If the area of the region \(x, y) : 1 - 2x leq y leq 4 - x^2, x geq 0, y geq 0\$\{(x, y) : 1 - 2x \leq y \leq 4 - x^{2}, x \geq 0, y \geq 0\}$ is fracalphabeta, alpha, beta, in N, gcd(alpha, beta) = 1$\frac{\alpha}{\beta}, \alpha, \beta, \in N, \gcd(\alpha, \beta) = 1$, then the value of (alpha + beta)$(\alpha + \beta)$ is :

### Related Formula textArea = int_x_1^x_2 (f(x) - g(x)) dx$$\text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) dx$$ textArea of Parabola piece: int_0^2 (4-x^2) dx$$\text{Area of Parabola piece: } \int_{0}^{2} (4-x^2) dx$$ ### Core Logic

The region is bounded above by y = 4 - x^2$y = 4 - x^2$, below by y = 1 - 2x$y = 1 - 2x$, and constrained to x geq 0, y geq 0$x \geq 0, y \geq 0$. The parabola intersects the x-axis at x=2$x=2$ (since 4-x^2=0, xgeq 0$4-x^2=0, x\geq 0$). The line intersects the x-axis at x=frac12$x=\frac{1}{2}$ (since 1-2x=0$1-2x=0$) and y-axis at y=1$y=1$. The required area is the area under the parabola in the first quadrant minus the small triangular region bounded by y = 1-2x, x=0, y=0$y = 1-2x, x=0, y=0$. ### Step 1: Calculate the Area textTotal area under parabola in 1st quadrant = int_0^2 (4 - x^2) dx = left[ 4x - fracx^33 right]_0^2 = 8 - frac83 = frac163$$\text{Total area under parabola in 1st quadrant} = \int_0^2 (4 - x^2) dx = \left[ 4x - \frac{x^3}{3} \right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$$ Area of the small right triangle formed by the line y=1-2x$y=1-2x$ in the first quadrant: Vertices are (0,0), (1/2,0), (0,1)$(0,0), (1/2,0), (0,1)$. textArea of triangle = frac12 times textbase times textheight = frac12 times frac12 times 1 = frac14$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$$ ### Step 2: Subtraction and Format Match textRequired Area = frac163 - frac14 = frac64 - 312 = frac6112$$\text{Required Area} = \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12}$$ Here, alpha = 61, beta = 12$\alpha = 61, \beta = 12$. Check gcd(61, 12) = 1$\gcd(61, 12) = 1$. This matches. So, alpha + beta = 61 + 12 = 73$\alpha + \beta = 61 + 12 = 73$. ### Pattern Recognition For areas defined by y geq g(x)$y \geq g(x)$ when g(x)$g(x)$ forms a simple geometric shape (like a line), subtract the geometric area directly rather than splitting the integral algebraically. It eliminates integration errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Integrals