The area of the region `{(x,y): xy le 8,1 le y le x^(2)}` is :

Options

A. `8 log_(e )2-(14)/(3)`
B. `8 log_(e )2-(7)/(3)`
C. `16 log_(e )2-6`
D. `16 log_(e )2-(14)/(3)`

Answer

To find the area of the region defined by the inequalities \( xy \leq 8 \), \( 1 \leq y \), and \( y \leq x^2 \), we can follow these steps:

### Step 1: Identify the curves
1. The first curve is given by \( xy = 8 \), which represents a rectangular hyperbola.
2. The second curve is \( y = 1 \), which is a horizontal line.
3. The third curve is \( y = x^2 \), which is an upward-opening parabola.

### Step 2: Sketch the curves
- Draw the rectangular hyperbola \( xy = 8 \), which approaches the axes but never touches them.
- Draw the line \( y = 1 \) horizontally across the graph.
- Draw the parabola \( y = x^2 \), which opens upwards.

### Step 3: Find the intersection points
To find the area of the region, we need to identify the points where these curves intersect.

1. Set \( y = x^2 \) into \( xy = 8 \):
\[ x \cdot x^2 = 8 \implies x^3 = 8 \implies x = 2 \]
Substituting \( x = 2 \) back into \( y = x^2 \):
\[ y = 2^2 = 4 \]
So, one intersection point is \( (2, 4) \).

2. The line \( y = 1 \) intersects the hyperbola \( xy = 8 \):
\[ x \cdot 1 = 8 \implies x = 8 \]
So, the intersection point is \( (8, 1) \).

### Step 4: Determine the area
The area we want to find is bounded by:
- The line \( y = 1 \)
- The hyperbola \( xy = 8 \)
- The parabola \( y = x^2 \)

The area can be calculated by integrating the difference of the functions from \( y = 1 \) to \( y = 4 \).

### Step 5: Set up the integral
The area \( A \) can be expressed as:
\[ A = \int_{1}^{4} \left( \frac{8}{y} - \sqrt{y} \right) dy \]
Where \( \frac{8}{y} \) is derived from rearranging \( xy = 8 \) to find \( x \) in terms of \( y \), and \( \sqrt{y} \) is derived from rearranging \( y = x^2 \).

### Step 6: Calculate the integral
Now we compute the integral:
\[ A = \int_{1}^{4} \left( \frac{8}{y} - y^{1/2} \right) dy \]
Calculating the two parts separately:
1. For \( \int \frac{8}{y} dy = 8 \ln |y| \)
2. For \( \int y^{1/2} dy = \frac{y^{3/2}}{3/2} = \frac{2}{3} y^{3/2} \)

Thus, we have:
\[ A = \left[ 8 \ln y - \frac{2}{3} y^{3/2} \right]_{1}^{4} \]

### Step 7: Evaluate the definite integral
Now we evaluate from \( 1 \) to \( 4 \):
\[ A = \left( 8 \ln 4 - \frac{2}{3} (4^{3/2}) \right) - \left( 8 \ln 1 - \frac{2}{3} (1^{3/2}) \right) \]
Since \( \ln 1 = 0 \):
\[ A = 8 \ln 4 - \frac{2}{3} \cdot 8 + 0 + \frac{2}{3} \]
\[ = 8 \ln 4 - \frac{16}{3} + \frac{2}{3} = 8 \ln 4 - \frac{14}{3} \]

### Final Answer
The area of the region is:
\[ A = 8 \ln 4 - \frac{14}{3} \]