Find the area of the region bounded by the parabola `x ^(2) =y,` the line `y = x +2` and the x-axis.

Answer

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

### Step 1: Find the points of intersection
We need to determine where the parabola and the line intersect. This can be done by setting the equations equal to each other:

\[ x^2 = x + 2 \]

Rearranging gives:

\[ x^2 - x - 2 = 0 \]

### Step 2: Factor the quadratic equation
We can factor the quadratic equation:

\[ (x - 2)(x + 1) = 0 \]

This gives us the solutions:

\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be found by integrating the difference between the upper function (the line) and the lower function (the parabola):

\[ A = \int_{-1}^{2} ((x + 2) - (x^2)) \, dx \]

### Step 4: Simplify the integrand
Now we simplify the integrand:

\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx = \int_{-1}^{2} (-x^2 + x + 2) \, dx \]

### Step 5: Calculate the integral
Now we compute the integral:

\[ A = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} \]

Calculating the upper limit \( x = 2 \):

\[ A = -\frac{2^3}{3} + \frac{2^2}{2} + 2(2) = -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = \frac{10}{3} \]

Now for the lower limit \( x = -1 \):

\[ A = -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) = \frac{1}{3} + \frac{1}{2} - 2 \]

Finding a common denominator (6):

\[ = \frac{2}{6} + \frac{3}{6} - \frac{12}{6} = \frac{5 - 12}{6} = -\frac{7}{6} \]

### Step 6: Combine the results
Now we combine the results from the upper and lower limits:

\[ A = \left( \frac{10}{3} - \left(-\frac{7}{6}\right) \right) \]

Finding a common denominator (6):

\[ = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \]

Thus, the area of the region bounded by the parabola \( y = x^2 \), the line \( y = x + 2 \), and the x-axis is:

\[ \boxed{\frac{9}{2}} \]