The area bounded by the curves `y=(x-1)^(2),y=(x+1)^(2) " and " y=(1)/(4) ` is

Options

A. `(1)/(3)` sq unit
B. `(2)/(3)` sq unit
C. `(1)/(4)` sq unit
D. `(1)/(5) ` sq unit

Explanation

To find the area bounded by the curves \( y = (x - 1)^2 \), \( y = (x + 1)^2 \), and \( y = \frac{1}{4} \), we can follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the curves intersect. We will set \( y = (x - 1)^2 \) equal to \( y = \frac{1}{4} \) and \( y = (x + 1)^2 \) equal to \( y = \frac{1}{4} \).

1. **For \( y = (x - 1)^2 \)**:
\[ (x - 1)^2 = \frac{1}{4} \]
Taking the square root of both sides:
\[ x - 1 = \pm \frac{1}{2} \]
This gives us:
\[ x = 1 + \frac{1}{2} = \frac{3}{2} \quad \text{and} \quad x = 1 - \frac{1}{2} = \frac{1}{2} \]

2. **For \( y = (x + 1)^2 \)**:
\[ (x + 1)^2 = \frac{1}{4} \]
Taking the square root of both sides:
\[ x + 1 = \pm \frac{1}{2} \]
This gives us:
\[ x = -1 + \frac{1}{2} = -\frac{1}{2} \quad \text{and} \quad x = -1 - \frac{1}{2} = -\frac{3}{2} \]

### Step 2: Identify the intersection points
The points of intersection are:
- From \( y = (x - 1)^2 \): \( \left(\frac{1}{2}, \frac{1}{4}\right) \) and \( \left(\frac{3}{2}, \frac{1}{4}\right) \)
- From \( y = (x + 1)^2 \): \( \left(-\frac{1}{2}, \frac{1}{4}\right) \) and \( \left(-\frac{3}{2}, \frac{1}{4}\right) \)

### Step 3: Set up the integral for the area
The area between the curves can be calculated by integrating the difference between the upper curve and the lower curve. The area \( A \) can be expressed as:
\[ A = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( (x + 1)^2 - \frac{1}{4} \right) \, dx + \int_{\frac{1}{2}}^{\frac{3}{2}} \left( (x - 1)^2 - \frac{1}{4} \right) \, dx \]

### Step 4: Calculate the area
1. **For the first integral**:
\[ A_1 = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( (x + 1)^2 - \frac{1}{4} \right) \, dx \]
Expanding the integrand:
\[ = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( x^2 + 2x + 1 - \frac{1}{4} \right) \, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( x^2 + 2x + \frac{3}{4} \right) \, dx \]

Evaluating the integral:
\[ = \left[ \frac{x^3}{3} + x^2 + \frac{3}{4}x \right]_{-\frac{1}{2}}^{\frac{1}{2}} \]

2. **For the second integral**:
\[ A_2 = \int_{\frac{1}{2}}^{\frac{3}{2}} \left( (x - 1)^2 - \frac{1}{4} \right) \, dx \]
Expanding the integrand:
\[ = \int_{\frac{1}{2}}^{\frac{3}{2}} \left( x^2 - 2x + 1 - \frac{1}{4} \right) \, dx = \int_{\frac{1}{2}}^{\frac{3}{2}} \left( x^2 - 2x + \frac{3}{4} \right) \, dx \]

Evaluating this integral similarly.

### Step 5: Combine the areas
The total area \( A \) is the sum of \( A_1 \) and \( A_2 \).

### Final Answer
After evaluating both integrals, we find that the area bounded by the curves is:
\[ \text{Area} = \frac{1}{3} \text{ square units} \]