize the region and set up the integrals.

The region is bounded by the y-axis (\(x=0\)), the line, and the parabola. The geometry suggests splitting the region into two parts at \(y=2\), the y-coordinate where the line intersects the y-axis.

Part 1 (Area A₁): For \(0 \le y \le 2\)

The region is bounded on the right by the parabola \(x_R = (y-2)^2 + 1\) and on the left by the y-axis \(x_L = 0\).

\[ A_1 = \int_{0}^{2} \left[ ((y-2)^2 + 1) - 0 \right] \, dy \]

Part 2 (Area A₂): For \(2 \le y \le 3\)

The region is bounded on the right by the parabola \(x_R = (y-2)^2 + 1\) and on the left by the line \(x_L = 2y - 4\).

\[ A_2 = \int_{2}^{3} \left[ ((y-2)^2 + 1) - (2y - 4) \right] \, dy \]

The total area is \(A = A_1 + A_2\).

Step 4: Calculate the first integral (\(A_1\)).

\[ A_1 = \int_{0}^{2} (y^2 - 4y + 4 + 1) \, dy = \int_{0}^{2} (y^2 - 4y + 5) \, dy \] \[ A_1 = \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_{0}^{2} \] \[ A_1 = \left( \frac{2^3}{3} - 2(2^2) + 5(2) \right) - (0) = \frac{8}{3} - 8 + 10 = \frac{8}{3} + 2 = \frac{14}{3} \]

Step 5: Calculate the second integral (\(A_2\)).

\[ A_2 = \int_{2}^{3} (y^2 - 4y + 5 - 2y + 4) \, dy = \int_{2}^{3} (y^2 - 6y + 9) \, dy \] \[ A_2 = \int_{2}^{3} (y-3)^2 \, dy \] \[ A_2 = \left[ \frac{(y-3)^3}{3} \right]_{2}^{3} = \frac{(3-3)^3}{3} - \frac{(2-3)^3}{3} = 0 - \frac{(-1)^3}{3} = \frac{1}{3} \]

Final Computation & Result:

The total area is the sum of \(A_1\) and \(A_2\).

\[ A = A_1 + A_2 = \frac{14}{3} + \frac{1}{3} = \frac{15}{3} = 5 \]

The area of the required region is 5 square units.