Area bounded by the curves `y=x^2 - 1 and x+y=3` is:

Options

A. (a) `(9)/(2)`
B. (b) `4`
C. (c) `(7sqrt17)/(2)`
D. (d) `(17sqrt17)/(6)`

Answer

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

### Step 1: Find the points of intersection
We need to find the points where the two curves intersect. To do this, we will set the equations equal to each other.

1. From the line equation \( x + y = 3 \), we can express \( y \) in terms of \( x \):
\[ y = 3 - x \]

2. Now, we set the two equations equal to each other:
\[ x^2 - 1 = 3 - x \]

3. Rearranging gives:
\[ x^2 + x - 4 = 0 \]

### Step 2: Solve the quadratic equation
We will use the quadratic formula to find the values of \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -4 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \]
Let \( \alpha = \frac{-1 + \sqrt{17}}{2} \) and \( \beta = \frac{-1 - \sqrt{17}}{2} \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = \beta \) to \( x = \alpha \) is given by:
\[ A = \int_{\beta}^{\alpha} \left( (3 - x) - (x^2 - 1) \right) \, dx \]

### Step 4: Simplify the integrand
Now simplify the integrand:
\[ A = \int_{\beta}^{\alpha} \left( 3 - x - x^2 + 1 \right) \, dx = \int_{\beta}^{\alpha} \left( -x^2 - x + 4 \right) \, dx \]

### Step 5: Integrate
Now we integrate:
\[ A = \int_{\beta}^{\alpha} \left( -x^2 - x + 4 \right) \, dx = \left[ -\frac{x^3}{3} - \frac{x^2}{2} + 4x \right]_{\beta}^{\alpha} \]

### Step 6: Evaluate the definite integral
Substituting the limits:
\[ A = \left( -\frac{\alpha^3}{3} - \frac{\alpha^2}{2} + 4\alpha \right) - \left( -\frac{\beta^3}{3} - \frac{\beta^2}{2} + 4\beta \right) \]

### Step 7: Substitute \( \alpha \) and \( \beta \)
Now we will compute \( \alpha^3, \alpha^2, \beta^3, \beta^2 \) using the values of \( \alpha \) and \( \beta \) found earlier.

### Step 8: Calculate the area
After substituting and simplifying, we find the area \( A \).

### Final Result
The area bounded by the curves \( y = x^2 - 1 \) and \( x + y = 3 \) is:
\[ A = \frac{17\sqrt{17}}{6} \]