If the area of the region {(x, y) : -2x + 1 l | vedclass

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(A) The region is bounded by the parabola $y = 4 - x^2$, the line $y = -2x + 1$, and the axes $x \ge 0, y \ge 0$.First, find the intersection of the c

Vedclass · Accepted Answer

$73$

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(A) The region is bounded by the parabola $y = 4 - x^2$, the line $y = -2x + 1$, and the axes $x \ge 0, y \ge 0$.First, find the intersection of the curves:$4 - x^2 = -2x + 1 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0$.Since $x \ge 0$, the intersection is at $x = 3$. However, the region is bounded by $x \ge 0$ and $y \ge 0$. The line $y = -2x + 1$ intersects the $x$-axis at $x = 0.5$ and the $y$-axis at $y = 1$.The area is the integral of the upper curve minus the lower curve from $x = 0$ to $x = 2$ (where $y = 4 - x^2$ meets the $x$-axis).Area $= \int_{0}^{0.5} (4 - x^2) dx + \int_{0.5}^{2} (4 - x^2 - (-2x + 1)) dx$$= \int_{0}^{0.5} (4 - x^2) dx + \int_{0.5}^{2} (3 + 2x - x^2) dx$$= [4x - \frac{x^3}{3}]_{0}^{0.5} + [3x + x^2 - \frac{x^3}{3}]_{0.5}^{2}$$= (2 - \frac{1}{24}) + [(6 + 4 - \frac{8}{3}) - (1.5 + 0.25 - \frac{0.125}{3})]$$= \frac{47}{24} + [\frac{22}{3} - \frac{21}{12} + \frac{1}{24}] = \frac{47}{24} + \frac{176 - 42 + 1}{24} = \frac{47 + 135}{24} = \frac{182}{24} = \frac{91}{12}$.Thus, $\alpha = 91, \beta = 12$, and $\alpha + \beta = 103$. Given the options, let's re-evaluate the region: The region is bounded by $y = 4 - x^2$ and $y = -2x + 1$ for $x \ge 0, y \ge 0$. The area is $\int_{0}^{2} (4 - x^2) dx - 	ext{Area of triangle with vertices } (0,0), (0.5,0), (0,1) = \frac{16}{3} - \frac{1}{4} = \frac{61}{12}$.$\alpha + \beta = 61 + 12 = 73$.

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

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