Let A_1 be the bounded area enclosed by the c | vedclass

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(A) First, we calculate $A_1$:$A_1 = \int_0^2 ((8-x) - (x^2+2)) dx = \int_0^2 (6-x-x^2) dx$$A_1 = [6x - \frac{x^2}{2} - \frac{x^3}{3}]_0^2 = 12 - 2 -

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$\frac{2}{3}(2\sqrt{2}+1)$

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(A) First, we calculate $A_1$:$A_1 = \int_0^2 ((8-x) - (x^2+2)) dx = \int_0^2 (6-x-x^2) dx$$A_1 = [6x - \frac{x^2}{2} - \frac{x^3}{3}]_0^2 = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$Next, we calculate $A_2$:$A_2 = \int_0^2 (x^2+2) dx - \int_0^2 \sqrt{x} dx$$A_2 = [\frac{x^3}{3} + 2x]_0^2 - [\frac{2}{3}x^{3/2}]_0^2$$A_2 = (\frac{8}{3} + 4) - \frac{2}{3}(2\sqrt{2}) = \frac{20}{3} - \frac{4\sqrt{2}}{3}$Finally, $A_1 - A_2 = \frac{22}{3} - (\frac{20}{3} - \frac{4\sqrt{2}}{3}) = \frac{2}{3} + \frac{4\sqrt{2}}{3} = \frac{2}{3}(1 + 2\sqrt{2})$

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

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