Let the area of the region {(x, y): x-2y+4 ge | vedclass

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(D) The region is bounded by $x = 2y-4$,$x = -2y^2$,$x = 8-4y^2$,and $y = 0$. First,find the intersection points: $1$) $2y-4 = -2y^2 \implies y^2+y-2=

Vedclass · Accepted Answer

$119$

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(D) The region is bounded by $x = 2y-4$,$x = -2y^2$,$x = 8-4y^2$,and $y = 0$. First,find the intersection points: $1$) $2y-4 = -2y^2 \implies y^2+y-2=0 \implies (y+2)(y-1)=0$. Since $y \geq 0$,$y=1$. $2$) $2y-4 = 8-4y^2 \implies 4y^2+2y-12=0 \implies 2y^2+y-6=0 \implies (2y-3)(y+2)=0$. Since $y \geq 0$,$y=3/2$. $3$) $-2y^2 = 8-4y^2 \implies 2y^2=8 \implies y^2=4 \implies y=2$. The area $A$ is given by: $A = \int_0^1 [(8-4y^2) - (-2y^2)] dy + \int_1^{3/2} [(8-4y^2) - (2y-4)] dy$ $A = \int_0^1 (8-2y^2) dy + \int_1^{3/2} (12-2y-4y^2) dy$ $A = [8y - \frac{2y^3}{3}]_0^1 + [12y - y^2 - \frac{4y^3}{3}]_1^{3/2}$ $A = (8 - \frac{2}{3}) + [(18 - \frac{9}{4} - \frac{4}{3} \cdot \frac{27}{8}) - (12 - 1 - \frac{4}{3})]$ $A = \frac{22}{3} + [(18 - 2.25 - 4.5) - (11 - 1.333)] = \frac{22}{3} + [11.25 - 9.666] = \frac{22}{3} + \frac{45}{4} - \frac{29}{3} = \frac{107}{12}$. Thus,$m=107, n=12$. Since $\gcd(107, 12)=1$,$m+n = 107+12 = 119$.

Let the area of the region $\{(x, y): x-2y+4 \geq 0, x+2y^2 \geq 0, x+4y^2 \leq 8, y \geq 0\}$ be $\frac{m}{n}$,where $m$ and $n$ are coprime numbers. Then $m+n$ is equal to

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