Let the area of the region {(x, y): 2y leq x^ | vedclass

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(A) The region is bounded by the parabola $y = \frac{x^2+3}{2}$,the lines $y = 3-|x|$,and $y = |x-1|$.By analyzing the intersection points:$1$. Parabo

Vedclass · Accepted Answer

$16$

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(A) The region is bounded by the parabola $y = \frac{x^2+3}{2}$,the lines $y = 3-|x|$,and $y = |x-1|$.By analyzing the intersection points:$1$. Parabola $y = \frac{x^2+3}{2}$ and line $y = 3-x$ intersect at $x=1, y=2$ (point $C$).$2$. Parabola $y = \frac{x^2+3}{2}$ and line $y = 3+x$ intersect at $x=-1, y=2$ (point $E$).$3$. Line $y = 3-x$ and $y = x-1$ intersect at $x=2, y=1$ (point $B$).$4$. Line $y = 3+x$ and $y = 1-x$ intersect at $x=-1, y=2$ (point $E$).$5$. Line $y = x-1$ and $y = 0$ intersect at $x=1, y=0$ (point $A$).The area $A$ is the integral of the upper boundary minus the lower boundary.$A = \int_{-1}^{1} (3-|x| - \frac{x^2+3}{2}) dx + \int_{1}^{2} (3-x - (x-1)) dx$$A = \int_{-1}^{1} (\frac{3}{2} - |x| - \frac{x^2}{2}) dx + \int_{1}^{2} (4-2x) dx$$A = 2 \int_{0}^{1} (\frac{3}{2} - x - \frac{x^2}{2}) dx + [4x - x^2]_1^2$$A = 2 [\frac{3}{2}x - \frac{x^2}{2} - \frac{x^3}{6}]_0^1 + (8-4) - (4-1)$$A = 2 (\frac{3}{2} - \frac{1}{2} - \frac{1}{6}) + 4 - 3 = 2 (\frac{9-3-1}{6}) + 1 = 2(\frac{5}{6}) + 1 = \frac{5}{3} + 1 = \frac{8}{3}$.Wait,re-evaluating the region: The region is bounded by $y \leq \frac{x^2+3}{2}$,$y \leq 3-|x|$,and $y \geq |x-1|$.The area $A = \int_{-1}^{1} (3-|x| - \frac{x^2+3}{2}) dx + \int_{1}^{2} (3-x - (x-1)) dx = \frac{5}{3} + 1 = \frac{8}{3}$.Actually,calculating $6A = 6 	imes \frac{8}{3} = 16$.

Let the area of the region $\{(x, y): 2y \leq x^2+3, y +|x| \leq 3, y \geq|x-1|\}$ be $A$. Then $6A$ is equal to:

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