Key Concepts and Formulas

• Area Between Curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by $\int_a^b |f(x) - g(x)| dx$.
• Definite Integral: The definite integral $\int_a^b f(x) dx$ represents the signed area under the curve $y = f(x)$ from $x = a$ to $x = b$.
• Given Functions: We are given f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right. and $g(x) = \left(x - \frac{1}{2}\right)^2$.

Step-by-Step Solution

Step 1: Determine the interval of integration.

We are asked to find the area between the curves $y = f(x)$ and $y = g(x)$ between the lines $2x = 1$ and $2x = \sqrt{3}$. This means we need to integrate from $x = \frac{1}{2}$ to $x = \frac{\sqrt{3}}{2}$.

Step 2: Determine which function is greater on the interval $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$.

On the interval $(\frac{1}{2}, 1]$, $f(x) = 1 - x$. Since $\frac{\sqrt{3}}{2} \in (\frac{1}{2}, 1]$, we consider $f(x) = 1 - x$ for $\frac{1}{2} \le x \le \frac{\sqrt{3}}{2}$.

Now we want to determine whether $f(x) > g(x)$ or $g(x) > f(x)$ on the interval $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$.

Let's consider the difference $f(x) - g(x) = (1 - x) - (x - \frac{1}{2})^2 = 1 - x - (x^2 - x + \frac{1}{4}) = 1 - x - x^2 + x - \frac{1}{4} = \frac{3}{4} - x^2$.

We want to see where $\frac{3}{4} - x^2 > 0$, which is equivalent to $x^2 < \frac{3}{4}$, or $-\frac{\sqrt{3}}{2} < x < \frac{\sqrt{3}}{2}$. Since our interval of integration is $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$, we have $f(x) > g(x)$ on this interval.

Step 3: Set up the integral for the area.

The area $A$ is given by $A = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} (f(x) - g(x)) dx = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} (1 - x - (x - \frac{1}{2})^2) dx = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} (\frac{3}{4} - x^2) dx$

Step 4: Evaluate the integral.

$A = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} (\frac{3}{4} - x^2) dx = \left[ \frac{3}{4}x - \frac{x^3}{3} \right]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} = \left( \frac{3}{4} \cdot \frac{\sqrt{3}}{2} - \frac{(\sqrt{3}/2)^3}{3} \right) - \left( \frac{3}{4} \cdot \frac{1}{2} - \frac{(1/2)^3}{3} \right)$ $A = \left( \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{24} \right) - \left( \frac{3}{8} - \frac{1}{24} \right) = \frac{3\sqrt{3}}{8} - \frac{\sqrt{3}}{8} - \frac{3}{8} + \frac{1}{24} = \frac{2\sqrt{3}}{8} - \frac{9}{24} + \frac{1}{24} = \frac{\sqrt{3}}{4} - \frac{8}{24} = \frac{\sqrt{3}}{4} - \frac{1}{3}$

Step 5: Compare to the options.

The area is $\frac{\sqrt{3}}{4} - \frac{1}{3}$. However, the "Correct Answer" given is ${1 \over 2} + {{\sqrt 3 } \over 4}$. There is an error in the problem statement. The correct answer to the problem as stated is $\frac{\sqrt{3}}{4} - \frac{1}{3}$.

Let's re-examine the problem and the given answer. If the correct answer is ${1 \over 2} + {{\sqrt 3 } \over 4}$ then there must be an error in our computation. Let's verify that $f(x) > g(x)$ on $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$: $f(x) - g(x) = (1-x) - (x - \frac{1}{2})^2 = 1 - x - x^2 + x - \frac{1}{4} = \frac{3}{4} - x^2$. At $x = \frac{1}{2}$, $f(x) - g(x) = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} > 0$. At $x = \frac{\sqrt{3}}{2}$, $f(x) - g(x) = \frac{3}{4} - \frac{3}{4} = 0$. So, our setup of the integral is correct.

Let's recalculate the definite integral: $A = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} (\frac{3}{4} - x^2) dx = \left[ \frac{3}{4}x - \frac{x^3}{3} \right]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} = \left( \frac{3}{4} \cdot \frac{\sqrt{3}}{2} - \frac{(\sqrt{3}/2)^3}{3} \right) - \left( \frac{3}{4} \cdot \frac{1}{2} - \frac{(1/2)^3}{3} \right)$ $A = \left( \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{8 \cdot 3} \right) - \left( \frac{3}{8} - \frac{1}{24} \right) = \left( \frac{3\sqrt{3}}{8} - \frac{\sqrt{3}}{8} \right) - \left( \frac{9}{24} - \frac{1}{24} \right) = \frac{2\sqrt{3}}{8} - \frac{8}{24} = \frac{\sqrt{3}}{4} - \frac{1}{3}$

Therefore, the area is $\frac{\sqrt{3}}{4} - \frac{1}{3}$.

Common Mistakes & Tips

• Incorrectly determining the interval: Make sure to correctly solve for the limits of integration based on the given lines.
• Incorrectly determining which function is greater: Carefully analyze the functions to determine which one is above the other on the interval of integration.
• Sign errors: Pay close attention to signs when evaluating the definite integral.

Summary

We found the area between the curves $f(x)$ and $g(x)$ by integrating the difference between the functions from $x = \frac{1}{2}$ to $x = \frac{\sqrt{3}}{2}$. We determined that $f(x) > g(x)$ on this interval and calculated the definite integral to be $\frac{\sqrt{3}}{4} - \frac{1}{3}$. However, the given correct answer is different. The correct answer to the problem as stated is $\frac{\sqrt{3}}{4} - \frac{1}{3}$.

Final Answer

The final answer is $\boxed{{\frac{\sqrt 3 } {4} - {1 \over 3}}}$, which corresponds to option (D).