Let $$A_{1}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,x+y=8$$ and 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 y-axis that lies in the first quadrant. Then $$A_{1}-A_{2}$$ is equal to

Given curves for $$A_1$$:

$$y = x^2 + 2, \quad x + y = 8 \implies y = 8 - x, \quad x = 0 \quad (\text{y-axis})$$

Finding the point of intersection of $$y = x^2 + 2$$ and $$y = 8 - x$$:

$$x^2 + 2 = 8 - x \implies x^2 + x - 6 = 0 \implies (x+3)(x-2) = 0 \implies x = 2 \quad (\text{since } x \ge 0)$$

$$A_1 = \int_{0}^{2} \left[ (8 - x) - (x^2 + 2) \right] dx = \int_{0}^{2} (6 - x - x^2) dx$$

$$A_1 = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$$

Given curves for $$A_2$$:

$$y = x^2 + 2, \quad y^2 = x \implies y = \sqrt{x}, \quad x = 2, \quad x = 0$$

$$A_2 = \int_{0}^{2} \left[ (x^2 + 2) - \sqrt{x} \right] dx$$

$$A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2} = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{20}{3} - \frac{4\sqrt{2}}{3}$$

$$A_1 - A_2 = \frac{22}{3} - \left( \frac{20}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{2}{3} + \frac{4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2} + 1)$$