Find the area enclosed between the curves
`y=x^(2)+1,y=2x-2` and the ordinates `x=-1` and x = 2

Answer

To find the area enclosed between the curves \( y = x^2 + 1 \) and \( y = 2x - 2 \) between the ordinates \( x = -1 \) and \( x = 2 \), we can follow these steps:

### Step 1: Find the points of intersection
To find the points where the curves intersect, we set the equations equal to each other:
\[ x^2 + 1 = 2x - 2 \]
Rearranging gives:
\[ x^2 - 2x + 3 = 0 \]
Calculating the discriminant:
\[ D = (-2)^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \]
Since the discriminant is negative, the curves do not intersect within the given range.

### Step 2: Evaluate the curves at the boundaries
Next, we evaluate the curves at the boundaries \( x = -1 \) and \( x = 2 \).

For \( x = -1 \):
\[ y_1 = (-1)^2 + 1 = 1 + 1 = 2 \]
\[ y_2 = 2(-1) - 2 = -2 - 2 = -4 \]

For \( x = 2 \):
\[ y_1 = (2)^2 + 1 = 4 + 1 = 5 \]
\[ y_2 = 2(2) - 2 = 4 - 2 = 2 \]

### Step 3: Determine the area between the curves
The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) can be calculated using the integral:
\[ A = \int_{-1}^{2} \left( (x^2 + 1) - (2x - 2) \right) \, dx \]

### Step 4: Simplify the integrand
Simplifying the integrand:
\[ A = \int_{-1}^{2} (x^2 + 1 - 2x + 2) \, dx = \int_{-1}^{2} (x^2 - 2x + 3) \, dx \]

### Step 5: Calculate the integral
Now we calculate the integral:
\[ A = \int_{-1}^{2} (x^2 - 2x + 3) \, dx \]
Calculating the antiderivative:
\[ = \left[ \frac{x^3}{3} - x^2 + 3x \right]_{-1}^{2} \]

Evaluating at the bounds:
\[ = \left( \frac{2^3}{3} - 2^2 + 3 \cdot 2 \right) - \left( \frac{(-1)^3}{3} - (-1)^2 + 3 \cdot (-1) \right) \]
\[ = \left( \frac{8}{3} - 4 + 6 \right) - \left( -\frac{1}{3} - 1 - 3 \right) \]
\[ = \left( \frac{8}{3} + 2 \right) - \left( -\frac{1}{3} - 4 \right) \]
\[ = \left( \frac{8}{3} + \frac{6}{3} \right) - \left( -\frac{1}{3} - \frac{12}{3} \right) \]
\[ = \frac{14}{3} + \frac{13}{3} = \frac{27}{3} = 9 \]

### Final Area
Thus, the area enclosed between the curves is:
\[ \boxed{9} \text{ square units} \]