Find the area enclosed by the curve `y=-x^(2)` and the straight line `x+y+2=0`.

Explanation

To find the area enclosed by the curve \( y = -x^2 \) and the straight line \( x + y + 2 = 0 \), we will follow these steps:

### Step 1: Find the points of intersection
We start by rewriting the equation of the line in terms of \( y \):
\[ y = -x - 2 \]

Now, we set the equations of the curve and the line equal to each other to find their points of intersection:
\[ -x^2 = -x - 2 \]
Rearranging gives us:
\[ x^2 - x - 2 = 0 \]

### Step 2: Solve the quadratic equation
We can factor the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 3: Find the corresponding \( y \) values
Now we substitute \( x = 2 \) and \( x = -1 \) back into the equation of the line to find the corresponding \( y \) values.

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

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

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be found using the integral:
\[ A = \int_{-1}^{2} \left[ \text{Upper Curve} - \text{Lower Curve} \right] \, dx \]
Here, the upper curve is the line \( y = -x - 2 \) and the lower curve is the parabola \( y = -x^2 \).

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

### Step 5: Evaluate the integral
Now we compute the integral:
\[ A = \int_{-1}^{2} (x^2 - x - 2) \, dx \]

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

Calculating at the upper limit \( x = 2 \):
\[ \frac{2^3}{3} - \frac{2^2}{2} - 2(2) = \frac{8}{3} - 2 - 4 = \frac{8}{3} - \frac{6}{3} - \frac{12}{3} = \frac{8 - 6 - 12}{3} = \frac{-10}{3} \]

Calculating at the lower limit \( x = -1 \):
\[ \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) = -\frac{1}{3} - \frac{1}{2} + 2 = -\frac{1}{3} - \frac{3}{6} + \frac{12}{6} = -\frac{1}{3} + \frac{9}{6} = -\frac{1}{3} + \frac{3}{2} = -\frac{1}{3} + \frac{9}{6} = \frac{5}{6} \]

### Step 6: Combine the results
Now we combine the results:
\[ A = \left( \frac{-10}{3} - \frac{5}{6} \right) \]
Finding a common denominator (6):
\[ A = \left( \frac{-20}{6} - \frac{5}{6} \right) = \frac{-25}{6} \]

Since area cannot be negative, we take the absolute value:
\[ A = \frac{25}{6} \]

### Final Result
Thus, the area enclosed by the curve and the line is:
\[ \boxed{\frac{25}{6}} \]