what is the area enclosed by the figure of equations y = 2/x, y=0.5x, y=8x where x>0. tell me the final area

To find the area enclosed by the curves \( y = \frac{2}{x} \), \( y = 0.5x \), and \( y = 8x \) where \( x > 0 \), we follow these steps:

### 1. **Find the Points of Intersection:**
We first need to determine where these curves intersect.

- **Intersection of \( y = \frac{2}{x} \) and \( y = 0.5x \):**
  \[
  \frac{2}{x} = 0.5x \quad \Rightarrow \quad 2 = 0.5x^2 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = 2
  \]
  So, the curves intersect at \( x = 2 \).

- **Intersection of \( y = \frac{2}{x} \) and \( y = 8x \):**
  \[
  \frac{2}{x} = 8x \quad \Rightarrow \quad 2 = 8x^2 \quad \Rightarrow \quad x^2 = \frac{1}{4} \quad \Rightarrow \quad x = \frac{1}{2}
  \]
  So, the curves intersect at \( x = \frac{1}{2} \).

- **Intersection of \( y = 0.5x \) and \( y = 8x \):**
  \[
  0.5x = 8x \quad \Rightarrow \quad 0.5x = 8x \quad \Rightarrow \quad \text{No real solution other than at } x = 0.
  \]

Thus, the area is enclosed between \( x = \frac{1}{2} \) and \( x = 2 \).

### 2. **Set Up the Integral:**
The area between two curves is given by the integral of the difference between the top and bottom functions.

For \( \frac{1}{2} \leq x \leq 2 \), the function on top is \( y = \frac{2}{x} \), and the function on the bottom is \( y = 0.5x \).

Thus, the area is:
\[
A = \int_{\frac{1}{2}}^{2} \left( \frac{2}{x} - 0.5x \right) \, dx
\]

### 3. **Evaluate the Integral:**
We calculate the integral:
\[
\int \left( \frac{2}{x} \right) \, dx = 2 \ln |x|
\]
\[
\int \left( 0.5x \right) \, dx = 0.25x^2
\]
Now, evaluate the definite integral:
\[
A = \left[ 2 \ln |x| - 0.25x^2 \right]_{\frac{1}{2}}^{2}
\]
\[
= \left( 2 \ln 2 - 0.25(2^2) \right) - \left( 2 \ln \frac{1}{2} - 0.25\left( \frac{1}{2}^2 \right) \right)
\]
\[
= \left( 2 \ln 2 - 1 \right) - \left( 2(-\ln 2) - \frac{1}{16} \right)
\]
\[
= (2 \ln 2 - 1) - (-2 \ln 2 - \frac{1}{16})
\]
\[
= 2 \ln 2 - 1 + 2 \ln 2 + \frac{1}{16}
\]
\[
= 4 \ln 2 - 1 + \frac{1}{16}
\]
\[
= 4 \ln 2 - \frac{16}{16} + \frac{1}{16} = 4 \ln 2 - \frac{15}{16}
\]

Thus, the final area is:
\[
\boxed{4 \ln 2 - \frac{15}{16}}
\]

Do you want a detailed explanation of any step or have any questions?

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Here are five related questions to deepen your understanding:
1. How do you find the area between two curves in general?
2. What is the significance of finding points of intersection when calculating areas between curves?
3. How does the natural logarithm \( \ln x \) appear in the result of this integral?
4. Can you describe the shape of the curves \( y = \frac{2}{x} \), \( y = 0.5x \), and \( y = 8x \)?
5. How does the choice of limits \( x = \frac{1}{2} \) and \( x = 2 \) affect the final area calculation?

**Tip:** When integrating to find the area between curves, always carefully identify which curve is on top and which is on the bottom to set up the correct integral.

What is the area enclosed by the figure of equations y = 2/x, y=0.5x, y=8x where x>0. Tell me the final area.

Calculate the area enclosed by the curves y = 2/x, y = 0.5x, and y = 8x where x > 0. This problem involves finding points of intersection, setting up integrals, and applying the fundamental theorem of calculus. Suitable for undergraduate students studying calculus.

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