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.

Answer

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

### Step 1: Identify the curves and their intersections
The curves we are dealing with are:
1. \( y = (x-1)^2 \)
2. \( y = (x+1)^2 \)
3. \( y = \frac{1}{4} \)

First, we need to find the points where these curves intersect.

### Step 2: Find the intersection points of \( y = (x-1)^2 \) and \( y = \frac{1}{4} \)
Set \( (x-1)^2 = \frac{1}{4} \):
\[ x - 1 = \frac{1}{2} \quad \text{or} \quad x - 1 = -\frac{1}{2} \]
This gives:
\[ x = \frac{3}{2} \quad \text{and} \quad x = \frac{1}{2} \]

### Step 3: Find the intersection points of \( y = (x+1)^2 \) and \( y = \frac{1}{4} \)
Set \( (x+1)^2 = \frac{1}{4} \):
\[ x + 1 = \frac{1}{2} \quad \text{or} \quad x + 1 = -\frac{1}{2} \]
This gives:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{3}{2} \]

### Step 4: Determine the area between the curves
The area we want to find is bounded by the curves from \( x = -\frac{1}{2} \) to \( x = \frac{3}{2} \).

### Step 5: Set up the integral for the area
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 6: Calculate the first integral
Calculate:
\[ \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( (x-1)^2 - \frac{1}{4} \right) \, dx \]
First, find \( (x-1)^2 \):
\[ (x-1)^2 = x^2 - 2x + 1 \]
So:
\[ \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( x^2 - 2x + 1 - \frac{1}{4} \right) \, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( x^2 - 2x + \frac{3}{4} \right) \, dx \]

### Step 7: Calculate the second integral
Calculate:
\[ \int_{\frac{1}{2}}^{\frac{3}{2}} \left( (x+1)^2 - \frac{1}{4} \right) \, dx \]
First, find \( (x+1)^2 \):
\[ (x+1)^2 = x^2 + 2x + 1 \]
So:
\[ \int_{\frac{1}{2}}^{\frac{3}{2}} \left( x^2 + 2x + 1 - \frac{1}{4} \right) \, dx = \int_{\frac{1}{2}}^{\frac{3}{2}} \left( x^2 + 2x + \frac{3}{4} \right) \, dx \]

### Step 8: Solve the integrals
1. For the first integral:
\[ \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( x^2 - 2x + \frac{3}{4} \right) \, dx = \left[ \frac{x^3}{3} - x^2 + \frac{3x}{4} \right]_{-\frac{1}{2}}^{\frac{1}{2}} \]

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

### Step 9: Combine the results
Add the results of both integrals to find the total area.

### Step 10: Final area calculation
After performing the calculations, we find that the total area is:
\[ \text{Total Area} = \frac{1}{3} \text{ square units} \]