The area (in sq. units) of the region `{(x,y):0leylex^(2)+1,0leylex+1,(1)/(2)lexle2}` is

Options

A. `(23)/(16)`
B. `(79)/(16)`
C. `(23)/(6)`
D. `(79)/(24)`

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 \(\frac{1}{2} \leq x \leq 2\), we will follow these steps:

### Step 1: Identify the curves and boundaries
The inequalities describe the following:
1. \(y = x^2 + 1\) is a parabola opening upwards.
2. \(y = x + 1\) is a straight line with a slope of 1.
3. The vertical lines \(x = \frac{1}{2}\) and \(x = 2\) define the left and right boundaries.

### Step 2: Find the intersection points of the curves
To find the intersection points between the parabola and the line, 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 solutions \(x = 0\) and \(x = 1\). Since we are only interested in the interval \(\frac{1}{2} \leq x \leq 2\), we will use \(x = 1\).

### Step 3: Set up the area calculation
The area can be calculated by integrating the difference between the upper function and the lower function over the defined intervals. We split the area into two parts:
1. From \(x = \frac{1}{2}\) to \(x = 1\), the upper function is \(y = x^2 + 1\) and the lower function is \(y = 0\).
2. From \(x = 1\) to \(x = 2\), the upper function is \(y = x + 1\) and the lower function is \(y = 0\).

### Step 4: Calculate the area using integration
The total area \(A\) can be expressed as:
\[ A = \int_{\frac{1}{2}}^{1} (x^2 + 1) \, dx + \int_{1}^{2} (x + 1) \, dx \]

#### Part 1: Calculate the first integral
\[ \int_{\frac{1}{2}}^{1} (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_{\frac{1}{2}}^{1} \]
Calculating the limits:
\[ = \left( \frac{1^3}{3} + 1 \right) - \left( \frac{(\frac{1}{2})^3}{3} + \frac{1}{2} \right) \]
\[ = \left( \frac{1}{3} + 1 \right) - \left( \frac{1}{24} + \frac{1}{2} \right) \]
\[ = \left( \frac{1}{3} + \frac{3}{3} \right) - \left( \frac{1}{24} + \frac{12}{24} \right) \]
\[ = \frac{4}{3} - \frac{13}{24} \]
Finding a common denominator (which is 24):
\[ = \frac{32}{24} - \frac{13}{24} = \frac{19}{24} \]

#### Part 2: Calculate the second integral
\[ \int_{1}^{2} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{1}^{2} \]
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 5: Combine the areas
Now we combine both areas:
\[ A = \frac{19}{24} + \frac{5}{2} \]
Converting \(\frac{5}{2}\) to a fraction with a denominator of 24:
\[ \frac{5}{2} = \frac{60}{24} \]
Thus,
\[ A = \frac{19}{24} + \frac{60}{24} = \frac{79}{24} \]

### Final Answer
The area of the region is \(\frac{79}{24}\) square units.

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