Using integration, find the area enclosed between the curve `y^(2) =2x+1`, and the `x-y-1=0`

Answer

To find the area enclosed between the curve \( y^2 = 2x + 1 \) and the line \( x - y - 1 = 0 \), we will follow these steps:

### Step 1: Find the intersection points of the curve and the line.

1. **Rewrite the line equation**: The line \( x - y - 1 = 0 \) can be rewritten as \( x = y + 1 \).
2. **Substitute into the curve equation**: Substitute \( x = y + 1 \) into the curve equation \( y^2 = 2x + 1 \):
\[ y^2 = 2(y + 1) + 1 \]
Simplifying this gives:
\[ y^2 = 2y + 2 + 1 \implies y^2 - 2y - 3 = 0 \]

### Step 2: Solve the quadratic equation.

3. **Factor the quadratic**: The equation \( y^2 - 2y - 3 = 0 \) can be factored as:
\[ (y - 3)(y + 1) = 0 \]
Thus, the solutions are:
\[ y = 3 \quad \text{and} \quad y = -1 \]

### Step 3: Find corresponding x-values for the intersection points.

4. **Find x-values**:
- For \( y = 3 \):
\[ x = 3 + 1 = 4 \quad \Rightarrow \quad (4, 3) \]
- For \( y = -1 \):
\[ x = -1 + 1 = 0 \quad \Rightarrow \quad (0, -1) \]

### Step 4: Set up the integral for the area.

5. **Identify the area to be integrated**: The area between the curves from \( y = -1 \) to \( y = 3 \) can be found using the formula:
\[ \text{Area} = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy \]
Here, \( x_{\text{right}} \) is given by the line \( x = y + 1 \) and \( x_{\text{left}} \) is given by the curve \( x = \frac{y^2 - 1}{2} \).

### Step 5: Write the integral.

6. **Set up the integral**:
\[ \text{Area} = \int_{-1}^{3} \left( (y + 1) - \frac{y^2 - 1}{2} \right) \, dy \]

### Step 6: Simplify the integrand.

7. **Simplify the expression**:
\[ \text{Area} = \int_{-1}^{3} \left( y + 1 - \frac{y^2}{2} + \frac{1}{2} \right) \, dy = \int_{-1}^{3} \left( -\frac{y^2}{2} + y + \frac{3}{2} \right) \, dy \]

### Step 7: Compute the integral.

8. **Integrate**:
\[ \text{Area} = \left[ -\frac{y^3}{6} + \frac{y^2}{2} + \frac{3y}{2} \right]_{-1}^{3} \]

### Step 8: Evaluate the definite integral.

9. **Evaluate at the limits**:
- At \( y = 3 \):
\[ -\frac{3^3}{6} + \frac{3^2}{2} + \frac{3 \cdot 3}{2} = -\frac{27}{6} + \frac{9}{2} + \frac{9}{2} = -\frac{27}{6} + \frac{18}{6} + \frac{18}{6} = -\frac{27}{6} + \frac{36}{6} = \frac{9}{6} = \frac{3}{2} \]
- At \( y = -1 \):
\[ -\frac{(-1)^3}{6} + \frac{(-1)^2}{2} + \frac{3(-1)}{2} = \frac{1}{6} + \frac{1}{2} - \frac{3}{2} = \frac{1}{6} + \frac{3}{6} - \frac{9}{6} = -\frac{5}{6} \]

### Step 9: Calculate the area.

10. **Final area calculation**:
\[ \text{Area} = \left( \frac{3}{2} - \left(-\frac{5}{6}\right) \right) = \frac{3}{2} + \frac{5}{6} = \frac{9}{6} + \frac{5}{6} = \frac{14}{6} = \frac{7}{3} \]

### Final Answer:
The area enclosed between the curve \( y^2 = 2x + 1 \) and the line \( x - y - 1 = 0 \) is \( \frac{7}{3} \).

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