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 \(xy \leq 8\), \(1 \leq y \leq x^2\), we will follow these steps:

### Step 1: Identify the curves
1. The curve \(xy = 8\) can be rewritten as \(y = \frac{8}{x}\), which represents a rectangular hyperbola.
2. The curve \(y = x^2\) is a parabola opening upwards.
3. The line \(y = 1\) is a horizontal line.

### Step 2: Find the points of intersection
To find the points of intersection of the curves, we need to solve the equations:

1. Set \(y = x^2\) equal to \(y = \frac{8}{x}\):
\[ x^2 = \frac{8}{x} \]
Multiplying both sides by \(x\) (assuming \(x \neq 0\)):
\[ x^3 = 8 \implies x = 2 \]
Substituting \(x = 2\) back into \(y = x^2\):
\[ y = 2^2 = 4 \]
So, one intersection point is \((2, 4)\).

2. Next, find the intersection of \(y = x^2\) and \(y = 1\):
\[ x^2 = 1 \implies x = \pm 1 \]
We only consider \(x = 1\) since \(y = x^2\) must be greater than or equal to 1. Thus, the point is \((1, 1)\).

3. Now, find the intersection of \(y = \frac{8}{x}\) and \(y = 1\):
\[ 1 = \frac{8}{x} \implies x = 8 \]
So, the intersection point is \((8, 1)\).

### Step 3: Set up the area integral
The area we want to find is bounded by the curves from \(x = 1\) to \(x = 2\) and from \(x = 2\) to \(x = 8\).

1. From \(x = 1\) to \(x = 2\), the upper curve is \(y = x^2\) and the lower curve is \(y = 1\).
\[ \text{Area}_1 = \int_{1}^{2} (x^2 - 1) \, dx \]

2. From \(x = 2\) to \(x = 8\), the upper curve is \(y = \frac{8}{x}\) and the lower curve is \(y = 1\).
\[ \text{Area}_2 = \int_{2}^{8} \left(\frac{8}{x} - 1\right) \, dx \]

### Step 4: Compute the integrals
1. Compute \(\text{Area}_1\):
\[ \text{Area}_1 = \int_{1}^{2} (x^2 - 1) \, dx = \left[\frac{x^3}{3} - x\right]_{1}^{2} \]
Evaluating:
\[ = \left(\frac{2^3}{3} - 2\right) - \left(\frac{1^3}{3} - 1\right) = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right) \]
\[ = \left(\frac{8}{3} - \frac{6}{3}\right) - \left(\frac{1}{3} - \frac{3}{3}\right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

2. Compute \(\text{Area}_2\):
\[ \text{Area}_2 = \int_{2}^{8} \left(\frac{8}{x} - 1\right) \, dx = \left[8 \ln x - x\right]_{2}^{8} \]
Evaluating:
\[ = \left(8 \ln 8 - 8\right) - \left(8 \ln 2 - 2\right) = 8 \ln 8 - 8 - 8 \ln 2 + 2 \]
\[ = 8 (\ln 8 - \ln 2) - 6 = 8 \ln 4 - 6 = 8 \cdot 2 \ln 2 - 6 = 16 \ln 2 - 6 \]

### Step 5: Combine the areas
The total area is:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{4}{3} + (16 \ln 2 - 6) \]
\[ = 16 \ln 2 - \frac{14}{3} \]

### Final Answer
Thus, the area of the region is:
\[ \boxed{16 \ln 2 - \frac{14}{3}} \]