Find the area enclosed between the straight line `y = x + 2 ` and the curve `x ^(2) =y.`

Answer

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

### Step 1: Identify the equations
The equations given are:
1. Straight line: \( y = x + 2 \)
2. Parabola: \( x^2 = y \)

### Step 2: Find the points of intersection
To find the area between the two curves, we first need to find the points where they intersect. We can do this by setting the equations equal to each other.

Substituting \( y \) from the line into the parabola:
\[ x^2 = x + 2 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]

### Step 3: Solve the quadratic equation
We can solve the quadratic equation \( x^2 - x - 2 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = -2 \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 + 8}}{2} \]
\[ x = \frac{1 \pm 3}{2} \]
This gives us two solutions:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 4: Find corresponding \( y \) values
Now we can find the corresponding \( y \) values for \( x = 2 \) and \( x = -1 \) using the line equation \( y = x + 2 \):
- For \( x = 2 \):
\[ y = 2 + 2 = 4 \quad \Rightarrow \quad (2, 4) \]
- For \( x = -1 \):
\[ y = -1 + 2 = 1 \quad \Rightarrow \quad (-1, 1) \]

### Step 5: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{-1}^{2} \left( (x + 2) - x^2 \right) \, dx \]

### Step 6: Calculate the integral
Now we compute the integral:
\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx \]

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

### Step 7: Evaluate the integral at the limits
First, evaluate at \( x = 2 \):
\[ = -\frac{2^3}{3} + \frac{2^2}{2} + 2 \cdot 2 \]
\[ = -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = \frac{10}{3} \]

Now, evaluate at \( x = -1 \):
\[ = -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2 \cdot (-1) \]
\[ = \frac{1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{3}{6} - \frac{12}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1 - 4.5}{3} = -\frac{3.5}{3} = -\frac{7}{6} \]

### Step 8: Combine results
Now subtract the two results:
\[ A = \left( \frac{10}{3} - \left( -\frac{7}{6} \right) \right) = \frac{10}{3} + \frac{7}{6} \]
Finding a common denominator (which is 6):
\[ = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \]

### Final Answer
Thus, the area enclosed between the line and the curve is:
\[ \boxed{\frac{9}{2}} \text{ square units} \]