The area of the figure bounded by the curves `y^(2)=2x+1` and `x-y-1=0` , is

Options

A. `2//3`
B. `4//3`
C. `8//3`
D. `16//3`

Answer

To find the area of the region bounded by the curves \( y^2 = 2x + 1 \) (a parabola) and \( x - y - 1 = 0 \) (a straight line), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by \( y^2 = 2x + 1 \). This can be rewritten as:
\[ y = \sqrt{2x + 1} \quad \text{and} \quad y = -\sqrt{2x + 1} \]
The second curve is a straight line given by \( x - y - 1 = 0 \), which can be rewritten as:
\[ y = x - 1 \]

### Step 2: Find the points of intersection
To find the points where these two curves intersect, we set \( y = \sqrt{2x + 1} \) equal to \( y = x - 1 \):
\[ \sqrt{2x + 1} = x - 1 \]
Squaring both sides gives:
\[ 2x + 1 = (x - 1)^2 \]
Expanding the right side:
\[ 2x + 1 = x^2 - 2x + 1 \]
Rearranging the equation:
\[ 0 = x^2 - 4x \]
Factoring:
\[ 0 = x(x - 4) \]
Thus, the solutions are \( x = 0 \) and \( x = 4 \).

### Step 3: Determine the corresponding y-values
For \( x = 0 \):
\[ y = 0 - 1 = -1 \]
For \( x = 4 \):
\[ y = 4 - 1 = 3 \]
Thus, the points of intersection are \( (0, -1) \) and \( (4, 3) \).

### Step 4: Set up the area integrals
The area between the curves can be found by integrating the top curve minus the bottom curve from \( x = 0 \) to \( x = 4 \).

The top curve is \( y = \sqrt{2x + 1} \) and the bottom curve is \( y = x - 1 \).

The area \( A \) is given by:
\[ A = \int_{0}^{4} \left( \sqrt{2x + 1} - (x - 1) \right) \, dx \]

### Step 5: Calculate the integral
We can simplify the integral:
\[ A = \int_{0}^{4} \left( \sqrt{2x + 1} - x + 1 \right) \, dx \]

Now, we will calculate each part of the integral separately.

1. **Integrate \( \sqrt{2x + 1} \)**:
Let \( u = 2x + 1 \), then \( du = 2dx \) or \( dx = \frac{du}{2} \).
When \( x = 0 \), \( u = 1 \) and when \( x = 4 \), \( u = 9 \):
\[ \int \sqrt{2x + 1} \, dx = \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} (2x + 1)^{3/2} \]

Evaluating from 0 to 4:
\[ \left[ \frac{1}{3} (2(4) + 1)^{3/2} - \frac{1}{3} (2(0) + 1)^{3/2} \right] = \frac{1}{3} (9^{3/2} - 1^{3/2}) = \frac{1}{3} (27 - 1) = \frac{26}{3} \]

2. **Integrate \( -x \)**:
\[ \int_{0}^{4} -x \, dx = -\left[ \frac{x^2}{2} \right]_{0}^{4} = -\left[ \frac{16}{2} - 0 \right] = -8 \]

3. **Integrate \( 1 \)**:
\[ \int_{0}^{4} 1 \, dx = [x]_{0}^{4} = 4 \]

### Step 6: Combine the results
Now, combine the results of the integrals:
\[ A = \frac{26}{3} - 8 + 4 = \frac{26}{3} - \frac{24}{3} + \frac{12}{3} = \frac{26 - 24 + 12}{3} = \frac{14}{3} \]

### Final Answer
The area of the figure bounded by the curves is:
\[ \boxed{\frac{14}{3}} \]