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

### Related Formula textArea enclosed integrating w.r.t y-axis: int (x_textright - x_textleft) dy$$\text{Area enclosed integrating w.r.t y-axis: } \int (x_{\text{right}} - x_{\text{left}}) dy$$ ### Core Logic Find intersection points between the parabola x = (y-2)^2 + 1$x = (y-2)^2 + 1$ and the line x = 2y - 4$x = 2y - 4$. Set them equal: (y-2)^2 + 1 = 2y - 4$$(y-2)^2 + 1 = 2y - 4$$ y^2 - 4y + 4 + 1 = 2y - 4$$y^2 - 4y + 4 + 1 = 2y - 4$$ y^2 - 6y + 9 = 0 Rightarrow (y-3)^2 = 0$$y^2 - 6y + 9 = 0 \Rightarrow (y-3)^2 = 0$$ The line is tangent to the parabola at y = 3$y = 3$. At y = 3$y = 3$, x = 2(3) - 4 = 2$x = 2(3) - 4 = 2$. We need the area enclosed by the parabola, the line, and the *positive coordinate axes*. Let's trace the boundary intercepts. The line x = 2y - 4$x = 2y - 4$ cuts the y-axis (x=0$x=0$) at y=2$y=2$. Parabola vertex is at (1, 2)$(1, 2)$. Parabola cuts y-axis (x=0$x=0$) at (y-2)^2 = -1$(y-2)^2 = -1$ (No real y-intercept). So the curve (y-2)^2 = x-1$(y-2)^2 = x-1$ only exists for x ge 1$x \ge 1$. ### Step 1: Setting up the Integral w.r.t y

Because the area is bound by the positive coordinate axes, we track integration strictly in the first quadrant. The required area is the area between the y-axis, the x-axis, the line, and the curve. Wait, the problem says "area enclosed by the parabola... line... and positive coordinate axes". Let's integrate with respect to y$y$. The boundary curves are x = (y-2)^2 + 1$x = (y-2)^2 + 1$ (which forms the rightmost boundary) and the axes. The area bounded by the curve with y-axis is from y=0$y=0$ to y=3$y=3$ minus the triangular region cut by the line below y=2$y=2$. The line intersects the x-axis (y=0$y=0$) at x=-4$x=-4$ and y-axis (x=0$x=0$) at y=2$y=2$. The positive coordinate axes enforce boundaries at x ge 0, y ge 0$x \ge 0, y \ge 0$. So the exact area is the integral of the parabola's x$x$-value from y=0$y=0$ to y=3$y=3$, minus the area of the small triangle outside the region bounded by the line x=2y-4$x=2y-4$ and the axes in the positive quadrant. The area under the parabola (to the left towards the y-axis) from y=0$y=0$ to y=3$y=3$ is int_0^3 x_textparabola dy$\int_0^3 x_{\text{parabola}} dy$. The line bounds the region on the left from y=2$y=2$ to y=3$y=3$. Between y=0$y=0$ and y=2$y=2$, the area goes fully to the y-axis. So, area = int_0^3 x_textparabola dy - textArea of Delta text bounded by line and y-axis$\int_0^3 x_{\text{parabola}} dy - \text{Area of } \Delta \text{ bounded by line and y-axis}$. The triangle bounded by x=2y-4, x=0, y=2, y=3$x=2y-4, x=0, y=2, y=3$ is: base is x=2(3)-4 = 2$x=2(3)-4 = 2$ at y=3$y=3$, height is 1$1$ (from y=2$y=2$ to y=3$y=3$). Triangle area = frac12 times 2 times 1 = 1$\frac{1}{2} \times 2 \times 1 = 1$. Hence, Enclosed Area = int_0^3 ((y-2)^2 + 1) dy - 1$\int_{0}^{3} ((y-2)^2 + 1) dy - 1$. ### Step 2: Evaluating the Integral textArea = int_0^3 (y^2 - 4y + 5) dy - 1$$\text{Area} = \int_{0}^{3} (y^2 - 4y + 5) dy - 1$$ = left[ fracy^33 - 2y^2 + 5y right]_0^3 - 1$$= \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^3 - 1$$ = left( frac273 - 2(9) + 5(3) right) - 0 - 1$$= \left( \frac{27}{3} - 2(9) + 5(3) \right) - 0 - 1$$ = (9 - 18 + 15) - 1 = 6 - 1 = 5$$= (9 - 18 + 15) - 1 = 6 - 1 = 5$$ ### Pattern Recognition Instead of evaluating horizontal strips with a broken left-bound, evaluate the total area mapped to the y-axis int_0^3 x dy$\int_0^3 x dy$ and simply subtract the basic geometric triangle removed by the straight line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integral Calculus