The area of the region bounded by y = x - 1 and `x=3-y^(2)` is

Options

A. `(5)/(2)`
B. `(7)/(2)`
C. `(9)/(2)`
D. 4

Answer

To find the area of the region bounded by the curves \( y = x - 1 \) and \( x = 3 - y^2 \), we will follow these steps:

### Step 1: Find the Intersection Points
We need to find the points where the two curves intersect. We can do this by substituting \( y = x - 1 \) into the equation \( x = 3 - y^2 \).

1. Substitute \( y = x - 1 \) into \( x = 3 - y^2 \):
\[ x = 3 - (x - 1)^2 \]
Expanding the equation:
\[ x = 3 - (x^2 - 2x + 1) \]
\[ x = 3 - x^2 + 2x - 1 \]
\[ x^2 - x - 2 = 0 \]

2. Factor the quadratic equation:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

3. Now, find the corresponding \( y \) values:
- For \( x = 2 \):
\[ y = 2 - 1 = 1 \]
- For \( x = -1 \):
\[ y = -1 - 1 = -2 \]

The intersection points are \( (2, 1) \) and \( (-1, -2) \).

### Step 2: Set Up the Integral
To find the area between the curves, we will integrate with respect to \( y \). The area \( A \) can be expressed as:
\[ A = \int_{y_1}^{y_2} (f_{\text{right}} - f_{\text{left}}) \, dy \]
where \( y_1 = -2 \) and \( y_2 = 1 \).

- The right curve is given by \( x = 3 - y^2 \).
- The left curve is given by \( x = y + 1 \).

Thus, the area can be expressed as:
\[ A = \int_{-2}^{1} \left( (3 - y^2) - (y + 1) \right) \, dy \]

### Step 3: Simplify the Integral
Now simplify the integrand:
\[ A = \int_{-2}^{1} (3 - y^2 - y - 1) \, dy = \int_{-2}^{1} (2 - y^2 - y) \, dy \]

### Step 4: Evaluate the Integral
Now we evaluate the integral:
\[ A = \int_{-2}^{1} (2 - y^2 - y) \, dy \]
This can be split into three separate integrals:
\[ A = \int_{-2}^{1} 2 \, dy - \int_{-2}^{1} y^2 \, dy - \int_{-2}^{1} y \, dy \]

Calculating each integral:
1. \(\int_{-2}^{1} 2 \, dy = 2[y]_{-2}^{1} = 2(1 - (-2)) = 2 \times 3 = 6\)
2. \(\int_{-2}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-2}^{1} = \left( \frac{1^3}{3} - \frac{(-2)^3}{3} \right) = \left( \frac{1}{3} - \frac{-8}{3} \right) = \frac{1 + 8}{3} = \frac{9}{3} = 3\)
3. \(\int_{-2}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{-2}^{1} = \left( \frac{1^2}{2} - \frac{(-2)^2}{2} \right) = \left( \frac{1}{2} - \frac{4}{2} \right) = \frac{1 - 4}{2} = -\frac{3}{2}\)

Putting it all together:
\[ A = 6 - 3 + \frac{3}{2} = 3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2} \]

### Final Answer
The area of the region bounded by the curves is:
\[ \boxed{\frac{9}{2}} \]