Find the area of the region enclosed by the parabola `x^(2)=y` and the line y = x+ 2.

Explanation

To find the area of the region enclosed by the parabola \( x^2 = y \) and the line \( y = x + 2 \), we will follow these steps:

### Step 1: Find the Points of Intersection
We start by setting the equations of the parabola and the line equal to each other to find their points of intersection.

1. The parabola is given by:
\[ y = x^2 \]
2. The line is given by:
\[ y = x + 2 \]

Setting these two equations equal to each other:
\[ x^2 = x + 2 \]

Rearranging gives:
\[ x^2 - x - 2 = 0 \]

### Step 2: Solve the Quadratic Equation
We can factor the quadratic:
\[ (x - 2)(x + 1) = 0 \]

Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Find the Corresponding y-coordinates
Now we substitute these x-values back into the equation of the line to find the corresponding y-coordinates.

1. For \( x = 2 \):
\[ y = 2 + 2 = 4 \quad \Rightarrow \quad (2, 4) \]
2. For \( x = -1 \):
\[ y = -1 + 2 = 1 \quad \Rightarrow \quad (-1, 1) \]

### Step 4: Set Up the Integral for Area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{-1}^{2} \left( \text{Upper curve} - \text{Lower curve} \right) \, dx \]

In this case, the upper curve is the line \( y = x + 2 \) and the lower curve is the parabola \( y = x^2 \).

Thus, we have:
\[ A = \int_{-1}^{2} \left( (x + 2) - x^2 \right) \, dx \]

### Step 5: Simplify the Integral
This simplifies to:
\[ A = \int_{-1}^{2} \left( -x^2 + x + 2 \right) \, dx \]

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

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

### Step 7: Evaluate the Integral at the Limits
1. Evaluate at \( x = 2 \):
\[ -\frac{2^3}{3} + \frac{2^2}{2} + 2 \cdot 2 = -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + 6 = \frac{10}{3} \]
2. Evaluate at \( x = -1 \):
\[ -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2 \cdot (-1) = \frac{1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{3}{6} - \frac{12}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1}{3} - \frac{9}{6} = \frac{1 - 9}{6} = -\frac{8}{6} = -\frac{4}{3} \]

### Step 8: Calculate the Area
Now, substituting back into the area formula:
\[ A = \left( \frac{10}{3} - \left(-\frac{4}{3}\right) \right) = \frac{10}{3} + \frac{4}{3} = \frac{14}{3} \]

### Final Answer
Thus, the area of the region enclosed by the parabola and the line is:
\[ \boxed{\frac{14}{3}} \text{ square units} \]