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

Question

Accepted Answer

A. 1. **Determine the effective lower boundary.** Since the region requires both $𝑦\ge 1-2𝑥$ and $𝑦\ge 0$ simultaneously, the actual lower bound is $\max(1-2𝑥,0)$. Since $1-2𝑥\ge 0$ only when $𝑥\le \frac{1}{2}$, the lower bound is $1-2𝑥$ for $𝑥\in\left[0,\frac{1}{2}ight]$ and $0$ for $𝑥\in\left[\frac{1}{2},2ight]$.

2. **Determine the upper limit of $𝑥$.** Since $𝑦\le 4-𝑥^2$ requires $4-𝑥^2\ge 0$ for the region to be non-empty, we need $𝑥\le 2$.

3. **Set up the total area as the full area under the parabola minus the small triangular strip cut off by the line near the origin.** Compute the area under $𝑦=4-𝑥^2$ over the entire range $[0,2]$ first, then subtract the extra piece where $𝑦=0$ would have applied but $1-2𝑥$ is actually higher (i.e., the small triangle between $𝑦=0$ and $𝑦=1-2𝑥$ over $\left[0,\frac{1}{2}ight]$):
$$
	ext{Area}=\int_0^2 \left(4-𝑥^2ight)d𝑥-\int_0^{1/2}(1-2𝑥)\,d𝑥
$$

4. **Evaluate the main integral**:
$$
\int_0^2 (4-𝑥^2)\,d𝑥=\left[4𝑥-\frac{𝑥^3}{3}ight]_0^2=8-\frac{8}{3}=\frac{16}{3}
$$

5. **Evaluate the correction integral**, since this is simply the small triangular strip with base $\frac{1}{2}$ and height $1$:
$$
\int_0^{1/2}(1-2𝑥)\,d𝑥=\left[𝑥-𝑥^2ight]_0^{1/2}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}
$$
This matches the geometric shortcut $\frac{1}{2}	imes 1	imes \frac{1}{2}=\frac{1}{4}$ (area of the right triangle with legs $1$ and $\frac{1}{2}$).

6. **Subtract to get the total area**:
$$
	ext{Area}=\frac{16}{3}-\frac{1}{4}=\frac{64-3}{12}=\frac{61}{12}
$$

7. **Identify $\alpha$ and $\beta$**, noting $\gcd(61,12)=1$:
$$
\alpha=61, \qquad \beta=12
$$

8. **Compute $\alpha+\beta$**:
$$
\alpha+\beta=61+12=73
$$

Hence, the answer is **Option A: 73**.