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

We have to find the area enclosed by all points that simultaneously satisfy the four inequalities $$x \ge 0,\; y \ge 0,\; y \ge x-2\; \text{and}\; y \le \sqrt{x}.$$

The curves that bound this region are the straight line $$y = x-2$$ and the right-opening parabola $$y = \sqrt{x}\; (\,\text{equivalently }x = y^{2}\,).$$ The additional conditions $$x \ge 0$$ and $$y \ge 0$$ keep us in the first quadrant.

First, we locate the point where the line and the parabola meet. We set $$x-2 = \sqrt{x}$$ and solve.

Let us write $$\sqrt{x}=t\;\Longrightarrow\;x=t^{2}.$$ Substituting this in $$x-2=\sqrt{x}$$ gives $$t^{2}-2=t.$$ Rearranging, $$t^{2}-t-2=0.$$ Using the quadratic formula $$t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,b=-1,c=-2,$$ we obtain $$t=\dfrac{1\pm\sqrt{1+8}}{2}=\dfrac{1\pm3}{2}.$$ The two roots are $$t=2$$ and $$t=-1.$$ Because $$t=\sqrt{x} \ge 0,$$ we keep only $$t=2.$$ Hence $$\sqrt{x}=2\;\Longrightarrow\;x=4,\quad y=2.$$ So, the curves meet at the point $$(4,\,2).$$

Next, we decide how the two curves sit relative to each other between $$x=0$$ and $$x=4.$$ For $$x \lt 4,$$ calculate the difference $$\sqrt{x}-(x-2).$$ At $$x=0,$$ it is $$0-(-2)=2\gt 0,$$ so the parabola is above the line. At $$x=4,$$ the difference is zero. Hence throughout $$0\le x\le4,$$ the curve $$y=\sqrt{x}$$ lies above $$y=x-2.$$ Beyond $$x=4,$$ the line rises faster and sits above the parabola, but those points cannot belong to the region because they would violate $$y\le\sqrt{x}.$$ Therefore the region is completely contained in $$0\le x\le4.$$

Because $$y \ge 0,$$ the actual lower boundary for $$y$$ is $$y=\max\{0,\;x-2\}.$$ Observe that

• For $$0\le x\le2,$$ we have $$x-2\le0,$$ so $$\max\{0,x-2\}=0.$$ Here the region stretches from the $$x$$-axis up to the parabola.
• For $$2\le x\le4,$$ the line is positive and becomes the lower boundary.

Hence we split the total area into two parts.

First part (from $$x=0$$ to $$x=2$$): The height of a vertical strip is $$\sqrt{x}-0.$$ Using the integral formula for area, $$A_{1}=\int_{0}^{2}\bigl(\sqrt{x}-0\bigr)\,dx=\int_{0}^{2}x^{1/2}\,dx.$$ We recall the power-rule antiderivative $$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C$$ for $$n\neq-1.$$ Taking $$n=\tfrac12,$$ $$\int x^{1/2}\,dx=\dfrac{x^{3/2}}{3/2}=\dfrac{2}{3}x^{3/2}.$$ Evaluating from $$0$$ to $$2,$$ $$A_{1}=\left.\dfrac{2}{3}x^{3/2}\right|_{0}^{2}=\dfrac{2}{3}\Bigl(2^{3/2}-0^{3/2}\Bigr)=\dfrac{2}{3}\cdot2\sqrt{2}=\dfrac{4\sqrt{2}}{3}.$$

Second part (from $$x=2$$ to $$x=4$$): Here the height of a strip is $$\sqrt{x}-(x-2).$$ Thus $$A_{2}=\int_{2}^{4}\Bigl(\sqrt{x}-(x-2)\Bigr)\,dx =\int_{2}^{4}x^{1/2}\,dx-\int_{2}^{4}(x-2)\,dx.$$

We have already integrated $$x^{1/2}$$, so $$\int_{2}^{4}x^{1/2}\,dx=\left.\dfrac{2}{3}x^{3/2}\right|_{2}^{4} =\dfrac{2}{3}\Bigl(4^{3/2}-2^{3/2}\Bigr) =\dfrac{2}{3}\Bigl(8-2\sqrt{2}\Bigr)=\dfrac{16}{3}-\dfrac{4\sqrt{2}}{3}.$$

Next, for the linear part, we use $$\int (x-2)\,dx=\dfrac{x^{2}}{2}-2x.$$ Therefore $$\int_{2}^{4}(x-2)\,dx=\left.\Bigl(\dfrac{x^{2}}{2}-2x\Bigr)\right|_{2}^{4} =\Bigl(8-8\Bigr)-\Bigl(2-4\Bigr)=0-(-2)=2.$$

Hence $$A_{2}=\Bigl(\dfrac{16}{3}-\dfrac{4\sqrt{2}}{3}\Bigr)-2 =\dfrac{16}{3}-\dfrac{4\sqrt{2}}{3}-\dfrac{6}{3} =\dfrac{10}{3}-\dfrac{4\sqrt{2}}{3}.$$

Finally, the total area is $$A=A_{1}+A_{2}= \dfrac{4\sqrt{2}}{3} + \Bigl(\dfrac{10}{3}-\dfrac{4\sqrt{2}}{3}\Bigr)=\dfrac{10}{3}.$$ All irrational terms cancel out, leaving a simple rational number.

Hence, the correct answer is Option B.