Indicate the region bounded by the curves `x^(2)=y,y=x+2` and x-axis and the area enclosed by them is …...........

Answer

To find the area enclosed by the curves \(x^2 = y\), \(y = x + 2\), and the x-axis, we will follow these steps:

### Step 1: Identify the curves
The first curve \(x^2 = y\) is a parabola that opens upwards. The second curve \(y = x + 2\) is a straight line with a slope of 1 and y-intercept of 2. The x-axis is represented by the line \(y = 0\).

### Step 2: Find the points of intersection
To find the area enclosed by these curves, we need to find the points where they intersect. We will set \(y = x + 2\) equal to \(y = x^2\):

\[ x^2 = x + 2 \]

Rearranging gives us:

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

### Step 3: Solve the quadratic equation
We can factor the quadratic equation:

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

Thus, the solutions are:

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

### Step 4: Determine the y-coordinates of the intersection points
Now we will find the corresponding y-coordinates for these x-values using either of the original equations. We will use \(y = x + 2\):

1. For \(x = 2\):

\[ y = 2 + 2 = 4 \]

2. For \(x = -1\):

\[ y = -1 + 2 = 1 \]

So the points of intersection are \((-1, 1)\) and \((2, 4)\).

### Step 5: Set up the integral for the area
The area between the curves from \(x = -1\) to \(x = 2\) can be calculated by integrating the upper curve minus the lower curve. The upper curve is the line \(y = x + 2\) and the lower curve is the parabola \(y = x^2\).

The area \(A\) is given by:

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

### Step 6: Evaluate the integral
Now we will evaluate the integral:

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

Calculating the integral:

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

Finding the antiderivative:

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

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

\[ A = \left[-\frac{2^3}{3} + \frac{2^2}{2} + 2(2)\right] = \left[-\frac{8}{3} + 2 + 4\right] = \left[-\frac{8}{3} + \frac{6}{3} + \frac{12}{3}\right] = \left[-\frac{8}{3} + \frac{18}{3}\right] = \frac{10}{3} \]

Calculating at the lower limit \(x = -1\):

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

Finding a common denominator (6):

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

### Step 7: Combine the results
Now we combine the results:

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

Finding a common denominator (6):

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

### Final Answer
The area enclosed by the curves is:

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