Sketch the region enclosed by the given curves.
y = 
3
x
,  y = 12x,  y = 
1
3
x,  x > 0 find the area of the region

We are given three curves:

1. \( y = 3x \)
2. \( y = 12x \)
3. \( y = \frac{1}{3}x \)
  
And the condition that \( x > 0 \).

### Step 1: Determine Points of Intersection

To find the region enclosed by these curves, we first need to find the points where the curves intersect each other.

1. **Intersection of \( y = 3x \) and \( y = 12x \):**
   \[
   3x = 12x \quad \Rightarrow \quad \text{These curves do not intersect as long as } x > 0.
   \]

2. **Intersection of \( y = 3x \) and \( y = \frac{1}{3}x \):**
   \[
   3x = \frac{1}{3}x \quad \Rightarrow \quad x = 0.
   \]
   But since \( x > 0 \), they do not intersect at a positive value of \( x \).

3. **Intersection of \( y = 12x \) and \( y = \frac{1}{3}x \):**
   \[
   12x = \frac{1}{3}x \quad \Rightarrow \quad x = 0 \text{ is the only solution.}
   \]

Thus, the curves don’t intersect each other in any region where \( x > 0 \), implying the curves are distinct and don't form a closed bounded area with one another directly, but we'll use their behavior to bound the area.

### Step 2: Define the Region

Since the curves don’t intersect in the usual sense, we will consider the region between the curve \( y = 12x \) (the highest line), and the curves \( y = 3x \) and \( y = \frac{1}{3}x \) (below it).

We'll calculate the area between these curves by splitting the problem into two parts:
1. The area between \( y = 12x \) and \( y = 3x \).
2. The area between \( y = 3x \) and \( y = \frac{1}{3}x \).

### Step 3: Set Integration Boundaries

We will set up the boundaries where \( x = 0 \) to \( x = 1 \), as these provide a reasonable area where the curves separate and the region is bounded.

1. **Area between \( y = 12x \) and \( y = 3x \):**
   \[
   \text{Area}_1 = \int_0^1 (12x - 3x) \, dx = \int_0^1 9x \, dx = \left[ \frac{9x^2}{2} \right]_0^1 = \frac{9}{2}.
   \]

2. **Area between \( y = 3x \) and \( y = \frac{1}{3}x \):**
   \[
   \text{Area}_2 = \int_0^1 \left( 3x - \frac{1}{3}x \right) dx = \int_0^1 \frac{8}{3}x \, dx = \left[ \frac{4x^2}{3} \right]_0^1 = \frac{4}{3}.
   \]

### Step 4: Total Area

The total enclosed area is:
\[
\text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{9}{2} + \frac{4}{3}.
\]

To add these:
\[
\text{Total Area} = \frac{27}{6} + \frac{8}{6} = \frac{35}{6}.
\]

### Final Answer:
The area of the region enclosed by the curves is \( \frac{35}{6} \) square units.

---

Would you like me to provide more details on how the integration was done or clarify any steps?

Here are some related questions:
1. How do we determine the limits of integration in such problems?
2. What happens if the curves intersect at multiple points?
3. How does the choice of integration limits affect the area calculation?
4. Can we use a graphical method to approximate the area before calculating?
5. What would change if the condition \( x > 0 \) was removed?

**Tip:** When curves do not intersect directly, always check for clear boundaries or split the region into simpler sections.

Sketch the region enclosed by the given curves: y = 3x, y = 12x, y = 1/3x, x > 0. Find the area of the region.

Discover how to calculate the area of the region enclosed by the curves y = 3x, y = 12x, and y = 1/3x when x > 0. This solution involves finding points of intersection, setting integration boundaries, and calculating the area using definite integrals. Suitable for advanced high school students or early college learners.

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