The area (in square units), bounded by `y=2-x^(2)` and `x+y=0` , is

Options

A. `(7)/(2)` sq. units
B. `(9)/(2)` sq. units
C. 9 sq. units
D. none of these

Answer

To find the area bounded by the curves \( y = 2 - x^2 \) and \( x + y = 0 \), 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 set the equations equal to each other.

1. From \( x + y = 0 \), we can express \( y \) as \( y = -x \).
2. Substitute \( y = -x \) into \( y = 2 - x^2 \):

\[ -x = 2 - x^2 \]

Rearranging gives:

\[ x^2 - x - 2 = 0 \]

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

1. The equation factors to:

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

2. Thus, the solutions are:

\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Find the corresponding y-values
Now we will find the y-values for the points of intersection.

1. For \( x = 2 \):

\[ y = -2 \]

2. For \( x = -1 \):

\[ y = 1 \]

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

### Step 4: Set up the integral for the area
The area \( A \) bounded by the curves can be found by integrating the difference of the functions from \( x = -1 \) to \( x = 2 \):

\[ A = \int_{-1}^{2} \left( (2 - x^2) - (-x) \right) dx \]

This simplifies to:

\[ A = \int_{-1}^{2} (2 - x^2 + x) \, dx \]

### Step 5: Calculate the integral
Now we will calculate the integral:

1. First, integrate the function:

\[ \int (2 - x^2 + x) \, dx = 2x - \frac{x^3}{3} + \frac{x^2}{2} \]

2. Evaluate from \( -1 \) to \( 2 \):

\[ A = \left[ 2(2) - \frac{(2)^3}{3} + \frac{(2)^2}{2} \right] - \left[ 2(-1) - \frac{(-1)^3}{3} + \frac{(-1)^2}{2} \right] \]

Calculating the first part:

\[ = 4 - \frac{8}{3} + 2 = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3} \]

Now for the second part:

\[ = -2 + \frac{1}{3} + \frac{1}{2} = -2 + \frac{3}{6} + \frac{2}{6} = -2 + \frac{5}{6} = -\frac{12}{6} + \frac{5}{6} = -\frac{7}{6} \]

3. Combine the results:

\[ A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} \]

To combine these fractions, find a common denominator (which is 6):

\[ A = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \]

### Final Answer
The area bounded by the curves is \( \frac{9}{2} \) square units.

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