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
To find the points where the curves intersect, we need to set the equations equal to each other.

The equation of the line can be rearranged as:
\[ y = 2 - x \]

Now, we set the two equations equal:
\[ x^2 = 2 - x \]

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

### Step 2: Factor the Quadratic Equation
Next, we will factor the quadratic equation:
\[ x^2 + x - 2 = (x - 1)(x + 2) = 0 \]

Setting each factor to zero gives us the points of intersection:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

Thus, the points of intersection are \( (-2, 4) \) and \( (1, 1) \).

### Step 3: Set Up the Integral for Area
The area \( A \) between the curves from \( x = -2 \) to \( x = 1 \) can be calculated using the integral:
\[ A = \int_{-2}^{1} [(2 - x) - (x^2)] \, dx \]

### Step 4: Simplify the Integral
Now, we simplify the integrand:
\[ A = \int_{-2}^{1} (2 - x - x^2) \, dx \]

### Step 5: Compute the Integral
We will compute the integral:
\[ A = \int_{-2}^{1} (2 - x - x^2) \, dx \]

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

### Step 6: Evaluate the Integral at the Limits
Now we will evaluate this at the limits:
1. For \( x = 1 \):
\[ = 2(1) - \frac{(1)^2}{2} - \frac{(1)^3}{3} = 2 - \frac{1}{2} - \frac{1}{3} = 2 - 0.5 - 0.3333 = 1.1667 \]

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

### Step 7: Combine the Results
Now we combine the results:
\[ A = \left(1.1667 - (-\frac{10}{3})\right) = 1.1667 + \frac{10}{3} \]
Converting \( 1.1667 \) to a fraction:
\[ = \frac{7}{6} + \frac{10}{3} = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2} \]

### Final Answer
Thus, the area bounded by the curves \( y = x^2 \) and \( x + y = 2 \) is:
\[ \boxed{\frac{9}{2}} \text{ square units} \]

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