If `f(x) = { x, 0 lt x lt 1/2, 1/2, x = 1/2, 1-x , 1/2 lt x lt 1}` and `g(x) = (x-1/2)^2` then find the area bounded by f(x) and g(x) from `x = 1/2` to `x= sqrt3/2`

Options

A. `sqrt3/4 - 1/3`
B. `sqrt3/4 + 1/3`
C. `2sqrt3`
D. `3sqrt3`

Answer

To find the area bounded by the functions \( f(x) \) and \( g(x) \) from \( x = \frac{1}{2} \) to \( x = \frac{\sqrt{3}}{2} \), we will follow these steps:

### Step 1: Understand the Functions
The piecewise function \( f(x) \) is defined as:
- \( f(x) = x \) for \( 0 < x < \frac{1}{2} \)
- \( f(x) = \frac{1}{2} \) for \( x = \frac{1}{2} \)
- \( f(x) = 1 - x \) for \( \frac{1}{2} < x < 1 \)

The function \( g(x) \) is given as:
\[ g(x) = (x - \frac{1}{2})^2 \]

### Step 2: Find Intersection Points
To find the area between the curves, we need to determine where they intersect between \( x = \frac{1}{2} \) and \( x = \frac{\sqrt{3}}{2} \).

Set \( f(x) = g(x) \):
1. For \( \frac{1}{2} < x < 1 \):
\[ 1 - x = (x - \frac{1}{2})^2 \]
Expanding the right side:
\[ 1 - x = x^2 - x + \frac{1}{4} \]
Rearranging gives:
\[ x^2 - \frac{3}{4} = 0 \]
Solving this, we find:
\[ x = \frac{\sqrt{3}}{2} \]

Thus, the intersection point in the interval is \( \left(\frac{\sqrt{3}}{2}, 1 - \frac{\sqrt{3}}{2}\right) \).

### Step 3: Set Up the Area Integral
The area \( A \) between the curves from \( x = \frac{1}{2} \) to \( x = \frac{\sqrt{3}}{2} \) is given by:
\[ A = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} [f(x) - g(x)] \, dx \]
Since \( f(x) = 1 - x \) and \( g(x) = (x - \frac{1}{2})^2 \) in this interval, we have:
\[ A = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \left[(1 - x) - (x - \frac{1}{2})^2\right] \, dx \]

### Step 4: Simplify the Integrand
We simplify the integrand:
\[ (1 - x) - (x - \frac{1}{2})^2 = 1 - x - \left(x^2 - x + \frac{1}{4}\right) = 1 - x - x^2 + x - \frac{1}{4} = \frac{3}{4} - x^2 \]

### Step 5: Evaluate the Integral
Now we compute the integral:
\[ A = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \left(\frac{3}{4} - x^2\right) \, dx \]
Calculating the integral:
\[ = \left[\frac{3}{4}x - \frac{x^3}{3}\right]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \]

Calculating the upper limit:
\[ = \frac{3}{4} \cdot \frac{\sqrt{3}}{2} - \frac{(\frac{\sqrt{3}}{2})^3}{3} = \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{24} = \frac{3\sqrt{3}}{8} - \frac{\sqrt{3}}{8} = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4} \]

Calculating the lower limit:
\[ = \frac{3}{4} \cdot \frac{1}{2} - \frac{(\frac{1}{2})^3}{3} = \frac{3}{8} - \frac{1}{24} = \frac{9}{24} - \frac{1}{24} = \frac{8}{24} = \frac{1}{3} \]

Now, subtract the lower limit from the upper limit:
\[ A = \left(\frac{\sqrt{3}}{4} - \frac{1}{3}\right) \]

### Final Step: Combine Results
To combine these results, we need a common denominator:
\[ = \frac{3\sqrt{3}}{12} - \frac{4}{12} = \frac{3\sqrt{3} - 4}{12} \]

Thus, the area bounded by \( f(x) \) and \( g(x) \) from \( x = \frac{1}{2} \) to \( x = \frac{\sqrt{3}}{2} \) is:
\[ \boxed{\frac{3\sqrt{3} - 4}{12}} \]