Find the area of the region `{(x,y)//x^(2)-x-1 le y le -1}`

Answer

To find the area of the region defined by the inequalities \( y \leq x^2 - x - 1 \) and \( y \geq -1 \), we will follow these steps:

### Step 1: Identify the curves and points of intersection
We have two curves:
1. \( y = x^2 - x - 1 \)
2. \( y = -1 \)

To find the points of intersection, we set the two equations equal to each other:
\[ x^2 - x - 1 = -1 \]
This simplifies to:
\[ x^2 - x = 0 \]
Factoring gives:
\[ x(x - 1) = 0 \]
Thus, the points of intersection are:
\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 2: Sketch the curves
The curve \( y = x^2 - x - 1 \) is a parabola that opens upwards, and it intersects the line \( y = -1 \) at the points \( (0, -1) \) and \( (1, -1) \).

### Step 3: Set up the area integral
The area we want to find is bounded above by the curve \( y = x^2 - x - 1 \) and below by the line \( y = -1 \) from \( x = 0 \) to \( x = 1 \).

The area \( A \) can be expressed as:
\[ A = \int_{0}^{1} \left( (x^2 - x - 1) - (-1) \right) \, dx \]
This simplifies to:
\[ A = \int_{0}^{1} (x^2 - x) \, dx \]

### Step 4: Evaluate the integral
Now we compute the integral:
\[ A = \int_{0}^{1} (x^2 - x) \, dx = \int_{0}^{1} x^2 \, dx - \int_{0}^{1} x \, dx \]
Calculating each integral separately:
\[ \int x^2 \, dx = \frac{x^3}{3} \quad \text{and} \quad \int x \, dx = \frac{x^2}{2} \]
Evaluating from 0 to 1:
\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]
\[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]
Thus, we have:
\[ A = \frac{1}{3} - \frac{1}{2} \]
Finding a common denominator (which is 6):
\[ A = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{1}{6} \]

### Final Answer
The area of the region is:
\[ \boxed{\frac{1}{6}} \text{ square units.} \]