The area bounded by `y = x^(2) , x + y = 2` is

Options

A. `(9)/(2)`
B. `(15)/(2)`
C. `9`
D. `(15`

Answer

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

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

\[ x + x^2 = 2 \]
Rearranging gives us:
\[ x^2 + x - 2 = 0 \]

### Step 2: Solve the Quadratic Equation
Now, we will solve the quadratic equation \( x^2 + x - 2 = 0 \) using the factorization method.

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

This gives us the solutions:
\[ x = 1 \quad \text{and} \quad x = -2 \]

### Step 3: Find Corresponding y-values
Now we will find the corresponding y-values for these x-values using \( y = x^2 \).

For \( x = 1 \):
\[ y = 1^2 = 1 \quad \Rightarrow \quad (1, 1) \]

For \( x = -2 \):
\[ y = (-2)^2 = 4 \quad \Rightarrow \quad (-2, 4) \]

### Step 4: Set Up the Integral
The area \( A \) between the curves from \( x = -2 \) to \( x = 1 \) can be found using the integral:
\[ A = \int_{-2}^{1} \left( (2 - x) - x^2 \right) \, dx \]

### Step 5: Simplify the Integrand
Now, we simplify the integrand:
\[ (2 - x) - x^2 = 2 - x - x^2 \]

### Step 6: Evaluate the Integral
Now we will evaluate the integral:
\[ A = \int_{-2}^{1} (2 - x - x^2) \, dx \]

Calculating the integral:
\[ A = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{1} \]

### Step 7: Calculate the Definite Integral
Now we will calculate the definite integral at the limits \( x = 1 \) and \( x = -2 \).

At \( x = 1 \):
\[ = 2(1) - \frac{1^2}{2} - \frac{1^3}{3} = 2 - \frac{1}{2} - \frac{1}{3} = 2 - \frac{3}{6} - \frac{2}{6} = 2 - \frac{5}{6} = \frac{12}{6} - \frac{5}{6} = \frac{7}{6} \]

At \( x = -2 \):
\[ = 2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{18}{3} + \frac{8}{3} = -\frac{10}{3} \]

### Step 8: Combine the Results
Now we combine the results:
\[ A = \left( \frac{7}{6} - \left(-\frac{10}{3}\right) \right) = \frac{7}{6} + \frac{10}{3} \]

To add these fractions, we find a common denominator (which is 6):
\[ \frac{10}{3} = \frac{20}{6} \]
Thus,
\[ A = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} \]

### Final Answer
The area bounded by the curves is:
\[ \boxed{\frac{27}{6}} \text{ square units.} \]