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To find the area of the enclosed region bounded by the curves \( y = \frac{2}{x} \), \( y = 8x \), and \( y = \frac{1}{2}x \), we will follow these steps:

1. **Find the points of intersection** of the curves. These points will serve as the limits for the integration.
2. **Set up integrals** that represent the area between the curves.
3. **Calculate the area** using definite integrals.

### Step 1: Find the points of intersection
We will find where the curves intersect by solving the equations pairwise:

#### Intersection of \( y = \frac{2}{x} \) and \( y = 8x \):
Set \( \frac{2}{x} = 8x \):
\[
\frac{2}{x} = 8x \implies 2 = 8x^2 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2}.
\]
At \( x = \frac{1}{2} \), the y-coordinate is \( y = 8\left(\frac{1}{2}\right) = 4 \). So, one intersection point is \( \left(\frac{1}{2}, 4\right) \).

#### Intersection of \( y = \frac{2}{x} \) and \( y = \frac{1}{2}x \):
Set \( \frac{2}{x} = \frac{1}{2}x \):
\[
\frac{2}{x} = \frac{1}{2}x \implies 2 = \frac{1}{2}x^2 \implies x^2 = 4 \implies x = 2.
\]
At \( x = 2 \), the y-coordinate is \( y = \frac{1}{2} \times 2 = 1 \). So, another intersection point is \( (2, 1) \).

#### Intersection of \( y = 8x \) and \( y = \frac{1}{2}x \):
Set \( 8x = \frac{1}{2}x \):
\[
8x = \frac{1}{2}x \implies 8 = \frac{1}{2} \implies \text{no solution}.
\]
So, these two lines do not intersect within the region \( x > 0 \).

### Step 2: Set up the integrals
We now have two points of intersection: \( \left(\frac{1}{2}, 4\right) \) and \( (2, 1) \). The region enclosed is between \( x = \frac{1}{2} \) and \( x = 2 \).

- From \( x = \frac{1}{2} \) to \( x = 1 \), the area between \( y = \frac{2}{x} \) and \( y = 8x \) is calculated.
- From \( x = 1 \) to \( x = 2 \), the area between \( y = \frac{2}{x} \) and \( y = \frac{1}{2}x \) is calculated.

So, the total area \( A \) is given by:
\[
A = \int_{\frac{1}{2}}^1 \left( \frac{2}{x} - 8x \right) dx + \int_1^2 \left( \frac{2}{x} - \frac{1}{2}x \right) dx.
\]

### Step 3: Calculate the area

#### First integral: \( \int_{\frac{1}{2}}^1 \left( \frac{2}{x} - 8x \right) dx \)
\[
\int_{\frac{1}{2}}^1 \frac{2}{x} dx = 2 \ln x \Big|_{\frac{1}{2}}^1 = 2 \ln 1 - 2 \ln \frac{1}{2} = 0 + 2 \ln 2 = 2 \ln 2,
\]
\[
\int_{\frac{1}{2}}^1 8x dx = 4x^2 \Big|_{\frac{1}{2}}^1 = 4(1^2) - 4\left(\frac{1}{2}\right)^2 = 4 - 1 = 3.
\]
Thus, the first integral is:
\[
2 \ln 2 - 3.
\]

#### Second integral: \( \int_1^2 \left( \frac{2}{x} - \frac{1}{2}x \right) dx \)
\[
\int_1^2 \frac{2}{x} dx = 2 \ln x \Big|_1^2 = 2 \ln 2 - 2 \ln 1 = 2 \ln 2,
\]
\[
\int_1^2 \frac{1}{2}x dx = \frac{1}{4}x^2 \Big|_1^2 = \frac{1}{4}(2^2) - \frac{1}{4}(1^2) = 1.
\]
Thus, the second integral is:
\[
2 \ln 2 - 1.
\]

### Final Area Calculation:
The total area is:
\[
A = (2 \ln 2 - 3) + (2 \ln 2 - 1) = 4 \ln 2 - 4.
\]

Thus, the area of the enclosed region is \( 4 \ln 2 - 4 \).

Would you like further details or clarification on any steps?

Here are 5 related questions to expand on this concept:
1. How do we determine which function is above the other in a given interval?
2. What happens if the limits of integration overlap or intersect?
3. Can you use geometry to find enclosed areas for simpler curves?
4. How do changes in the functions' slopes affect the shape of the enclosed region?
5. How would we handle finding the area if the functions were non-continuous?

**Tip:** Always sketch the curves to visually understand the region you are integrating over. It helps in determining the limits and the correct functions for integration.

Find the area of the enclosed region bounded by the curves y = 2/x, y = 8x, and y = 1/2x, x > 0.

This detailed solution guides you through finding the area of the region enclosed by the curves y = 2/x, y = 8x, and y = 1/2x using definite integrals. It covers key calculus concepts like curve intersection and setting up integrals between specific limits. Suitable for advanced high school or early university students.

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