Find the area enclosed between the curves
`y^(2)=2x+6` and `y=x-1`

Answer

To find the area enclosed between the curves \( y^2 = 2x + 6 \) and \( y = x - 1 \), we can follow these steps:

### Step 1: Find the points of intersection
We will set the equations equal to each other to find the points of intersection.

1. From the equation \( y = x - 1 \), we can express \( x \) in terms of \( y \):
\[ x = y + 1 \]

2. Substitute \( x = y + 1 \) into the equation \( y^2 = 2x + 6 \):
\[ y^2 = 2(y + 1) + 6 \]
Simplifying this gives:
\[ y^2 = 2y + 2 + 6 \]
\[ y^2 - 2y - 8 = 0 \]

3. Factor the quadratic:
\[ (y - 4)(y + 2) = 0 \]
Thus, the solutions for \( y \) are:
\[ y = 4 \quad \text{and} \quad y = -2 \]

### Step 2: Find corresponding \( x \) values
Now, we will find the \( x \) values corresponding to the \( y \) values we found.

1. For \( y = 4 \):
\[ x = 4 + 1 = 5 \]

2. For \( y = -2 \):
\[ x = -2 + 1 = -1 \]

### Step 3: Set up the integral for the area
The area \( A \) enclosed between the curves from \( y = -2 \) to \( y = 4 \) can be calculated using the integral:
\[ A = \int_{-2}^{4} (x_{\text{right}} - x_{\text{left}}) \, dy \]
Where \( x_{\text{right}} \) is from the line \( y = x - 1 \) and \( x_{\text{left}} \) is from the parabola \( y^2 = 2x + 6 \).

1. From \( y = x - 1 \):
\[ x_{\text{right}} = y + 1 \]

2. From \( y^2 = 2x + 6 \):
\[ x_{\text{left}} = \frac{y^2 - 6}{2} \]

Thus, the area becomes:
\[ A = \int_{-2}^{4} \left( (y + 1) - \frac{y^2 - 6}{2} \right) dy \]

### Step 4: Simplify the integrand
Now, simplify the expression inside the integral:
\[ A = \int_{-2}^{4} \left( y + 1 - \frac{y^2}{2} + 3 \right) dy \]
\[ = \int_{-2}^{4} \left( -\frac{y^2}{2} + y + 4 \right) dy \]

### Step 5: Calculate the integral
Now we can compute the integral:
\[ A = \int_{-2}^{4} \left( -\frac{y^2}{2} + y + 4 \right) dy \]
Calculating the integral:
\[ = \left[ -\frac{y^3}{6} + \frac{y^2}{2} + 4y \right]_{-2}^{4} \]

### Step 6: Evaluate the definite integral
1. Evaluate at \( y = 4 \):
\[ -\frac{4^3}{6} + \frac{4^2}{2} + 4 \cdot 4 = -\frac{64}{6} + 8 + 16 = -\frac{64}{6} + 24 = -\frac{64}{6} + \frac{144}{6} = \frac{80}{6} = \frac{40}{3} \]

2. Evaluate at \( y = -2 \):
\[ -\frac{(-2)^3}{6} + \frac{(-2)^2}{2} + 4 \cdot (-2) = \frac{8}{6} + 2 - 8 = \frac{8}{6} + 2 - \frac{48}{6} = \frac{8 + 12 - 48}{6} = \frac{-28}{6} = -\frac{14}{3} \]

### Step 7: Find the area
Now, subtract the two evaluations:
\[ A = \left( \frac{40}{3} - \left(-\frac{14}{3}\right) \right) = \frac{40}{3} + \frac{14}{3} = \frac{54}{3} = 18 \]

Thus, the area enclosed between the curves is:
\[ \boxed{18} \text{ square units} \]