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

Re-express equations: Parabola:

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

Line:

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

Determine intersection points: Equate the expressions:

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

Rearrange and square to obtain:

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

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

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

Define the area: Integrate:

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

Evaluate the integral:

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

Sum the 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

The computed result is: 5