Find the area bounded by the curves `0leylex^2+1,0le ylex+1,1/2le xle2`

Answer

To find the area bounded by the curves \(0 \leq y \leq x^2 + 1\), \(0 \leq y \leq x + 1\), and \(\frac{1}{2} \leq x \leq 2\), we will follow these steps:

### Step 1: Identify the curves and the region of integration
The curves we are dealing with are:
1. \(y = x^2 + 1\)
2. \(y = x + 1\)

We also have the vertical lines \(x = \frac{1}{2}\) and \(x = 2\) that bound our region horizontally.

### Step 2: Find the points of intersection
To determine where the curves intersect, we set \(x^2 + 1 = x + 1\):
\[ x^2 + 1 = x + 1 \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
This gives us \(x = 0\) and \(x = 1\).

### Step 3: Determine the area between the curves
We need to evaluate the area in two segments:
1. From \(x = \frac{1}{2}\) to \(x = 1\), where \(y = x^2 + 1\) is above \(y = 0\).
2. From \(x = 1\) to \(x = 2\), where \(y = x + 1\) is above \(y = 0\).

### Step 4: Set up the integrals
1. For the first segment \(\left[\frac{1}{2}, 1\right]\):
\[ \text{Area}_1 = \int_{\frac{1}{2}}^{1} (x^2 + 1 - 0) \, dx \]
2. For the second segment \(\left[1, 2\right]\):
\[ \text{Area}_2 = \int_{1}^{2} (x + 1 - 0) \, dx \]

### Step 5: Calculate the integrals
1. Calculate \(\text{Area}_1\):
\[ \text{Area}_1 = \int_{\frac{1}{2}}^{1} (x^2 + 1) \, dx = \left[\frac{x^3}{3} + x\right]_{\frac{1}{2}}^{1} \]
Evaluating this:
\[ = \left(\frac{1^3}{3} + 1\right) - \left(\frac{(\frac{1}{2})^3}{3} + \frac{1}{2}\right) = \left(\frac{1}{3} + 1\right) - \left(\frac{1}{24} + \frac{1}{2}\right) \]
\[ = \frac{4}{3} - \left(\frac{1}{24} + \frac{12}{24}\right) = \frac{4}{3} - \frac{13}{24} = \frac{32}{24} - \frac{13}{24} = \frac{19}{24} \]

2. Calculate \(\text{Area}_2\):
\[ \text{Area}_2 = \int_{1}^{2} (x + 1) \, dx = \left[\frac{x^2}{2} + x\right]_{1}^{2} \]
Evaluating this:
\[ = \left(\frac{2^2}{2} + 2\right) - \left(\frac{1^2}{2} + 1\right) = \left(2 + 2\right) - \left(\frac{1}{2} + 1\right) \]
\[ = 4 - \frac{3}{2} = \frac{8}{2} - \frac{3}{2} = \frac{5}{2} \]

### Step 6: Combine the areas
The total area \(A\) is given by:
\[ A = \text{Area}_1 + \text{Area}_2 = \frac{19}{24} + \frac{5}{2} = \frac{19}{24} + \frac{60}{24} = \frac{79}{24} \]

### Final Answer
The area bounded by the curves is \(\frac{79}{24}\) square units.