y = 2/x,  y = 8x,  y = (1/8)x,  x > 0
Find the area of the region.

To find the area of the region enclosed by the curves \( y = \frac{2}{x} \), \( y = 8x \), and \( y = \frac{1}{8}x \) with the restriction \( x > 0 \), we need to follow these steps:

1. **Find the points of intersection** of the curves to determine the boundaries of the region.
2. **Set up the integral(s)** to calculate the area between these curves.

### Step 1: Find the Points of Intersection

We need to find the \( x \)-coordinates where the curves intersect by solving the following equations:

1. **Intersection of \( y = \frac{2}{x} \) and \( y = 8x \):**

   \[
   \frac{2}{x} = 8x
   \]

   Multiplying both sides by \( x \) gives:

   \[
   2 = 8x^2 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2}
   \]

2. **Intersection of \( y = \frac{2}{x} \) and \( y = \frac{1}{8}x \):**

   \[
   \frac{2}{x} = \frac{1}{8}x
   \]

   Multiplying both sides by \( 8x \) gives:

   \[
   16 = x^2 \implies x = 4
   \]

3. **Intersection of \( y = 8x \) and \( y = \frac{1}{8}x \):**

   \[
   8x = \frac{1}{8}x
   \]

   This equation simplifies to:

   \[
   x = 0
   \]

   Since \( x > 0 \), there is no intersection in this region, so we do not consider this case.

Thus, the curves intersect at \( x = \frac{1}{2} \) and \( x = 4 \).

### Step 2: Set Up the Integral to Calculate the Area

To find the area between the curves, we evaluate the integral of the difference between the top and bottom functions from \( x = \frac{1}{2} \) to \( x = 4 \).

- For \( \frac{1}{2} \leq x \leq 4 \):
  - The top curve is \( y = \frac{2}{x} \).
  - The bottom curve is the maximum of \( y = 8x \) and \( y = \frac{1}{8}x \).

However, \( y = 8x \) dominates \( y = \frac{1}{8}x \) in this interval, so we subtract \( y = 8x \) from \( y = \frac{2}{x} \).

The area \( A \) is then given by:

\[
A = \int_{\frac{1}{2}}^{4} \left( \frac{2}{x} - 8x \right) \, dx
\]

### Step 3: Evaluate the Integral

Let's compute the integral:

\[
A = \int_{\frac{1}{2}}^{4} \left( \frac{2}{x} - 8x \right) \, dx
\]

\[
A = \left[ 2 \ln |x| - 4x^2 \right]_{\frac{1}{2}}^{4}
\]

Evaluate at the bounds:

\[
A = \left( 2 \ln 4 - 4(4^2) \right) - \left( 2 \ln \frac{1}{2} - 4\left(\frac{1}{2}\right)^2 \right)
\]

\[
A = \left( 2 \ln 4 - 64 \right) - \left( 2 \ln \frac{1}{2} - 1 \right)
\]

\[
A = \left( 2 \times 2 \ln 2 - 64 \right) - \left( 2 \times (-\ln 2) - 1 \right)
\]

\[
A = \left( 4 \ln 2 - 64 \right) - \left( -2 \ln 2 - 1 \right)
\]

\[
A = 4 \ln 2 - 64 + 2 \ln 2 + 1 = 6 \ln 2 - 63
\]

Thus, the area of the region is \( \boxed{6 \ln 2 - 63} \).

Do you want details or have any questions?

### Related Questions:
1. How would the area change if the interval were extended?
2. What is the geometric interpretation of the logarithmic terms in the area calculation?
3. How would the solution change if \( x < 0 \) were included?
4. What is the physical significance of such an area in real-world contexts?
5. How do the points of intersection change if the curves are shifted?

**Tip:** Always sketch the region when solving area problems to better visualize the integrals required.

Learn how to calculate the area of the region enclosed by the curves y = 2/x, y = 8x, and y = 1/8x with x > 0. Follow a detailed step-by-step solution involving finding points of intersection and setting up integrals. Suitable for advanced college-level students and anyone interested in calculus and integration.

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