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

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