Find the area of the region bounded by y `=x^(2)+1`,y=x ,x=0 and y=2

Answer

To find the area of the region bounded by the curves \( y = x^2 + 1 \), \( y = x \), \( x = 0 \), and \( y = 2 \), we can follow these steps:

### Step 1: Identify the curves and their intersections
We have the following curves:
1. \( y = x^2 + 1 \) (a parabola)
2. \( y = x \) (a straight line)
3. \( x = 0 \) (the y-axis)
4. \( y = 2 \) (a horizontal line)

First, we need to find the points of intersection of these curves.

**Finding the intersection of \( y = x \) and \( y = 2 \):**
Set \( x = 2 \).

**Finding the intersection of \( y = x^2 + 1 \) and \( y = 2 \):**
Set \( x^2 + 1 = 2 \):
\[ x^2 = 1 \implies x = \pm 1 \]
So, the points of intersection are:
- \( (1, 2) \)
- \( (-1, 2) \)

### Step 2: Set up the area calculation
The area we want to find is bounded by the curves from \( x = 0 \) to \( x = 1 \) (the intersection points with \( y = 2 \)).

### Step 3: Calculate the area
The area \( A \) can be calculated as:
\[ A = \int_{0}^{1} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]
Where:
- \( y_{\text{top}} = 2 \) (the line \( y = 2 \))
- \( y_{\text{bottom}} = x^2 + 1 \) (the parabola)

Thus, the integral becomes:
\[ A = \int_{0}^{1} (2 - (x^2 + 1)) \, dx \]
\[ A = \int_{0}^{1} (1 - x^2) \, dx \]

### Step 4: Evaluate the integral
Now, we compute the integral:
\[ A = \left[ x - \frac{x^3}{3} \right]_{0}^{1} \]
Calculating the limits:
\[ A = \left( 1 - \frac{1^3}{3} \right) - \left( 0 - 0 \right) \]
\[ A = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 5: Final area calculation
Now, we need to consider the area above the line \( y = x \) and below the line \( y = 2 \) from \( x = 1 \) to \( x = 2 \):
\[ A_2 = \int_{1}^{2} (2 - x) \, dx \]
Calculating this integral:
\[ A_2 = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2} \]
Calculating the limits:
\[ A_2 = \left( 4 - 2 \right) - \left( 2 - \frac{1}{2} \right) \]
\[ A_2 = 2 - \frac{3}{2} = \frac{1}{2} \]

### Step 6: Combine the areas
Now, we combine both areas:
\[ \text{Total Area} = A + A_2 = \frac{2}{3} + \frac{1}{2} \]
Finding a common denominator (6):
\[ \text{Total Area} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} \]

Thus, the area of the region bounded by the curves is:
\[ \boxed{\frac{7}{6}} \]