Given: $$f(x) = \begin{cases} x, & 0 \le x \lt \frac{1}{2} \\ \frac{1}{2}, & x = \frac{1}{2} \\ 1-x, & \frac{1}{2} \lt x \le 1 \end{cases}$$
and $$g(x) = \left(x - \frac{1}{2}\right)^2, x \in R$$. Then, the area (in sq. units) of the region bounded by the curves, $$y = f(x)$$ and $$y = g(x)$$ between the lines $$2x = 1$$ and $$2x = \sqrt{3}$$, is:

We are asked to find the planar region enclosed by the two curves $$y = f(x)$$ and $$y = g(x)$$ between the vertical lines $$2x = 1$$ and $$2x = \sqrt{3}$$.

First convert the given vertical lines into the usual $$x = \text{constant}$$ form. From $$2x = 1$$ we obtain $$x = \dfrac{1}{2}$$, and from $$2x = \sqrt{3}$$ we get $$x = \dfrac{\sqrt{3}}{2}$$. Hence the entire region of interest lies on the interval $$\dfrac{1}{2} \le x \le \dfrac{\sqrt{3}}{2}$$.

Inside this interval, the definition of $$f(x)$$ that applies is $$f(x) = 1 - x$$ because $$x \gt \dfrac{1}{2}$$ but $$x \le 1$$. The function $$g(x)$$ is always $$g(x) = \left(x - \dfrac{1}{2}\right)^{2}$$. Thus, on the segment we care about, the two competing functions are

$$f(x) = 1 - x, \qquad g(x) = \left(x - \dfrac{1}{2}\right)^2.$$

To know which one is above the other, subtract:

$$f(x) - g(x) \;=\; (1 - x) - (x - \tfrac{1}{2})^{2}.$$

Expand the square:

$$(x - \tfrac{1}{2})^{2} \;=\; x^{2} - x + \tfrac14.$$

Substituting this, we get

$$f(x) - g(x) = (1 - x) - \bigl(x^{2} - x + \tfrac14\bigr) = 1 - x - x^{2} + x - \tfrac14 = \tfrac34 - x^{2}.$$

We clearly see that $$\tfrac34 - x^{2} \gt 0$$ as long as $$x^{2} \lt \tfrac34 \Longleftrightarrow x \lt \dfrac{\sqrt3}{2}$$, which indeed covers our entire interval except its right‐hand end, where equality holds. Thus, for every $$x$$ with $$\dfrac12 \le x \lt \dfrac{\sqrt3}{2}$$ we have $$f(x) \gt g(x)$$, and at $$x = \dfrac{\sqrt3}{2}$$ the two curves meet. Therefore the required bounded region is sandwiched between $$y = f(x)$$ on the top and $$y = g(x)$$ on the bottom.

The area between two curves from $$x = a$$ to $$x = b$$ is given by the integral formula

$$\text{Area} \;=\; \int_{a}^{b} \bigl[\;y_{\text{upper}} \;-\; y_{\text{lower}}\bigr] \,dx.$$

Here $$a = \dfrac12$$, $$b = \dfrac{\sqrt3}{2}$$, $$y_{\text{upper}} = f(x) = 1 - x$$ and $$y_{\text{lower}} = g(x) = (x - \tfrac12)^2$$. Hence,

$$\text{Area} = \int_{1/2}^{\sqrt3/2} \Bigl[(1 - x) - \bigl(x - \tfrac12\bigr)^2\Bigr] \,dx.$$

We have already simplified the integrand to $$\tfrac34 - x^{2}$$, so the integral becomes

$$\text{Area} = \int_{1/2}^{\sqrt3/2} \left(\tfrac34 - x^{2}\right)\,dx.$$

We now integrate term by term. The antiderivative of a constant $$k$$ is $$kx$$, and the antiderivative of $$x^{2}$$ is $$\dfrac{x^{3}}{3}$$. Therefore,

$$\int \left(\tfrac34 - x^{2}\right)\,dx \;=\; \tfrac34\,x \;-\; \dfrac{x^{3}}{3} + C.$$

Applying the limits $$x = \dfrac12$$ and $$x = \dfrac{\sqrt3}{2}$$ we write

$$\text{Area} = \left[\tfrac34\,x - \dfrac{x^{3}}{3}\right]_{\,x=\tfrac12}^{\,x=\tfrac{\sqrt3}{2}}.$$

First evaluate at the upper limit $$x = \dfrac{\sqrt3}{2}$$:

$$\tfrac34\!\left(\dfrac{\sqrt3}{2}\right) = \dfrac{3\sqrt3}{8},$$

$${x^{3}} = \left(\dfrac{\sqrt3}{2}\right)^{3} = \dfrac{(\sqrt3)^{3}}{8} = \dfrac{3\sqrt3}{8},$$

$$\dfrac{x^{3}}{3} = \dfrac{3\sqrt3/8}{3} = \dfrac{\sqrt3}{8}.$$

Thus the expression at the upper limit equals

$$\dfrac{3\sqrt3}{8} - \dfrac{\sqrt3}{8} = \dfrac{2\sqrt3}{8} = \dfrac{\sqrt3}{4}.$$

Now evaluate at the lower limit $$x = \dfrac12$$:

$$\tfrac34\!\left(\dfrac12\right) = \dfrac34 \times \dfrac12 = \dfrac38,$$

$${x^{3}} = \left(\dfrac12\right)^{3} = \dfrac18,$$

$$\dfrac{x^{3}}{3} = \dfrac{1/8}{3} = \dfrac1{24}.$$

Hence the expression at the lower limit equals

$$\dfrac38 - \dfrac1{24} = \dfrac{9}{24} - \dfrac1{24} = \dfrac8{24} = \dfrac13.$$

Finally subtract the two results to get the required area:

$$\text{Area} = \left(\dfrac{\sqrt3}{4}\right) - \left(\dfrac13\right) = \dfrac{\sqrt3}{4} - \dfrac13.$$

This value matches Option B in the list provided.

Hence, the correct answer is Option B.