The area (in sq. units) of the region {x in R | vedclass

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(B) The given region is bounded by $y = \sqrt{x}$,$y = x - 2$,$x = 0$,and $y = 0$.To find the intersection of $y = \sqrt{x}$ and $y = x - 2$,we set $\

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$\frac{10}{3}$

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(B) The given region is bounded by $y = \sqrt{x}$,$y = x - 2$,$x = 0$,and $y = 0$.To find the intersection of $y = \sqrt{x}$ and $y = x - 2$,we set $\sqrt{x} = x - 2$.Squaring both sides,we get $x = (x - 2)^2 = x^2 - 4x + 4$,which simplifies to $x^2 - 5x + 4 = 0$.Factoring gives $(x - 4)(x - 1) = 0$,so $x = 4$ or $x = 1$.Since $y = \sqrt{x}$ must be non-negative,at $x = 1$,$y = 1$ but $x - 2 = -1$,which is not equal. Thus,the intersection is at $(4, 2)$.The area is the integral of the upper curve minus the lower curve.For $0 \le x \le 2$,the region is bounded by $y = \sqrt{x}$ and $y = 0$. Area $A_1 = \int_{0}^{2} \sqrt{x} \, dx = [\frac{2}{3}x^{3/2}]_{0}^{2} = \frac{2}{3}(2\sqrt{2}) = \frac{4\sqrt{2}}{3}$.For $2 \le x \le 4$,the region is bounded by $y = \sqrt{x}$ and $y = x - 2$. Area $A_2 = \int_{2}^{4} (\sqrt{x} - (x - 2)) \, dx = [\frac{2}{3}x^{3/2} - \frac{x^2}{2} + 2x]_{2}^{4}$.$A_2 = (\frac{2}{3}(8) - \frac{16}{2} + 8) - (\frac{2}{3}(2\sqrt{2}) - \frac{4}{2} + 4) = (\frac{16}{3}) - (\frac{4\sqrt{2}}{3} + 2) = \frac{10}{3} - \frac{4\sqrt{2}}{3}$.Total Area = $A_1 + A_2 = \frac{4\sqrt{2}}{3} + \frac{10}{3} - \frac{4\sqrt{2}}{3} = \frac{10}{3}$ sq. units.

The area (in sq. units) of the region $\{x \in R : x \ge 0, y \ge 0, y \ge x - 2, \text{ and } y \le \sqrt{x}\}$ is

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