The area (in square units) bounded by`y=x^(2)+x+1 and x+y=2` is

Options

A. `2sqrt2`
B. `(4sqrt2)/(3)`
C. `(8sqrt2)/(3)`
D. `(sqrt2)/(3)`

Answer

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

### Step 1: Find the points of intersection
We start by equating the two equations to find the points where they intersect.

1. Rewrite the line equation \( x + y = 2 \) in terms of \( y \):
\[ y = 2 - x \]

2. Set the two equations equal to each other:
\[ x^2 + x + 1 = 2 - x \]

3. Rearranging gives:
\[ x^2 + 2x - 1 = 0 \]

4. Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -1 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]

5. Thus, the points of intersection are:
\[ x_1 = -1 - \sqrt{2}, \quad x_2 = -1 + \sqrt{2} \]

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x_1 \) to \( x_2 \) is given by:
\[ A = \int_{x_1}^{x_2} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]
Where \( y_{\text{top}} = 2 - x \) (the line) and \( y_{\text{bottom}} = x^2 + x + 1 \) (the parabola).

Thus, we have:
\[ A = \int_{x_1}^{x_2} \left( (2 - x) - (x^2 + x + 1) \right) \, dx \]

### Step 3: Simplify the integrand
Now simplify the integrand:
\[ A = \int_{x_1}^{x_2} \left( 2 - x - x^2 - x - 1 \right) \, dx = \int_{x_1}^{x_2} \left( 1 - 2x - x^2 \right) \, dx \]

### Step 4: Compute the integral
Now we compute the integral:
\[ A = \int_{x_1}^{x_2} (1 - 2x - x^2) \, dx \]

The antiderivative is:
\[ x - x^2 - \frac{x^3}{3} \]

### Step 5: Evaluate the definite integral
Now, we evaluate the integral from \( x_1 \) to \( x_2 \):
\[ A = \left[ x - x^2 - \frac{x^3}{3} \right]_{x_1}^{x_2} \]

Calculating this will involve substituting \( x_2 \) and \( x_1 \) into the antiderivative and finding the difference.

### Step 6: Substitute the limits
Substituting \( x_2 = -1 + \sqrt{2} \) and \( x_1 = -1 - \sqrt{2} \):
1. Calculate \( A(x_2) \):
\[ A(-1 + \sqrt{2}) = (-1 + \sqrt{2}) - (-1 + \sqrt{2})^2 - \frac{(-1 + \sqrt{2})^3}{3} \]
2. Calculate \( A(x_1) \):
\[ A(-1 - \sqrt{2}) = (-1 - \sqrt{2}) - (-1 - \sqrt{2})^2 - \frac{(-1 - \sqrt{2})^3}{3} \]

### Step 7: Final calculation
After substituting and simplifying, you will find the area \( A \).

### Final Area Calculation
After performing all calculations, the area bounded by the curves is:
\[ \text{Area} = \frac{8\sqrt{2}}{3} \text{ square units} \]