If the area of the region { (x, y) : 1 − 2x ≤ y ≤ 4 − x², x ≥ 0, y ≥ 0 } is α β , α, β ∈ N, gcd(α, β) = 1, then the value of (α + β) is :

The correct answer is 73 (Option 2).

Step 1: Understand the Given Region

We are given the region bounded by:

$R=\left\{\right(x,y):1-2x\le y\le 4-x^2,x\ge 0,y\ge 0\}$

This region lies entirely in the first quadrant since $x\ge 0$ and $y\ge 0$.

The upper boundary is the downward-opening parabola $y=4-x^2$.

The lower boundary is determined by $y\ge 1-2x$ and $y\ge 0$.

Step 2: Determine Key Boundary Points

1. The parabola $y=4-x^2$ intersects the x-axis ($y=0$) in the first quadrant at $x=2$.

2. The line $y=1-2x$ intersects the x-axis ($y=0$) at $x=\frac{1}{2}$ and the y-axis ($x=0$) at $y=1$.

Step 3: Calculate the Total Area

The region $R$ is formed by taking the total area under the parabola in the first quadrant and subtracting the area of the small triangular region under the line $y=1-2x$ in the first quadrant.

Area under the parabola from $x=0$ to $x=2$:

$A_1={\int }_0^2(4-x^2)dx$

$A_1={[4x-\frac{x^3}{3}]}_0^2=4\left(2\right)-\frac{2^3}{3}=8-\frac{8}{3}=\frac{16}{3}$

Area of the triangular region under the line in the first quadrant:

$A_2=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times \frac{1}{2}\times 1=\frac{1}{4}$

Therefore, the area of the region $R$ is:

$\text{Area}=A_1-A_2=\frac{16}{3}-\frac{1}{4}=\frac{64-3}{12}=\frac{61}{12}$

Step 4: Find $\alpha +\beta$

Comparing with $\frac{\alpha }{\beta }$, we have:

$\alpha =61$ and $\beta =12$

Since $\gcd =1$ and $61,12\in {\rm N}$:

$\alpha +\beta =61+12=73$