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 \(x + y = 2\), we can follow these steps:

### Step 1: Rewrite the equations
We have the first curve as:
\[ y = x^2 + x + 1 \]
The second equation can be rewritten as:
\[ y = 2 - x \]

### Step 2: Find the points of intersection
To find the area between the two curves, we first need to determine the points where they intersect. We set the two equations equal to each other:
\[ x^2 + x + 1 = 2 - x \]
Rearranging gives:
\[ x^2 + 2x - 1 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula to solve for \(x\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 2\), and \(c = -1\):
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]
Thus, the points of intersection are:
\[ x_1 = -1 - \sqrt{2}, \quad x_2 = -1 + \sqrt{2} \]

### Step 4: Set up the integral for the area
The area \(A\) between the curves from \(x_1\) to \(x_2\) is given by the integral:
\[ A = \int_{x_1}^{x_2} ((2 - x) - (x^2 + x + 1)) \, dx \]
This simplifies to:
\[ A = \int_{x_1}^{x_2} (1 - x - x^2) \, dx \]

### Step 5: Calculate the integral
Now we compute the integral:
\[ A = \int (1 - x - x^2) \, dx = x - \frac{x^2}{2} - \frac{x^3}{3} + C \]
Evaluating this from \(x_1\) to \(x_2\):
\[ A = \left[ x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{x_1}^{x_2} \]

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

### Step 7: Simplify and calculate the area
After substituting and simplifying, we find that:
\[ A = \frac{8\sqrt{2}}{3} \]

### Final Answer
Thus, the area bounded by the curves is:
\[ \boxed{\frac{8\sqrt{2}}{3}} \text{ square units} \]