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 __________.

The given curves are the parabola $$(y-2)^2 = x - 1$$ and the line $$x - 2y + 4 = 0$$.
Rewriting them in $$x = \dots$$ form:

Parabola : $$x = (y-2)^2 + 1 = y^2 - 4y + 5$$ $$-(1)$$
Line       : $$x = 2y - 4$$ $$-(2)$$

To find their intersection, equate $$(1)$$ and $$(2)$$:

$$(y-2)^2 + 1 = 2y - 4$$

$$y^2 - 4y + 4 + 1 = 2y - 4$$

$$y^2 - 6y + 9 = 0 \;\;\Rightarrow\;\; (y-3)^2 = 0$$

Hence $$y = 3$$ and, from $$(2)$$, $$x = 2(3) - 4 = 2$$.
The curves meet at $$P(2,\,3)$$.

Intercepts with the positive axes:

• Line with $$y$$-axis: put $$x = 0$$ in $$(2)$$ → $$-2y + 4 = 0 \Rightarrow y = 2$$ ⇒ point $$A(0,\,2)$$.
• Parabola with $$x$$-axis: put $$y = 0$$ in $$(1)$$ → $$x = (-2)^2 + 1 = 5$$ ⇒ point $$B(5,\,0)$$.

Tracing the boundary in the first quadrant gives the closed path
$$O(0,0) \rightarrow A(0,2) \rightarrow P(2,3) \rightarrow B(5,0) \rightarrow O(0,0)$$.
A horizontal slice at height $$y$$ therefore starts at the left boundary and ends at the right boundary as follows:

For $$0 \le y \le 2$$, the left boundary is the $$y$$-axis $$x = 0$$ and the right boundary is the parabola $$(1)$$.
Area $$A_1 = \displaystyle\int_{0}^{2} \bigl(y^2 - 4y + 5\bigr)\,dy$$

Integrating term by term,

$$\int y^2\,dy = \frac{y^3}{3},\;\; \int (-4y)\,dy = -2y^2,\;\; \int 5\,dy = 5y$$

Evaluated from $$0$$ to $$2$$:

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

For $$2 \le y \le 3$$, the left boundary is the line $$(2)$$ and the right boundary is the parabola $$(1)$$.
Area $$A_2 = \displaystyle\int_{2}^{3} \Bigl[(y^2 - 4y + 5) - (2y - 4)\Bigr]\,dy$$

Simplify the integrand:

$$(y^2 - 4y + 5) - (2y - 4) = y^2 - 6y + 9 = (y-3)^2$$

Integrate:

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

At $$y = 3$$: $$\frac{27}{3} - 27 + 27 = 9$$
At $$y = 2$$: $$\frac{8}{3} - 12 + 18 = \frac{26}{3}$$

Hence $$A_2 = 9 - \frac{26}{3} = \frac{1}{3}$$

Total required area:

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

Area of the enclosed region = 5 square units.