1. Understand the curves and interval

The given functions are

and

The vertical lines are

So we need the area between the curves for

$x\in\left[\frac12,\frac{\sqrt3}{2}\right].$

2. Use the correct branch of $f(x)$

Since on this interval we have $x\ge \frac12$, the relevant branch is

$f(x)=1-x \quad \text{for } \frac12 < x \le 1.$

At the single point $x=\frac12$, $f\left(\frac12\right)=\frac12$, but a single point does not affect area.

Thus area is computed between

$y=1-x$

and

$y=\left(x-\frac12\right)^2.$

3. Find which curve lies above the other

Compare

$1-x \quad \text{and} \quad \left(x-\frac12\right)^2.$

Set them equal:

Expand:

So,

In our interval $\left[\frac12,\frac{\sqrt3}{2}\right]$, the curves meet at

Test a point, say $x=\frac12$:

Hence $f(x)>g(x)$ on the interval, so

$\text{Area}=\int_{1/2}^{\sqrt3/2}\left[(1-x)-\left(x-\frac12\right)^2\right]dx.$

4. Simplify the integrand

Therefore,

$(1-x)-\left(x^2-x+\frac14\right)=1-x-x^2+x-\frac14=\frac34-x^2.$

So,

$\text{Area}=\int_{1/2}^{\sqrt3/2}\left(\frac34-x^2\right)dx.$

5. Integrate

Thus,

$\text{Area}=\left[\frac34x-\frac{x^3}{3}\right]_{1/2}^{\sqrt3/2}.$

Now evaluate.

At $x=\frac{\sqrt3}{2}$:

At $x=\frac12$:

Hence,

$\text{Area}=\frac{\sqrt3}{4}-\frac13.$

6. Match with options

$\boxed{\frac{\sqrt3}{4}-\frac13}$

This is Option D.

7. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the answer agrees with the stored answer.