Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$,$x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$,$y^2=x$,$x=2$,and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to
- A$\frac{2}{3}(2\sqrt{2}+1)$
- B$\frac{2}{3}(4\sqrt{2}+1)$
- C$\frac{2}{3}(\sqrt{2}+1)$
- D$\frac{2}{3}(3\sqrt{2}+1)$
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$A_1 = \{(x, y) : x \geq 0, y \geq 0, 2x + 2y - x^2 - y^2 > 1 > x + y\}$
$A_2 = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^2 + y^2\}$
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$A_1 = \{(x, y) : x \geq 0, y \geq 0, 2x + 2y - x^2 - y^2 > 1 > x + y\}$
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