Find the area of the region in the first quadrant enclosed by `x=y^(2) and x =y+2`

Answer

To find the area of the region in the first quadrant enclosed by the curves \( x = y^2 \) and \( x = y + 2 \), we can follow these steps:

### Step 1: Sketch the curves
First, we need to understand the shapes of the curves we are dealing with.

- The curve \( x = y^2 \) is a parabola that opens to the right.
- The line \( x = y + 2 \) is a straight line with a y-intercept of 2 and a slope of 1.

### Step 2: Find the points of intersection
To find the area between the curves, we need to determine where they intersect. We set the equations equal to each other:

\[ y^2 = y + 2 \]

Rearranging gives:

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

Now we can factor this quadratic:

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

Thus, the solutions are:

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

Since we are interested in the first quadrant, we will take \( y = 2 \).

### Step 3: Set up the integral
Next, we set up the integral to calculate the area. We will use horizontal strips (dy) to find the area between the curves from \( y = 0 \) to \( y = 2 \).

The area \( A \) can be expressed as:

\[ A = \int_{0}^{2} (x_{\text{right}} - x_{\text{left}}) \, dy \]

Here, \( x_{\text{right}} = y + 2 \) and \( x_{\text{left}} = y^2 \). Thus, we have:

\[ A = \int_{0}^{2} ((y + 2) - (y^2)) \, dy \]

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

\[ A = \int_{0}^{2} (y + 2 - y^2) \, dy \]

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

\[ A = \int_{0}^{2} (-y^2 + y + 2) \, dy \]

Calculating the integral term by term:

\[ A = \left[ -\frac{y^3}{3} + \frac{y^2}{2} + 2y \right]_{0}^{2} \]

Evaluating at the bounds:

\[ = \left( -\frac{2^3}{3} + \frac{2^2}{2} + 2 \cdot 2 \right) - \left( -\frac{0^3}{3} + \frac{0^2}{2} + 2 \cdot 0 \right) \]

Calculating the terms:

\[ = \left( -\frac{8}{3} + 2 + 4 \right) \]

Converting 2 and 4 to fractions with a common denominator:

\[ = \left( -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} \right) \]

Combining the fractions:

\[ = \left( \frac{-8 + 6 + 12}{3} \right) = \frac{10}{3} \]

### Final Answer
Thus, the area of the region in the first quadrant enclosed by the curves is:

\[ \boxed{\frac{10}{3}} \]