The area of the region bounded by the curves `x=y^(2)-2` and x=y is

Options

A. `9/4`
B. 9
C. `9/2`
D. `9/7`

Answer

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

### Step 1: Find the Points of Intersection
To find the points of intersection of the curves \( x = y^2 - 2 \) and \( x = y \), we set them equal to each other:

\[ y^2 - 2 = y \]

Rearranging gives us:

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

Factoring the quadratic equation:

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

Thus, the solutions are:

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

### Step 2: Determine the Corresponding x-values
Now we substitute these y-values back into either equation to find the corresponding x-values.

For \( y = 2 \):

\[ x = 2^2 - 2 = 4 - 2 = 2 \]

For \( y = -1 \):

\[ x = (-1)^2 - 2 = 1 - 2 = -1 \]

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

### Step 3: Set Up the Integral
The area \( A \) between the curves from \( y = -1 \) to \( y = 2 \) is given by the integral of the difference of the functions:

\[ A = \int_{-1}^{2} \left( (y^2 - 2) - y \right) dy \]

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

\[ A = \int_{-1}^{2} (y^2 - y - 2) \, dy \]

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

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

Calculating the integral at the upper limit \( y = 2 \):

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

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

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

Finding a common denominator (6):

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

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

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

Finding a common denominator (6):

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

Since area cannot be negative, we take the absolute value:

\[ A = \frac{9}{2} \]

### Final Answer
Thus, the area of the region bounded by the curves \( x = y^2 - 2 \) and \( x = y \) is:

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