If the area of the region (x,y) : 1 - 2x y 4 - x^2, x 0, y 0 is , where , N, (,) = 1, then…

Question

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

Accepted Answer

For $0 \le x \le \tfrac12$: $1-2x \ge 0$, so the lower bound is $y = 1-2x$; for $\tfrac12 \le x \le 2$: $1-2x < 0$, so (combined with $y\ge0$) the lower bound is $y=0$. The upper bound $4-x^2 \ge 0$ only for $x \le 2$.

$$\text{Area} = \int_0^{1/2}\left[(4-x^2)-(1-2x)\right]dx + \int_{1/2}^{2}(4-x^2)\,dx$$

$$= \int_0^{1/2}(3+2x-x^2)\,dx + \int_{1/2}^{2}(4-x^2)\,dx = \frac{41}{24} + \frac{27}{8} = \frac{41}{24}+\frac{81}{24} = \frac{122}{24} = \frac{61}{12}$$

So $\alpha = 61$, $\beta = 12$ (coprime), and $\alpha + \beta = 73$.

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