Find the area of region : `{(x,y) : 0 le y le x^(2) + 1, 0 le y le x + 1,0 le x le 2}`.

Answer

To find the area of the region defined by the inequalities \(0 \leq y \leq x^2 + 1\), \(0 \leq y \leq x + 1\), and \(0 \leq x \leq 2\), we will follow these steps:

### Step 1: Identify the curves and their intersection points
The curves involved are:
1. \(y = x^2 + 1\) (an upward parabola)
2. \(y = x + 1\) (a straight line)

We need to find the points where these two curves intersect. To do this, we set them equal to each other:
\[ x^2 + 1 = x + 1 \]
Simplifying this gives:
\[ x^2 - x = 0 \]
Factoring out \(x\):
\[ x(x - 1) = 0 \]
Thus, \(x = 0\) and \(x = 1\) are the points of intersection.

### Step 2: Determine the y-coordinates of the intersection points
Now we can find the corresponding y-coordinates for these x-values:
- For \(x = 0\):
\[ y = 0^2 + 1 = 1 \quad \text{(Point: (0, 1))} \]
- For \(x = 1\):
\[ y = 1^2 + 1 = 2 \quad \text{(Point: (1, 2))} \]

### Step 3: Identify the area segments
The area we need to calculate is bounded by:
- From \(x = 0\) to \(x = 1\): The area under the curve \(y = x^2 + 1\)
- From \(x = 1\) to \(x = 2\): The area under the line \(y = x + 1\)

### Step 4: Set up the integral for the area
The area \(A\) can be expressed as:
\[ A = \int_{0}^{1} (x^2 + 1) \, dx + \int_{1}^{2} (x + 1) \, dx \]

### Step 5: Calculate the first integral
Calculating the first integral:
\[ \int_{0}^{1} (x^2 + 1) \, dx = \left[\frac{x^3}{3} + x\right]_{0}^{1} = \left(\frac{1^3}{3} + 1\right) - \left(0 + 0\right) = \frac{1}{3} + 1 = \frac{4}{3} \]

### Step 6: Calculate the second integral
Calculating the second integral:
\[ \int_{1}^{2} (x + 1) \, dx = \left[\frac{x^2}{2} + x\right]_{1}^{2} = \left(\frac{2^2}{2} + 2\right) - \left(\frac{1^2}{2} + 1\right) = \left(2 + 2\right) - \left(\frac{1}{2} + 1\right) = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2} \]

### Step 7: Combine the areas
Now, we can combine both areas:
\[ A = \frac{4}{3} + \frac{5}{2} \]
To add these fractions, we need a common denominator. The least common multiple of 3 and 2 is 6:
\[ A = \frac{4 \cdot 2}{6} + \frac{5 \cdot 3}{6} = \frac{8}{6} + \frac{15}{6} = \frac{23}{6} \]

### Final Answer
Thus, the area of the region is:
\[ \boxed{\frac{23}{6}} \text{ square units} \]