STATEMENT-1 : The area bounded by the region `{(x,y) : 0 le y le x^(2) + 1 , 0 le y le x + 1 , 0 le x le 2}` is `(23)/(9)` sq. units
and
STATEMENT-2 : Required Area is ` int_(a)^(b) (y_(2) - y_(1)) `dx

Options

A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True , Statement-2 is False
D. Statement-1 is False , Statement-2 is True

Answer

To solve the problem of finding the area bounded by the curves defined in the question, we will follow these steps:

### Step 1: Understand the curves
We have two curves:
1. \( y = x^2 + 1 \)
2. \( y = x + 1 \)

We also have the limits for \( x \) as \( 0 \leq x \leq 2 \).

### Step 2: Find the intersection points
To find the area between these two curves, we first need to find their intersection points. We set the equations equal to each other:

\[ x^2 + 1 = x + 1 \]

Subtracting \( x + 1 \) from both sides gives:

\[ x^2 - x = 0 \]

Factoring out \( x \):

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

This gives us the intersection points:

\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 3: Set up the integral for the area
The area between the curves from \( x = 0 \) to \( x = 1 \) will be given by:

\[ \text{Area} = \int_{0}^{1} ((x + 1) - (x^2 + 1)) \, dx \]

From \( x = 1 \) to \( x = 2 \), the area will be:

\[ \text{Area} = \int_{1}^{2} ((x + 1) - 0) \, dx \]

### Step 4: Calculate the area from \( x = 0 \) to \( x = 1 \)
Calculating the integral:

\[ \int_{0}^{1} ((x + 1) - (x^2 + 1)) \, dx = \int_{0}^{1} (x - x^2) \, dx \]

Now, integrating:

\[ = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} \]

Calculating the limits:

\[ = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \]
\[ = \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Step 5: Calculate the area from \( x = 1 \) to \( x = 2 \)
Now, calculating the integral:

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

Integrating:

\[ = \left[ \frac{x^2}{2} + x \right]_{1}^{2} \]

Calculating the limits:

\[ = \left( \frac{2^2}{2} + 2 \right) - \left( \frac{1^2}{2} + 1 \right) \]
\[ = \left( 2 + 2 \right) - \left( \frac{1}{2} + 1 \right) = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2} \]

### Step 6: Total area
Now, we sum the areas from both intervals:

\[ \text{Total Area} = \frac{1}{6} + \frac{5}{2} \]

To add these fractions, we find a common denominator:

\[ \frac{1}{6} + \frac{15}{6} = \frac{16}{6} = \frac{8}{3} \]

### Final Result
Thus, the total area bounded by the curves is:

\[ \text{Area} = \frac{8}{3} \text{ square units} \]

### Conclusion
Now, let's evaluate the statements:
- **Statement 1** is incorrect as the area is \( \frac{8}{3} \) and not \( \frac{23}{9} \).
- **Statement 2** is correct as the area can be calculated using the integral \( \int_{a}^{b} (y_2 - y_1) \, dx \).